Question

Evaluate the integral: \[ \int_{-a}^{a} x^2 \left( \frac{e^{x^3} - e^{-x^3}}{e^{x^3} + e^{-x^3}} \right) dx \]

• A. \( a^2 \)
• B. \( 0 \)
• C. \( a \)
• D. \( 2 \displaystyle \int_{0}^{a} x^2 \left( \frac{e^{x^3} - e^{-x^3}}{e^{x^3} + e^{-x^3}} \right) dx \)

Hint:

Always check whether the integrand is an even or odd function before evaluating a definite integral with symmetric limits. This can save a lot of calculation time.

Answer 1

The Correct Option is B

Step 1: Identify the nature of the function.
Consider the integrand \[ f(x) = x^2 \left( \frac{e^{x^3} - e^{-x^3}}{e^{x^3} + e^{-x^3}} \right) \] Here, \( x^2 \) is an even function, while \[ \frac{e^{x^3} - e^{-x^3}}{e^{x^3} + e^{-x^3}} \] is an odd function because replacing \( x \) by \( -x \) changes the sign of the numerator but not the denominator.

Step 2: Determine the parity of the integrand.
The product of an even function and an odd function is an odd function. Hence, \( f(x) \) is an odd function.

Step 3: Use the property of definite integrals.
For any odd function \( f(x) \), \[ \int_{-a}^{a} f(x)\,dx = 0 \] Therefore, \[ \int_{-a}^{a} x^2 \left( \frac{e^{x^3} - e^{-x^3}}{e^{x^3} + e^{-x^3}} \right) dx = 0 \]

Step 4: Final conclusion.
Using the symmetry property of odd functions over symmetric limits, the value of the given integral is zero.