Question

A function f(x) satisfies the relation ${\int }_2^{x}f\left(t\right)=\frac{x^2}{2}+{\int }_x^2{t}^{2}f\left(t\right)dt$
Then answer the following
The value of ${\int }_{-2}^{2}f\left(x\right)$ dx is

• A. $0$
• B. $log2$
• C. $\frac{1}{2}log2$
• D. $1$

Answer 1

The correct option is A $0$

${\int }_2^{x}f\left(t\right)dt=\frac{x^2}{2}+{\int }_x^2{t}^{2}f\left(t\right)dtDiff.Wrt{}^{\prime }{x}^{\prime }f\left(x\right)=x-x^{2}f\left(x\right)\Rightarrow f\left(x\right)=\frac{x}{1+x^2}$

f(x) is odd function.