Math 108
Sample Final Test

1. f ′ (x)=3 x 2 −4x+1 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaGaeyypa0JaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI0aGaamiEaiabgUcaRiaaigdaaaa@4133@ . So, f(x)= ∫ f ′ (x) dx = ∫ (3 x 2 −4x+1) dx= x 3 −2 x 2 +x+C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9maapeaabaGabmOzayaafaGaaiikaiaadIhacaGGPaGaaGjbVlaadsgacaWG4baaleqabeqdcqGHRiI8aOGaeyypa0Zaa8qaaeaacaGGOaGaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI0aGaamiEaiabgUcaRiaaigdacaGGPaGaaGjbVlaadsgacaWG4bGaeyypa0JaamiEamaaCaaaleqabaGaaG4maaaakiabgkHiTiaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamiEaiabgUcaRiaadoeaaSqabeqaniabgUIiYdaaaa@5BD2@ .

Since f(2)=5 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaaIYaGaaiykaiabg2da9iaaiwdaaaa@3AB3@ , we have (2) 3 −2 (2) 2 +2+C=5 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaaikdacaGGPaWaaWbaaSqabeaacaaIZaaaaOGaeyOeI0IaaGOmaiaacIcacaaIYaGaaiykamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaikdacqGHRaWkcaWGdbGaeyypa0JaaGynaaaa@42B5@  or, 8−8+2+C=5 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGioaiabgkHiTiaaiIdacqGHRaWkcaaIYaGaey4kaSIaam4qaiabg2da9iaaiwdaaaa@3D6C@  or, C=3 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qaiabg2da9iaaiodaaaa@3879@ .

Hence f(x)= x 3 −2 x 2 +x+3 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIYaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadIhacqGHRaWkcaaIZaaaaa@433D@

 

2. f ′ (x)= x 2 −3x 2x = x 2 2x − 3x 2x = 1 2 x− 3 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaGaeyypa0ZaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaG4maiaadIhaaeaacaaIYaGaamiEaaaacqGH9aqpdaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaGaamiEaaaacqGHsisldaWcaaqaaiaaiodacaWG4baabaGaaGOmaiaadIhaaaGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaacaWG4bGaeyOeI0YaaSaaaeaacaaIZaaabaGaaGOmaaaaaaa@4FD0@ . So, f(x)= ∫ f ′ (x) dx= ∫ ( 1 2 x− 3 2 ) dx= 1 4 x 2 − 3 2 x+C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9maapeaabaGabmOzayaafaGaaiikaiaadIhacaGGPaGaaGjbVlaadsgacaWG4bGaeyypa0Zaa8qaaeaadaqadaqaamaalaaabaGaaGymaaqaaiaaikdaaaGaamiEaiabgkHiTmaalaaabaGaaG4maaqaaiaaikdaaaaacaGLOaGaayzkaaGaaGjbVlaadsgacaWG4bGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGinaaaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0YaaSaaaeaacaaIZaaabaGaaGOmaaaacaWG4bGaey4kaSIaam4qaaWcbeqab0Gaey4kIipaaSqabeqaniabgUIiYdaaaa@5983@

Since f(−2)=1 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacqGHsislcaaIYaGaaiykaiabg2da9iaaigdaaaa@3B9C@ , we have 1 4 (−2) 2 − 3 2 (−2)+C=1+3+C=1 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaaabaGaaGinaaaacaGGOaGaeyOeI0IaaGOmaiaacMcadaahaaWcbeqaaiaaikdaaaGccqGHsisldaWcaaqaaiaaiodaaeaacaaIYaaaaiaacIcacqGHsislcaaIYaGaaiykaiabgUcaRiaadoeacqGH9aqpcaaIXaGaey4kaSIaaG4maiabgUcaRiaadoeacqGH9aqpcaaIXaaaaa@4959@  or, C=−3 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qaiabg2da9iabgkHiTiaaiodaaaa@3966@ .

Hence f(x)= 1 4 x 2 − 3 2 x−3 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9maalaaabaGaaGymaaqaaiaaisdaaaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTmaalaaabaGaaG4maaqaaiaaikdaaaGaamiEaiabgkHiTiaaiodaaaa@42CB@ .

 

3. f ′ (x)= x 2 +2x+3 x = x 2 x + 2x x + 3 x =x+2+ 3 x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaGaeyypa0ZaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaadIhacqGHRaWkcaaIZaaabaGaamiEaaaacqGH9aqpdaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaaakeaacaWG4baaaiabgUcaRmaalaaabaGaaGOmaiaadIhaaeaacaWG4baaaiabgUcaRmaalaaabaGaaG4maaqaaiaadIhaaaGaeyypa0JaamiEaiabgUcaRiaaikdacqGHRaWkdaWcaaqaaiaaiodaaeaacaWG4baaaaaa@521C@

So, f(x)= ∫ f ′ (x) dx= ∫ ( x+2+ 3 x ) dx= 1 2 x 2 +2x+3ln⁡x+C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9maapeaabaGabmOzayaafaGaaiikaiaadIhacaGGPaGaaGjbVlaadsgacaWG4bGaeyypa0Zaa8qaaeaadaqadaqaaiaadIhacqGHRaWkcaaIYaGaey4kaSYaaSaaaeaacaaIZaaabaGaamiEaaaaaiaawIcacaGLPaaacaaMe8UaamizaiaadIhacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaGaamiEaiabgUcaRiaaiodaciGGSbGaaiOBaiaadIhacqGHRaWkcaWGdbaaleqabeqdcqGHRiI8aaWcbeqab0Gaey4kIipaaaa@5D76@

Since f(1)=−4 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaaIXaGaaiykaiabg2da9iabgkHiTiaaisdaaaa@3B9E@ , we have 1 2 (1) 2 +2(1)+3ln⁡(1)+C=−4 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaaabaGaaGOmaaaacaGGOaGaaGymaiaacMcadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaGaaiikaiaaigdacaGGPaGaey4kaSIaaG4maiGacYgacaGGUbGaaiikaiaaigdacaGGPaGaey4kaSIaam4qaiabg2da9iabgkHiTiaaisdaaaa@4820@  or, 1 2 +2+0+C=−4 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaaabaGaaGOmaaaacqGHRaWkcaaIYaGaey4kaSIaaGimaiabgUcaRiaadoeacqGH9aqpcqGHsislcaaI0aaaaa@3F0A@  or, C=− 13 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qaiabg2da9iabgkHiTmaalaaabaGaaGymaiaaiodaaeaacaaIYaaaaaaa@3AED@ .

So, f(x)= 1 2 x 2 +2x+3ln⁡x− 13 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaikdacaWG4bGaey4kaSIaaG4maiGacYgacaGGUbGaamiEaiabgkHiTmaalaaabaGaaGymaiaaiodaaeaacaaIYaaaaaaa@47F8@ .

 

4. ∫ (3 x 2 +1)(x−2) dx = ∫ (3 x 3 −6 x 2 +x−2)dx =3 x 4 4 −6 x 3 3 + x 2 2 −2x+C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6601@

= 3 4 x 4 −2 x 3 + 1 2 x 2 −2x+C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIZaaabaGaaGinaaaacaWG4bWaaWbaaSqabeaacaaI0aaaaOGaeyOeI0IaaGOmaiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkdaWcaaqaaiaaigdaaeaacaaIYaaaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIYaGaamiEaiabgUcaRiaadoeaaaa@45AE@

 

5.  ∫ (3 x 2 +1) e x 3 +x−7  dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qaaeaacaGGOaGaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIXaGaaiykaiaadwgadaahaaWcbeqaaiaadIhadaahaaadbeqaaiaaiodaaaWccqGHRaWkcaWG4bGaeyOeI0IaaG4naaaakiaaysW7caWGKbGaamiEaaWcbeqab0Gaey4kIipaaaa@47A0@ . Put u= x 3 +x−7 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDaiabg2da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaWG4bGaeyOeI0IaaG4naaaa@3D6C@ , then du=(3 x 2 +1) dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadwhacqGH9aqpcaGGOaGaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIXaGaaiykaiaaysW7caWGKbGaamiEaaaa@41ED@ .

So, ∫ (3 x 2 +1) e x 3 +x−7  dx= ∫ e x 3 +x−7 (3 x 2 +1)dx= ∫ e u du= e u +C = e x 3 +x−7 +C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qaaeaacaGGOaGaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIXaGaaiykaiaadwgadaahaaWcbeqaaiaadIhadaahaaadbeqaaiaaiodaaaWccqGHRaWkcaWG4bGaeyOeI0IaaG4naaaakiaaysW7caWGKbGaamiEaiabg2da9maapeaabaGaamyzamaaCaaaleqabaGaamiEamaaCaaameqabaGaaG4maaaaliabgUcaRiaadIhacqGHsislcaaI3aaaaOGaaiikaiaaiodacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGymaiaacMcacaWGKbGaamiEaiabg2da9maapeaabaGaamyzamaaCaaaleqabaGaamyDaaaakiaadsgacaWG1bGaeyypa0JaamyzamaaCaaaleqabaGaamyDaaaakiabgUcaRiaadoeaaSqabeqaniabgUIiYdaaleqabeqdcqGHRiI8aaWcbeqab0Gaey4kIipakiabg2da9iaadwgadaahaaWcbeqaaiaadIhadaahaaadbeqaaiaaiodaaaWccqGHRaWkcaWG4bGaeyOeI0IaaG4naaaakiabgUcaRiaadoeaaaa@6DF0@ .

 

6. ∫ 1+ (ln⁡x) 2 x  dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qaaeaadaWcaaqaaiaaigdacqGHRaWkcaGGOaGaciiBaiaac6gacaWG4bGaaiykamaaCaaaleqabaGaaGOmaaaaaOqaaiaadIhaaaaaleqabeqdcqGHRiI8aOGaaGjbVlaadsgacaWG4baaaa@433D@ . Put u=ln⁡x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDaiabg2da9iGacYgacaGGUbGaamiEaaaa@3ACF@ , then du= 1 x dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadwhacqGH9aqpdaWcaaqaaiaaigdaaeaacaWG4baaaiaadsgacaWG4baaaa@3C85@ .

So, ∫ 1+ (ln⁡x) 2 x dx= ∫ ( 1+ (ln⁡x) 2 ) 1 x dx = ∫ (1+ u 2 )du=u+ u 3 3 +C =(ln⁡x)+ (ln⁡x) 3 3 +C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6F9B@

 

7.   I= ∫ (2x+3) e 4x+1 dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2da9maapeaabaGaaiikaiaaikdacaWG4bGaey4kaSIaaG4maiaacMcacaWGLbWaaWbaaSqabeaacaaI0aGaamiEaiabgUcaRiaaigdaaaGccaWGKbGaamiEaaWcbeqab0Gaey4kIipaaaa@44CD@ .  Put f(x)=2x+3 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9iaaikdacaWG4bGaey4kaSIaaG4maaaa@3D8D@  and g ′ (x)= e 4x+1 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4zayaafaGaaiikaiaadIhacaGGPaGaeyypa0JaamyzamaaCaaaleqabaGaaGinaiaadIhacqGHRaWkcaaIXaaaaaaa@3EB1@ . Then f ′ (x)=2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafaGaaiikaiaadIhacaGGPaGaeyypa0JaaGOmaaaa@3AFD@  and g(x)= ∫ e 4x+1 dx = 1 4 e 4x+1 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zaiaacIcacaWG4bGaaiykaiabg2da9maapeaabaGaamyzamaaCaaaleqabaGaaGinaiaadIhacqGHRaWkcaaIXaaaaOGaamizaiaadIhaaSqabeqaniabgUIiYdGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaI0aaaaiaadwgadaahaaWcbeqaaiaaisdacaWG4bGaey4kaSIaaGymaaaaaaa@4998@ . Due to integration by parts we have, ∫ f(x) g ′ (x)dx=f(x)g(x)− ∫ f ′ (x)g(x)dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qaaeaacaWGMbGaaiikaiaadIhacaGGPaGabm4zayaafaGaaiikaiaadIhacaGGPaGaamizaiaadIhacqGH9aqpcaWGMbGaaiikaiaadIhacaGGPaGaam4zaiaacIcacaWG4bGaaiykaiabgkHiTmaapeaabaGabmOzayaafaGaaiikaiaadIhacaGGPaGaam4zaiaacIcacaWG4bGaaiykaiaadsgacaWG4baaleqabeqdcqGHRiI8aaWcbeqab0Gaey4kIipaaaa@5344@ .

So, I=(2x+3)( 1 4 e 4x+1 )− ∫ (2)( 1 4 e 4x+1 ) dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2da9iaacIcacaaIYaGaamiEaiabgUcaRiaaiodacaGGPaWaaeWaaeaadaWcaaqaaiaaigdaaeaacaaI0aaaaiaadwgadaahaaWcbeqaaiaaisdacaWG4bGaey4kaSIaaGymaaaaaOGaayjkaiaawMcaaiabgkHiTmaapeaabaGaaiikaiaaikdacaGGPaWaaeWaaeaadaWcaaqaaiaaigdaaeaacaaI0aaaaiaadwgadaahaaWcbeqaaiaaisdacaWG4bGaey4kaSIaaGymaaaaaOGaayjkaiaawMcaaaWcbeqab0Gaey4kIipakiaadsgacaWG4baaaa@5276@  = 1 4 (2x+3) e 4x+1 − 1 2 ∫ e 4x+1 dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaaabaGaaGinaaaacaGGOaGaaGOmaiaadIhacqGHRaWkcaaIZaGaaiykaiaadwgadaahaaWcbeqaaiaaisdacaWG4bGaey4kaSIaaGymaaaakiabgkHiTmaalaaabaGaaGymaaqaaiaaikdaaaWaa8qaaeaacaWGLbWaaWbaaSqabeaacaaI0aGaamiEaiabgUcaRiaaigdaaaGccaWGKbGaamiEaaWcbeqab0Gaey4kIipaaaa@4B6F@

 = 1 4 (2x+3) e 4x+1 − 1 8 e 4x+1 +C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaaabaGaaGinaaaacaGGOaGaaGOmaiaadIhacqGHRaWkcaaIZaGaaiykaiaadwgadaahaaWcbeqaaiaaisdacaWG4bGaey4kaSIaaGymaaaakiabgkHiTmaalaaabaGaaGymaaqaaiaaiIdaaaGaamyzamaaCaaaleqabaGaaGinaiaadIhacqGHRaWkcaaIXaaaaOGaey4kaSIaam4qaaaa@493E@  = 1 8 e 4x+1 [ 2(2x+3)−1 ]+C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaaabaGaaGioaaaacaWGLbWaaWbaaSqabeaacaaI0aGaamiEaiabgUcaRiaaigdaaaGcdaWadaqaaiaaikdacaGGOaGaaGOmaiaadIhacqGHRaWkcaaIZaGaaiykaiabgkHiTiaaigdaaiaawUfacaGLDbaacqGHRaWkcaWGdbaaaa@46A5@  = 1 8 e 4x+1 (4x+6−1)+C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaaabaGaaGioaaaacaWGLbWaaWbaaSqabeaacaaI0aGaamiEaiabgUcaRiaaigdaaaGccaGGOaGaaGinaiaadIhacqGHRaWkcaaI2aGaeyOeI0IaaGymaiaacMcacqGHRaWkcaWGdbaaaa@43FC@

= 1 8 e 4x+1 (4x+5)+C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaaabaGaaGioaaaacaWGLbWaaWbaaSqabeaacaaI0aGaamiEaiabgUcaRiaaigdaaaGccaGGOaGaaGinaiaadIhacqGHRaWkcaaI1aGaaiykaiabgUcaRiaadoeaaaa@4253@ .

 



8. The graph of f(x)=− x 2 +5x−6 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiykaiabg2da9iabgkHiTiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI1aGaamiEaiabgkHiTiaaiAdaaaa@415D@  is an inverted parabola. It intersects the x-axis when y = 0 or, − x 2 +5x−6=0 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiwdacaWG4bGaeyOeI0IaaGOnaiabg2da9iaaicdaaaa@3ED6@  or, x 2 −5x+6=0 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiwdacaWG4bGaey4kaSIaaGOnaiabg2da9iaaicdaaaa@3DE9@  or, (x−2)(x−3)=0 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaadIhacqGHsislcaaIYaGaaiykaiaacIcacaWG4bGaeyOeI0IaaG4maiaacMcacqGH9aqpcaaIWaaaaa@3FAD@  or, x=2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2da9iaaikdaaaa@38AD@  and x=3 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2da9iaaiodaaaa@38AE@ . The graph is below the x-axis on the left of the line x = 2 and to the right of the line x = 3. And, the graph is above the x-axis in between the line x = 2 and x = 3. So, due to Fundamental Theorem of Calculus, the area bounded by the graph of the parabola, the x-axis, the lines x = 1 and x = 4 is given by

− ∫ 1 2 (− x 2 +5x−6) dx+ ∫ 2 3 (− x 2 +5x−6)dx− ∫ 3 4 (− x 2 +5x−6)dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@638F@

= ( 1 3 x 3 − 5 2 x 2 +6x ) | 1 2 − ( 1 3 x 3 − 5 2 x 2 +6x ) | 2 3 + ( 1 3 x 3 − 5 2 x 2 +6x ) | 3 4 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqGaaeaadaqadaqaamaalaaabaGaaGymaaqaaiaaiodaaaGaamiEamaaCaaaleqabaGaaG4maaaakiabgkHiTmaalaaabaGaaGynaaqaaiaaikdaaaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiAdacaWG4baacaGLOaGaayzkaaaacaGLiWoadaqhaaWcbaGaaGymaaqaaiaaikdaaaGccqGHsisldaabcaqaamaabmaabaWaaSaaaeaacaaIXaaabaGaaG4maaaacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaeyOeI0YaaSaaaeaacaaI1aaabaGaaGOmaaaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOnaiaadIhaaiaawIcacaGLPaaaaiaawIa7amaaDaaaleaacaaIYaaabaGaaG4maaaakiabgUcaRmaaeiaabaWaaeWaaeaadaWcaaqaaiaaigdaaeaacaaIZaaaaiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsisldaWcaaqaaiaaiwdaaeaacaaIYaaaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI2aGaamiEaaGaayjkaiaawMcaaaGaayjcSdWaa0baaSqaaiaaiodaaeaacaaI0aaaaaaa@65A0@

= [ 1 3 (8−1)− 5 2 (4−1)+6(2−1) ]−[ 1 3 (27−8)− 5 2 (9−4)+6(3−2) ]+[ 1 3 (64−27)− 5 2 (16−9)+6(4−3) ] MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaadaWcaaqaaiaaigdaaeaacaaIZaaaaiaacIcacaaI4aGaeyOeI0IaaGymaiaacMcacqGHsisldaWcaaqaaiaaiwdaaeaacaaIYaaaaiaacIcacaaI0aGaeyOeI0IaaGymaiaacMcacqGHRaWkcaaI2aGaaiikaiaaikdacqGHsislcaaIXaGaaiykaaGaay5waiaaw2faaiabgkHiTmaadmaabaWaaSaaaeaacaaIXaaabaGaaG4maaaacaGGOaGaaGOmaiaaiEdacqGHsislcaaI4aGaaiykaiabgkHiTmaalaaabaGaaGynaaqaaiaaikdaaaGaaiikaiaaiMdacqGHsislcaaI0aGaaiykaiabgUcaRiaaiAdacaGGOaGaaG4maiabgkHiTiaaikdacaGGPaaacaGLBbGaayzxaaGaey4kaSYaamWaaeaadaWcaaqaaiaaigdaaeaacaaIZaaaaiaacIcacaaI2aGaaGinaiabgkHiTiaaikdacaaI3aGaaiykaiabgkHiTmaalaaabaGaaGynaaqaaiaaikdaaaGaaiikaiaaigdacaaI2aGaeyOeI0IaaGyoaiaacMcacqGHRaWkcaaI2aGaaiikaiaaisdacqGHsislcaaIZaGaaiykaaGaay5waiaaw2faaaaa@7349@  = 7 3 − 15 2 +6− 19 3 + 25 2 −6+ 37 3 − 35 2 +6 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaI3aaabaGaaG4maaaacqGHsisldaWcaaqaaiaaigdacaaI1aaabaGaaGOmaaaacqGHRaWkcaaI2aGaeyOeI0YaaSaaaeaacaaIXaGaaGyoaaqaaiaaiodaaaGaey4kaSYaaSaaaeaacaaIYaGaaGynaaqaaiaaikdaaaGaeyOeI0IaaGOnaiabgUcaRmaalaaabaGaaG4maiaaiEdaaeaacaaIZaaaaiabgkHiTmaalaaabaGaaG4maiaaiwdaaeaacaaIYaaaaiabgUcaRiaaiAdaaaa@4C63@  = 1 3 (7−19+37)+ 1 2 (−15+25−35)+6 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaaabaGaaG4maaaacaGGOaGaaG4naiabgkHiTiaaigdacaaI5aGaey4kaSIaaG4maiaaiEdacaGGPaGaey4kaSYaaSaaaeaacaaIXaaabaGaaGOmaaaacaGGOaGaeyOeI0IaaGymaiaaiwdacqGHRaWkcaaIYaGaaGynaiabgkHiTiaaiodacaaI1aGaaiykaiabgUcaRiaaiAdaaaa@4AEC@

= 25 3 − 25 2 +6=− 25 6 +6= 11 6 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIYaGaaGynaaqaaiaaiodaaaGaeyOeI0YaaSaaaeaacaaIYaGaaGynaaqaaiaaikdaaaGaey4kaSIaaGOnaiabg2da9iabgkHiTmaalaaabaGaaGOmaiaaiwdaaeaacaaI2aaaaiabgUcaRiaaiAdacqGH9aqpdaWcaaqaaiaaigdacaaIXaaabaGaaGOnaaaaaaa@4638@

 

9. ∫ 2 4 ∫ 1 x x e y  dy dx = ∫ 2 4 x e y | y=1 y=x  dx= ∫ 2 4 ( x e x −xe ) dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6ACD@

= x e x − e x −e x 2 2 | x=2 x=4 =(4 e 4 − e 4 −8e)−(2 e 2 − e 2 −2e) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@5C91@  = 3 e 4 − e 2 −6e MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaG4maiaadwgadaahaaWcbeqaaiaaisdaaaGccqGHsislcaWGLbWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOnaiaadwgaaaa@3DEB@

In the above, we have written ∫ x e x dx=x e x − e x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qaaeaacaWG4bGaamyzamaaCaaaleqabaGaamiEaaaakiaadsgacaWG4bGaeyypa0JaamiEaiaadwgadaahaaWcbeqaaiaadIhaaaGccqGHsislcaWGLbWaaWbaaSqabeaacaWG4baaaaqabeqaniabgUIiYdaaaa@4401@ . This can be verified by using the integration by parts techniques. (Note: I may have written something erroneous in the blackboard in the last class)

 

10. The graph of y=1− x 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2da9iaaigdacqGHsislcaWG4bWaaWbaaSqabeaacaaIYaaaaaaa@3B80@  is an inverted parabola which intersects the x-axis at x = -1 and x = 1. So, the region bounded by the parabola and the x-axis is just the region bounded by the same and the lines x = -1 and x = 1. So,

∬ R (x+2y)dxdy = ∫ −1 1 ∫ 0 1− x 2 (x+2y)dydx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8GeaeaacaGGOaGaamiEaiabgUcaRiaaikdacaWG5bGaaiykaiaadsgacaWG4bGaamizaiaadMhaaSqaaiaadkfaaeqaniabgUIiYlabgUIiYdGccqGH9aqpdaWdXbqaamaapehabaGaaiikaiaadIhacqGHRaWkcaaIYaGaamyEaiaacMcacaWGKbGaamyEaiaadsgacaWG4baaleaacaaIWaaabaGaaGymaiabgkHiTiaadIhadaahaaadbeqaaiaaikdaaaaaniabgUIiYdaaleaacqGHsislcaaIXaaabaGaaGymaaqdcqGHRiI8aaaa@58A7@  = ∫ −1 1 ( xy+ y 2 ) | y=0 y=1− x 2  dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaadaabcaqaamaabmaabaGaamiEaiaadMhacqGHRaWkcaWG5bWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaaacaGLiWoadaqhaaWcbaGaamyEaiabg2da9iaaicdaaeaacaWG5bGaeyypa0JaaGymaiabgkHiTiaadIhadaahaaadbeqaaiaaikdaaaaaaaWcbaGaeyOeI0IaaGymaaqaaiaaigdaa0Gaey4kIipakiaaysW7caWGKbGaamiEaaaa@4EA2@  = ∫ −1 1 [ x(1− x 2 )+ (1− x 2 ) 2 ] dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaadaWadaqaaiaadIhacaGGOaGaaGymaiabgkHiTiaadIhadaahaaWcbeqaaiaaikdaaaGccaGGPaGaey4kaSIaaiikaiaaigdacqGHsislcaWG4bWaaWbaaSqabeaacaaIYaaaaOGaaiykamaaCaaaleqabaGaaGOmaaaaaOGaay5waiaaw2faaiaaysW7caWGKbGaamiEaaWcbaGaeyOeI0IaaGymaaqaaiaaigdaa0Gaey4kIipaaaa@4CD3@

= ∫ −1 1 (x− x 3 +1−2 x 2 + x 4 ) dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaacaGGOaGaamiEaiabgkHiTiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaIXaGaeyOeI0IaaGOmaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWG4bWaaWbaaSqabeaacaaI0aaaaOGaaiykaiaaysW7caWGKbGaamiEaaWcbaGaeyOeI0IaaGymaaqaaiaaigdaa0Gaey4kIipaaaa@4B6B@  = ∫ −1 1 ( x 4 − x 3 −2 x 2 +x+1) dx MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaacaGGOaGaamiEamaaCaaaleqabaGaaGinaaaakiabgkHiTiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIYaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaadIhacqGHRaWkcaaIXaGaaiykaiaaysW7caWGKbGaamiEaaWcbaGaeyOeI0IaaGymaaqaaiaaigdaa0Gaey4kIipaaaa@4B6B@  = ( x 5 5 − x 4 4 − 2 x 3 3 + x 2 2 +x ) | x=−1 x=1 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqGaaeaadaqadaqaamaalaaabaGaamiEamaaCaaaleqabaGaaGynaaaaaOqaaiaaiwdaaaGaeyOeI0YaaSaaaeaacaWG4bWaaWbaaSqabeaacaaI0aaaaaGcbaGaaGinaaaacqGHsisldaWcaaqaaiaaikdacaWG4bWaaWbaaSqabeaacaaIZaaaaaGcbaGaaG4maaaacqGHRaWkdaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaaaaiabgUcaRiaadIhaaiaawIcacaGLPaaaaiaawIa7amaaDaaaleaacaWG4bGaeyypa0JaeyOeI0IaaGymaaqaaiaadIhacqGH9aqpcaaIXaaaaaaa@4FF6@

= 1 5 [ (1) 5 − (−1) 5 ]− 1 4 [ (1) 4 − (−1) 4 ]− 2 3 [ (1) 3 − (−1) 3 ]+ 1 2 [ (1) 2 − (−1) 2 ]+[ (1)−(−1) ] MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@6F17@

= 1 5 (2)− 1 4 (0)− 2 3 (2)+ 1 2 (0)+(2) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaaabaGaaGOmaaaacaGGOaGaaGOmaiaacMcacqGHsisldaWcaaqaaiaaigdaaeaacaaI0aaaaiaacIcacaaIWaGaaiykaiabgkHiTmaalaaabaGaaGOmaaqaaiaaiodaaaGaaiikaiaaikdacaGGPaGaey4kaSYaaSaaaeaacaaIXaaabaGaaGOmaaaacaGGOaGaaGimaiaacMcacqGHRaWkcaGGOaGaaGOmaiaacMcaaaa@4A11@  = 2 5 − 4 3 +2= 6−20+30 15 = 16 15 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIYaaabaGaaGynaaaacqGHsisldaWcaaqaaiaaisdaaeaacaaIZaaaaiabgUcaRiaaikdacqGH9aqpdaWcaaqaaiaaiAdacqGHsislcaaIYaGaaGimaiabgUcaRiaaiodacaaIWaaabaGaaGymaiaaiwdaaaGaeyypa0ZaaSaaaeaacaaIXaGaaGOnaaqaaiaaigdacaaI1aaaaaaa@47A6@ .

 

11. dy dx = y 2 +3 y(x+3) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG4baaaiabg2da9maalaaabaGaamyEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiodaaeaacaWG5bGaaiikaiaadIhacqGHRaWkcaaIZaGaaiykaaaaaaa@4364@  or, ydy y 2 +3 = dx x+3 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWG5bGaamizaiaadMhaaeaacaWG5bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaG4maaaacqGH9aqpdaWcaaqaaiaadsgacaWG4baabaGaamiEaiabgUcaRiaaiodaaaaaaa@420B@ . Integrating both sides, we get 1 2 ln⁡( y 2 +3)=ln⁡(x+3)+C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIXaaabaGaaGOmaaaaciGGSbGaaiOBaiaacIcacaWG5bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaG4maiaacMcacqGH9aqpciGGSbGaaiOBaiaacIcacaWG4bGaey4kaSIaaG4maiaacMcacqGHRaWkcaWGdbaaaa@46CB@

Or, ln⁡( y 2 +3)=2ln⁡(x+3)+C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaGGOaGaamyEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiodacaGGPaGaeyypa0JaaGOmaiGacYgacaGGUbGaaiikaiaadIhacqGHRaWkcaaIXaGaaiykaiabgUcaRiaadoeaaaa@45FE@

Or, ln⁡( y 2 +3)=ln⁡ (x+3) 2 +C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6gacaGGOaGaamyEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiodacaGGPaGaeyypa0JaciiBaiaac6gacaGGOaGaamiEaiabgUcaRiaaigdacaGGPaWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaam4qaaaa@4635@

Or, y 2 +3= e ln (x+3) 2 . e C =K (x+3) 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiodacqGH9aqpcaWGLbWaaWbaaSqabeaacaGGOaGaamiEaiabgUcaRiaaigdacaGGPaWaaWbaaWqabeaacaaIYaaaaaaakiaac6cacaWGLbWaaWbaaSqabeaacaWGdbaaaOGaeyypa0Jaam4saiaacIcacaWG4bGaey4kaSIaaGymaiaacMcadaahaaWcbeqaaiaaikdaaaaaaa@49CF@  ( where K= e C MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4saiabg2da9iaadwgadaahaaWcbeqaaiaadoeaaaaaaa@39A3@  )

 

12. 2 d 2 y d x 2 −5 dy dx +3y=0 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmamaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadMhaaeaacaWGKbGaamiEamaaCaaaleqabaGaaGOmaaaaaaGccqGHsislcaaI1aWaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG4baaaiabgUcaRiaaiodacaWG5bGaeyypa0JaaGimaaaa@4653@

The auxillary equation of the above linear homogenous differential equation is 2 m 2 −5m+3=0 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaad2gadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI1aGaamyBaiabgUcaRiaaiodacqGH9aqpcaaIWaaaaa@3E8C@  or, (2m−3)(m−1)=0 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaaikdacaWGTbGaeyOeI0IaaG4maiaacMcacaGGOaGaamyBaiabgkHiTiaaigdacaGGPaGaeyypa0JaaGimaaaa@4052@  or, m= 3 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabg2da9maalaaabaGaaG4maaqaaiaaikdaaaaaaa@396F@  and m=1 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabg2da9iaaigdaaaa@38A1@ . So, the general solution of the given differential equation is given by y=A e 3 2 x +B e x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2da9iaadgeacaWGLbWaaWbaaSqabeaadaWcaaqaaiaaiodaaeaacaaIYaaaaiaadIhaaaGccqGHRaWkcaWGcbGaamyzamaaCaaaleqabaGaamiEaaaaaaa@401C@ .

To verify, we find dy dx = 3 2 A e 3 2 x +B e x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG4baaaiabg2da9maalaaabaGaaG4maaqaaiaaikdaaaGaamyqaiaadwgadaahaaWcbeqaamaalaaabaGaaG4maaqaaiaaikdaaaGaamiEaaaakiabgUcaRiaadkeacaWGLbWaaWbaaSqabeaacaWG4baaaaaa@4484@  and d 2 y d x 2 = 9 4 A e 3 2 x +B e x MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGKbWaaWbaaSqabeaacaaIYaaaaOGaamyEaaqaaiaadsgacaWG4bWaaWbaaSqabeaacaaIYaaaaaaakiabg2da9maalaaabaGaaGyoaaqaaiaaisdaaaGaamyqaiaadwgadaahaaWcbeqaamaalaaabaGaaG4maaqaaiaaikdaaaGaamiEaaaakiabgUcaRiaadkeacaWGLbWaaWbaaSqabeaacaWG4baaaaaa@4672@ .

So, 2 d 2 y d x 2 −5 dy dx +3y MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmamaalaaabaGaamizamaaCaaaleqabaGaaGOmaaaakiaadMhaaeaacaWGKbGaamiEamaaCaaaleqabaGaaGOmaaaaaaGccqGHsislcaaI1aWaaSaaaeaacaWGKbGaamyEaaqaaiaadsgacaWG4baaaiabgUcaRiaaiodacaWG5baaaa@4493@  = 2( 9 4 A e 3 2 x +B e x )−5( 3 2 A e 3 2 x +B e x )+3( A e 3 2 x +B e x ) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@5A46@

= A e 3 2 x ( 9 2 − 15 2 +3 )+B e x (2−5+3) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadwgadaahaaWcbeqaamaalaaabaGaaG4maaqaaiaaikdaaaGaamiEaaaakmaabmaabaWaaSaaaeaacaaI5aaabaGaaGOmaaaacqGHsisldaWcaaqaaiaaigdacaaI1aaabaGaaGOmaaaacqGHRaWkcaaIZaaacaGLOaGaayzkaaGaey4kaSIaamOqaiaadwgadaahaaWcbeqaaiaadIhaaaGccaGGOaGaaGOmaiabgkHiTiaaiwdacqGHRaWkcaaIZaGaaiykaaaa@4B6C@  = A e 3 2 x (0)+B e x (0) MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadwgadaahaaWcbeqaamaalaaabaGaaG4maaqaaiaaikdaaaGaamiEaaaakiaacIcacaaIWaGaaiykaiabgUcaRiaadkeacaWGLbWaaWbaaSqabeaacaWG4baaaOGaaiikaiaaicdacaGGPaaaaa@4248@  = 0 + 0 = 0.

 

13. f(x,y)= x+y x−y MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacIcacaWG4bGaaiilaiaadMhacaGGPaGaeyypa0ZaaSaaaeaacaWG4bGaey4kaSIaamyEaaqaaiaadIhacqGHsislcaWG5baaaaaa@41B8@ ,

So, ∂f ∂x = (1)(x−y)−(x+y)(1) (x−y) 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacqGHciITcaWGMbaabaGaeyOaIyRaamiEaaaacqGH9aqpdaWcaaqaaiaacIcacaaIXaGaaiykaiaacIcacaWG4bGaeyOeI0IaamyEaiaacMcacqGHsislcaGGOaGaamiEaiabgUcaRiaadMhacaGGPaGaaiikaiaaigdacaGGPaaabaGaaiikaiaadIhacqGHsislcaWG5bGaaiykamaaCaaaleqabaGaaGOmaaaaaaaaaa@4E7E@  = −2y (x−y) 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacqGHsislcaaIYaGaamyEaaqaaiaacIcacaWG4bGaeyOeI0IaamyEaiaacMcadaahaaWcbeqaaiaaikdaaaaaaaaa@3DCF@  

And

  ∂f ∂y = (1)(x−y)−(x+y)(−1) (x−y) 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacqGHciITcaWGMbaabaGaeyOaIyRaamyEaaaacqGH9aqpdaWcaaqaaiaacIcacaaIXaGaaiykaiaacIcacaWG4bGaeyOeI0IaamyEaiaacMcacqGHsislcaGGOaGaamiEaiabgUcaRiaadMhacaGGPaGaaiikaiabgkHiTiaaigdacaGGPaaabaGaaiikaiaadIhacqGHsislcaWG5bGaaiykamaaCaaaleqabaGaaGOmaaaaaaaaaa@4F6C@  = 2x (x−y) 2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacaaIYaGaamiEaaqaaiaacIcacaWG4bGaeyOeI0IaamyEaiaacMcadaahaaWcbeqaaiaaikdaaaaaaaaa@3CE1@

So,

∂ 2 f ∂ x 2 = ∂ ∂x ( ∂f ∂x )= ∂ ∂x ( −2y (x−y) 2 )=(−2y) ∂ ∂x ( x−y ) −2 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=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@653D@  = (−2y)(−2) (x−y) −3 (1)= 4y (x−y) 3 MathType@MTEF@5@5@+=feaagaart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiabgkHiTiaaikdacaWG5bGaaiykaiaacIcacqGHsislcaaIYaGaaiykaiaacIcacaWG4bGaeyOeI0IaamyEaiaacMcadaahaaWcbeqaaiabgkHiTiaaiodaaaGccaGGOaGaaGymaiaacMcacqGH9aqpdaWcaaqaaiaaisdacaWG5baabaGaaiikaiaadIhacqGHsislcaWG5bGaaiykamaaCaaaleqabaGaaG4maaaaaaaaaa@4D23@

∂ 2 f ∂ y 2 = ∂ ∂y ( ∂f ∂y )= ∂ ∂y (