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Step-by-Step Solution
Step 1: Express the integral in a convenient form
We want to find the value of
I = \displaystyle \int_{-4}^4 f\bigl(x^2\bigr)\,dx.
Because f\bigl(x^2\bigr) depends on x^2 , the integrand is an even function with respect to x . Therefore, we can use the property of even functions to convert the integral from [-4,4] to [0,4] and multiply by 2 :
I
= \int_{-4}^4 f(x^2)\,dx
= 2 \int_{0}^4 f(x^2)\,dx.
Step 2: Use the given functional equation
We have the functional relationship:
f\bigl(x^2\bigr) + g(4 - x) = 4x^3.
We also know that
g(4 - x) + g(x) = 0 \quad \Longrightarrow \quad g(4 - x) = -g(x).
Step 3: Substitute x by (4 - x) in the functional equation
Replacing x by (4 - x) in
f\bigl(x^2\bigr) + g(4 - x) = 4x^3,
gives
f\bigl((4 - x)^2\bigr) + g(x) = 4\bigl(4 - x\bigr)^3.
Step 4: Add the two functional equations and simplify
Adding
f\bigl(x^2\bigr) + g(4 - x) = 4x^3
and
f\bigl((4 - x)^2\bigr) + g(x) = 4\bigl(4 - x\bigr)^3
results in:
f\bigl(x^2\bigr) + f\bigl((4 - x)^2\bigr) + g(4 - x) + g(x)
= 4\bigl[x^3 + (4 - x)^3\bigr].
Since g(4 - x) + g(x) = 0, we get:
f\bigl(x^2\bigr) + f\bigl((4 - x)^2\bigr)
= 4\bigl[x^3 + (4 - x)^3\bigr].
Step 5: Form the integral for f(x^2) + f((4 - x)^2)
From Step 1,
I
= 2\int_{0}^4 f(x^2)\,dx
\quad\text{and similarly}\quad
I
= 2\int_{0}^4 f\bigl((4 - x)^2\bigr)\,dx.
Adding these two expressions for I gives:
2I
= 2 \int_{0}^4 \Bigl[f\bigl(x^2\bigr) + f\bigl((4 - x)^2\bigr)\Bigr] dx.
Hence,
2I
= 2 \int_{0}^4 4\bigl[x^3 + (4 - x)^3\bigr] \, dx
\quad\Longrightarrow\quad
2I
= 8 \int_{0}^4 \bigl[x^3 + (4 - x)^3\bigr] \, dx.
Therefore,
I
= 4 \int_{0}^4 \bigl[x^3 + (4 - x)^3\bigr] \, dx.
Step 6: Compute the integral
We evaluate
\int_{0}^4 x^3 \, dx
= \left[ \frac{x^4}{4} \right]_0^4
= \frac{4^4}{4}
= \frac{256}{4}
= 64.
Next, let us compute
\int_{0}^4 (4 - x)^3 \, dx.
By substituting t = 4 - x, we get dt = -dx. When x = 0, t = 4, and when x = 4, t = 0. Thus,
\int_{0}^4 (4 - x)^3 \, dx
= \int_{4}^0 t^3 \,(-dt)
= \int_{0}^4 t^3 \, dt
= \left[ \frac{t^4}{4} \right]_0^4
= 64.
Therefore,
\int_{0}^4 \bigl[x^3 + (4 - x)^3\bigr] \, dx
= 64 + 64
= 128.
Step 7: Final value of the integral
From Step 5,
I
= 4 \times 128
= 512.
Hence, the value of
\int_{-4}^4 f\bigl(x^2\bigr)\, dx
is \boxed{512} .
