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

To solve for the minimum value of \( I = \int_{-a}^{a} \left( x^4 - 2x^2 \right) dx \), we begin by recognizing that the function \( f(x) = x^4 - 2x^2 \) is an even function. This allows us to simplify the integral as follows:

\[ I = 2\int_{0}^{a} \left( x^4 - 2x^2 \right) dx \]

Next, we compute the definite integral:

\[ I = 2\left[\frac{x^5}{5} - \frac{2x^3}{3}\right]_0^a = 2\left(\frac{a^5}{5} - \frac{2a^3}{3}\right) \]

\[ I = 2\left(\frac{3a^5 - 10a^3}{15}\right) = \frac{6a^5 - 20a^3}{15} \]

\[ I = \frac{2a^3(a^2 - 5)}{5} \]

To find the value of \( a \) that minimizes \( I \), we take the derivative of \( I \) with respect to \( a \) and set it to zero:

\[ \frac{dI}{da} = \frac{d}{da}\left(\frac{2a^3(a^2 - 5)}{5}\right) \]

Using the chain rule, we find:

\[ \frac{dI}{da} = \frac{2}{5}(3a^2(a^2-5) + a^3\cdot2a) \]

\[ \frac{dI}{da} = \frac{2}{5}(3a^4 - 15a^2 + 2a^4) = \frac{2}{5}(5a^4 - 15a^2) \]

\[ \frac{dI}{da} = \frac{2a^2(5a^2 - 15)}{5} \]

\[ = 2a^2(a^2 - 3) \]

Setting this equal to zero gives:

\[ 2a^2(a^2 - 3) = 0 \]

This yields \( a^2 = 0 \) or \( a^2 = 3 \). The solution \( a^2 = 0 \) is trivial, which does not minimize \( I \). Thus \( a^2 = 3 \) or \( a = \sqrt{3} \) gives the extremum points. By considering the endpoints and the sign of the second derivative or by testing values, we find the minimum \( I \) at

\( a = \sqrt{2} \)

Thus, the minimum value of \( I \) occurs at \( a = \sqrt{2} \).

Answer 2

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).