Question

Given that \( \frac{d}{dx} \left[ \int_0^{\phi(x)} f(t) dt \right] = \phi'(x) f(\phi(x)) \). If \( \int_0^{x^3} f(t) dt = x^2 \sin 2\pi x \), then the value of \( f(8) \) is

• A. \( \frac{2\pi}{3} \)
• B. \( \frac{4\pi}{3} \)
• C. \( \frac{\pi}{3} \)
• D. \( \frac{\pi}{12} \)

Hint:

Use the Fundamental Theorem of Calculus Part 1 along with the chain rule to differentiate the integral \( \int_0^{\phi(x)} f(t) dt \). Remember the derivative of \( \sin(ax) \) is \( a \cos(ax) \) and the derivative of \( \cos(ax) \) is \( -a \sin(ax) \). Solve for \( f(\phi(x)) \) and then substitute the value of \( x \) that makes \( \phi(x) = 8 \).

Answer 1

The Correct Option is A

We are given \( \int_0^{x^3} f(t) dt = x^2 \sin 2\pi x \).
Differentiate both sides with respect to \( x \) using the given formula for the derivative of a definite integral: $$ \frac{d}{dx} \left[ \int_0^{x^3} f(t) dt \right] = \frac{d}{dx} (x^2 \sin 2\pi x) $$ Here, \( \phi(x) = x^3 \), so \( \phi'(x) = 3x^2 \).
Using the formula, the left side becomes \( \phi'(x) f(\phi(x)) = 3x^2 f(x^3) \).
Now, differentiate the right side using the product rule: $$ \frac{d}{dx} (x^2 \sin 2\pi x) = (2x) \sin 2\pi x + x^2 (\cos 2\pi x \cdot 2\pi) = 2x \sin 2\pi x + 2\pi x^2 \cos 2\pi x $$ Equating both sides: $$ 3x^2 f(x^3) = 2x \sin 2\pi x + 2\pi x^2 \cos 2\pi x $$ We need to find \( f(8) \).
To get \( x^3 = 8 \), we set \( x = 2 \).
Substitute \( x = 2 \) into the equation: $$ 3(2)^2 f(2^3) = 2(2) \sin (2\pi \cdot 2) + 2\pi (2)^2 \cos (2\pi \cdot 2) $$ $$ 3(4) f(8) = 4 \sin (4\pi) + 8\pi \cos (4\pi) $$ We know that \( \sin(4\pi) = 0 \) and \( \cos(4\pi) = 1 \).
$$ 12 f(8) = 4(0) + 8\pi (1) $$ $$ 12 f(8) = 8\pi $$ $$ f(8) = \frac{8\pi}{12} = \frac{2\pi}{3} $$