First slide
Evaluation of definite integrals
Question

Let f:(0,)R and F(x)=0xf(t)dt. If Fx2=x2(1+x) then f(4) equals

Moderate
Solution

Fx2=0x2f(t) dt , there fore, x2(1+x)=

0x2f(t)dt.  Differentiating both sides w.r.t. x using Property

17, we have

2x+3x2=fx2·2xfx2=1+(3/2)x

Putting x=2 we have f(4)=1+3=4.

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