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Question

Let f:RR be a differentiable function such that f(x)=x2+x0etf(xt)dt. Then, the value of 10f(x)dx is

A
14
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B
112
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C
512
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D
127
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Solution

The correct option is C 512
f(x)=x2+x0etf(xt)dt(i)
Using the property:
baf(x)dx=baf(a+bx)dx
f(x)=x2+x0e(xt)f(x(xt))dt
=x2+exx0etf(t)dt(ii)
Differentiating w.r.t. x, we get
f(x)=2xexx0etf(t)dt+exexf(x)
f(x)=2xexx0etf(t)dt+f(x)
f(x)=2x+x2 [Using equation (ii)]
f(x)=x33+x2+c
Also, f(0)=0 [From equation (i)]
f(x)=x33+x2
10f(x)dx=10(x33+x2)dx
=[x412+x33]10=112+13=512

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