Question

Let $f\left(x\right)$ and $g\left(x\right)$ be two function satisfying $f\left(x^2\right)+g(4-x)=4x^3,g(4-x)+g\left(x\right)=0$, then the value of ${\int }_{-4}^{4}f\left(x^2\right)dx$ is:

• A. 512
• B. 64
• C. 256
• D. 0

Answer 1

The correct option is A 512

Let $I={\int }_{-4}^{4}f\left(x^2\right)dx$

$\therefore I={\int }_{-4}^{0}f\left(x^2\right)dx+{\int }_0^{4}f\left(x^2\right)dx$

$\therefore I={\int }_4^{0}f\left(\right(-x{)}^2)dx+{\int }_0^{4}f(x^2)dx$

$\therefore I=2{\int }_0^{4}f\left(x^2\right)dx$

$\therefore I=2{\int }_0^4(4x^3-g(4-x\left)\right)dx$ (from given equation)

$\therefore I=2{\int }_0^{4}4x^{3}dx-2{\int }_0^{4}g(4-x)dx$

$\therefore I=2{\int }_0^{4}4x^{3}dx-2{\int }_0^{2}g(4-x)dx-2{\int }_2^{4}g(4-x)dx$

$\therefore I=2{\int }_0^{4}4x^{3}dx-2{\int }_0^{2}g(4-x)dx-2{\int }_2^{4}g(4-x)dx$ ... (1)

Let $I_1={\int }_2^{4}g(4-x)dx$

Now, consider $(4-x)=y$ which is substituted in $I_1$

$\therefore I_1={\int }_0^{2}g\left(y\right)dy$ (inverting limits and $dy$ both provide negative signs)

Substituting $I_1$ in (1), we have

$I=2{\int }_0^{4}4x^{3}dx-2{\int }_0^{2}g(4-x)dx-2{\int }_0^{2}g\left(y\right)dy$

Also, $g(4-x)=-g\left(x\right)$ from given equation and $g\left(y\right)=g\left(x\right)$ from variable substitution

$\therefore I=2{\int }_0^{4}4x^{3}dx-2{\int }_0^2-g\left(x\right)dx-2{\int }_0^{2}g\left(x\right)dx$

$\therefore I=2{\int }_0^{4}4x^{3}dx$

$\therefore I=512$

This is the required answer.