Question

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

Answer 1

Finding the value of ${\int }_{-4}^{4}f\left(x^2\right)dx$:

Given, $f\left(x^2\right)+g(4-x)=4x^3$ and $g\left(4-x\right)+g\left(x\right)=0$

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

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

$\Rightarrow I=2{\int }_0^{4}f\left(x^2\right)dx\dots \left(1\right)\left({\int }_{-4}^{0}f\left(x^2\right)dx={\int }_0^{4}f\left(x^2\right)dx\right)$

Now replace $x$ with $(4-x)$ ,

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

Adding equation $\left(1\right)$ and $\left(2\right)$, we get,

$\Rightarrow 2I=2{\int }_0^4\left[f\left(x^2\right)+f\left({\left(4-x\right)}^2\right)\right]dx\dots \left(3\right)$

Now using, $f\left(x^2\right)+g(4-x)=4x^3\dots \left(4\right)$

Replace $x$ with $(4-x)$.

$\Rightarrow f\left({\left(4-x\right)}^2\right)+g(4-x)=4{\left(4-x\right)}^3\dots \left(5\right)$

Adding equation $\left(4\right)$ and $\left(5\right)$, we get,

$$\begin{gathered} \Rightarrow f\left({\left(4-x\right)}^2\right)+f\left(x^2\right)+g\left(x\right)+g(4-x)=4x^3+4{(4-x)}^3 \\ \Rightarrow f\left({\left(4-x\right)}^2\right)+f\left(x^2\right)=4\left(x^3+{\left(4-x\right)}^3\right) \end{gathered}$$

Now, $I={\int }_0^4\left(x^3+{\left(4-x\right)}^3\right)dx=512$.

Hence, the value of ${\int }_{-4}^{4}f\left(x^2\right)dx$ is $512$.