Question

If \( I = \int_{-a}^{a} \left( x^4 - 2x^2 \right) \, dx \), then \( I \) is minimum at \( a = \):

• A. \( 2 \)
• B. \( -\sqrt{2} \)
• C. \( \sqrt{2} \)
• D. \( -2 \)

Hint:

When solving integrals involving even powers of \( x \), use symmetry to simplify your calculations. For symmetric intervals, odd powers of \( x \) will cancel out, leaving only the even powers.

Answer 1

The Correct Option is C

The integral given is \( I = \int_{-a}^{a} \left( x^4 - 2x^2 \right) \, dx \). First, calculate the integral: \[ I = \int_{-a}^{a} x^4 \, dx - 2 \int_{-a}^{a} x^2 \, dx \] The integral of \( x^4 \) is: \[ \int_{-a}^{a} x^4 \, dx = \left[ \frac{x^5}{5} \right]_{-a}^{a} = \frac{a^5}{5} - \left( -\frac{a^5}{5} \right) = \frac{2a^5}{5} \] The integral of \( x^2 \) is: \[ \int_{-a}^{a} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-a}^{a} = \frac{a^3}{3} - \left( -\frac{a^3}{3} \right) = \frac{2a^3}{3} \] Thus, the integral \( I \) becomes: \[ I = \frac{2a^5}{5} - 2 \times \frac{2a^3}{3} = \frac{2a^5}{5} - \frac{4a^3}{3} \] Now, to find the minimum of \( I \), differentiate \( I \) with respect to \( a \) and set the derivative equal to zero. \[ \frac{dI}{da} = \frac{10a^4}{5} - \frac{12a^2}{3} = 2a^4 - 4a^2 \] Set the derivative equal to zero: \[ 2a^4 - 4a^2 = 0 \quad \Rightarrow \quad 2a^2(a^2 - 2) = 0 \] This gives \( a^2 = 2 \), so \( a = \pm \sqrt{2} \). Thus, \( I \) is minimum at \( a = \sqrt{2} \). Hence, the correct answer is option (3).