`f''(t) = 2e^t + 3sin(t), f(0) = 0, f(pi) = 0` Find `f`.

Textbook Question

Chapter 4, 4.9 - Problem 46 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`f''(t)=2e^t+3sin(t)`

`f'(t)=int(2e^t+3sin(t))dt`

`f'(t)=2e^t-3cos(t)+C_1`

`f(t)=int(2e^t-3cos(t)+C_1)dt`

`f(t)=2e^t-3sin(t)+C_1t+C_2`

Now let's find constants C_1 and C_2 , given f(0)=0 and f(pi)=0

`f(0)=0=2e^0-3sin(0)+C_1(0)+C_2`

`0=2(1)-3(0)+0+C_2`

`C_2=-2`

`f(pi)=0=2e^pi-3sin(pi)+C_1(pi)+C_2`

`0=2e^pi-3(0)+C_1(pi)-2`

`piC_1=2-2e^pi`

`C_1=(2(1-e^pi))/pi`

`:.f(t)=2e^t-3sin(t)+(2(1-e^pi)t)/pi-2`

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