Animated Solution for Mathematics - Definite Integration: Let $f(x)$ and $g(x)$ be two functions satisfying $f(x^2)+g(4-x)=4x^3$ and $g(4-x)+g(x)=0$, then the value of ${\int }_{-4}^{4}f(x^2)dx$ is ____.

Unlocking the Integral

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that, at first glance, looks like a tangled mess of unknown functions.

We have $f(x^2)$ and $g(4-x)$ dancing around in a functional equation, and we are asked to find the value of a definite integral. It feels daunting, doesn't it?

But here is the secret: in the world of calculus, complexity is often just a mask for hidden symmetry. Let us peel back that mask together.

The Power of Even Functions

We begin with the integral $I={\int }_{-4}^{4}f(x^2)dx$. Before we even touch a pen to paper, look at the integrand. The argument is $x^2$.

This is our first clue. If we replace $x$ with $-x$, the term $x^2$ remains $x^2$. This means $f((-x{)}^2)=f(x^2)$.

Mathematically, this confirms that our function is even. Because even functions are symmetric about the y-axis, the area under the curve from $-4$ to $0$ is an exact mirror image of the area from $0$ to $4$.

Therefore, we can simplify our integral to:

This is a powerful simplification. We have halved our work and focused our attention on the positive domain, where things are usually much easier to handle.

The Functional Equation as a Key

Now, we look at the given equation: $f(x^2)+g(4-x)=4x^3$. We don't know $f$, and we don't know $g$.

But we have a relationship between them. Let us isolate $f(x^2)$:

Substituting this into our integral, we get:

By the linearity of integration, we can split this into two parts: $I=2{\int }_0^{4}4x^{3}dx-2{\int }_0^{4}g(4-x)dx$. Let us call these $I_1$ and $I_2$.

The first part, $I_1$, is a straightforward polynomial integral. The second part, $I_2$, is where the mystery lies.

The Vanishing Act of $g(x)$

Let us evaluate $I_1$ first. The integral of $4x^3$ is $x^4$. Evaluating from $0$ to $4$, we get $4^4-0^4=256$.

Multiplying by the $2$ outside, we get $I_1=512$. Now, for $I_2=2{\int }_0^{4}g(4-x)dx$.

The substitution $t=4-x$ transforms this integral into $2{\int }_0^{4}g(t)dt$, which is just $2{\int }_0^{4}g(x)dx$.

But wait! We have another piece of information: $g(4-x)+g(x)=0$. This implies $g(4-x)=-g(x)$.

If we integrate this relation from $0$ to $4$, we get:

Since we know ${\int }_0^{4}g(4-x)dx={\int }_0^{4}g(x)dx$, this equation becomes $2{\int }_0^{4}g(x)dx=0$.

This means the integral of $g(x)$ over this interval is exactly zero! The positive and negative areas have perfectly cancelled each other out. The mystery function $g(x)$ has vanished into thin air.

We are left with $I=I_1-I_2=512-0=512$. Look at that!

We didn't need to know the explicit form of $f(x)$ or $g(x)$. We only needed to understand the symmetry and the properties of the definite integral.

This is the beauty of JEE mathematics—it rewards conceptual clarity over brute-force calculation. You have successfully navigated the trap and arrived at the solution: 512.