Question:easy

If $f(x)$ is an even function, then $\int_{-a}^{a} f(x)dx=$

Show Hint

Always check the parity first! If $f(x)$ was an odd function ($f(-x) = -f(x)$), the integral over a symmetric interval $[-a, a]$ would be exactly 0.
  • $\int_{0}^{a} f(x)dx$
  • $2 \int_{0}^{a} f(x)dx$
  • $2a$
  • $0$
Show Solution

The Correct Option is B

Solution and Explanation

1. Definition of an Even Function: A function $f(x)$ is considered even if $f(-x) = f(x)$ for all $x$ in its domain. Geometrically, even functions are symmetric with respect to the $y$-axis.

2. Splitting the Integral: We can split the integral over the symmetric interval $[-a, a]$ into two parts: $$\int_{-a}^{a} f(x)dx = \int_{-a}^{0} f(x)dx + \int_{0}^{a} f(x)dx$$

3. Property Transformation: For the first part, let $x = -t$, then $dx = -dt$. When $x = -a$, $t = a$; when $x = 0$, $t = 0$: $$\int_{-a}^{0} f(x)dx = \int_{a}^{0} f(-t)(-dt) = \int_{0}^{a} f(-t)dt$$ Since the function is even, $f(-t) = f(t)$: $$\int_{0}^{a} f(-t)dt = \int_{0}^{a} f(t)dt$$
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