`f''(x) = 2 + cos(x), f(0) = -1, f(pi/2) = 0` Find `f`.

Expert Answers
gsarora17 eNotes educator| Certified Educator

`f''(x)=2+cos(x)`

`f'(x)=int(2+cos(x)dx`

`f'(x)=2x+sin(x)+C_1`

`f(x)=int(2x+sin(x)+C_1)dx`

`f(x)=2(x^2/2)-cos(x)+C_1x+C_2`

`f(x)=x^2-cos(x)+C_1x+C_2`

Now lets find constants C_1 and C_2 , given f(0)=-1 and f(pi/2)=0

`f(0)=-1=0^2-cos(0)+C_1(0)+C_2`

`-1=-1+C_2`

`C_2=0`

`f(pi/2)=0=(pi/2)^2-cos(pi/2)+C1(pi/2)+C_2`

`0=pi^2/4-0+pi/2C_1`

`C_1=-pi/2`

`:.f(x)=x^2-cos(x)-(pi/2)x`

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