Question

\(\int_{-a}^{a} |x| \,dx =\)

• A. a
• B. 2a
• C. 0
• D. \(a^2\)

Hint:

Recognizing whether a function is even or odd is a powerful shortcut for definite integrals over symmetric intervals like \([-a, a]\).

• If \(f(x)\) is even, \(\int_{-a}^{a} f(x) \,dx = 2 \int_{0}^{a} f(x) \,dx\).

• If \(f(x)\) is odd, \(\int_{-a}^{a} f(x) \,dx = 0\).
This can simplify the calculation significantly.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to evaluate the definite integral of the absolute value function, \(|x|\), over a symmetric interval \([-a, a]\).

Step 2: Key Formula or Approach:
The function \(f(x) = |x|\) is an even function, because \(f(-x) = |-x| = |x| = f(x)\).
For any even function, we have the property:
\[ \int_{-a}^{a} f(x) \,dx = 2 \int_{0}^{a} f(x) \,dx \]
We also need the definition of \(|x|\): \(|x| = x\) for \(x \ge 0\).

Step 3: Detailed Explanation:
Let \(I = \int_{-a}^{a} |x| \,dx\).
Since \(|x|\) is an even function, we can simplify the integral:
\[ I = 2 \int_{0}^{a} |x| \,dx \]
In the interval \([0, a]\), \(x\) is non-negative, so \(|x| = x\).
The integral becomes:
\[ I = 2 \int_{0}^{a} x \,dx \]
Now, we evaluate this simple integral:
\[ I = 2 \left[ \frac{x^2}{2} \right]_0^a \]
\[ I = 2 \left( \frac{a^2}{2} - \frac{0^2}{2} \right) \]
\[ I = 2 \left( \frac{a^2}{2} \right) = a^2 \]

Alternative Method (Splitting the integral):
We can split the integral based on the definition of \(|x|\):
\(|x| = -x\) for \(x \lt 0\) and \(|x| = x\) for \(x \ge 0\).
\[ I = \int_{-a}^{0} |x| \,dx + \int_{0}^{a} |x| \,dx \]
\[ I = \int_{-a}^{0} (-x) \,dx + \int_{0}^{a} x \,dx \]
\[ I = \left[ -\frac{x^2}{2} \right]_{-a}^{0} + \left[ \frac{x^2}{2} \right]_{0}^{a} \]
\[ I = \left( -\frac{0^2}{2} - \left(-\frac{(-a)^2}{2}\right) \right) + \left( \frac{a^2}{2} - \frac{0^2}{2} \right) \]
\[ I = \left( 0 + \frac{a^2}{2} \right) + \left( \frac{a^2}{2} - 0 \right) = \frac{a^2}{2} + \frac{a^2}{2} = a^2 \]

Step 4: Final Answer:
The value of the integral is \(a^2\).