Let's solve the problem step by step: --- ### **Step 1: Use the property of even functions** Given: \( f(x) \) is an even function, so \( f(-x) = f(x) \) for all \( x \). Consider the integral: \[ \int_{-a}^{a} f(x) \, dx \] Let's split this integral at zero: \[ \int_{-a}^{a} f(x) \, dx = \int_{-a}^{0} f(x) \, dx + \int_{0}^{a} f(x) \, dx \] **Explanation of Step 1:** We use the property of definite integrals to split the interval at 0, since the function is even and the interval is symmetric about 0. --- ### **Step 2: Change of variable in the first integral** Let \( u = -x \). Then when \( x = -a, u = a \) and when \( x = 0, u = 0 \). Also, \( dx = -du \). So, \[ \int_{-a}^{0} f(x) \, dx = \int_{a}^{0} f(-u) (-du) = \int_{0}^{a} f(-u) du \] Since \( f \) is even, \( f(-u) = f(u) \): \[ \int_{-a}^{0} f(x) \, dx = \int_{0}^{a} f(u) du \] So, the total integral is: \[ \int_{-a}^{a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx \] **Explanation of Step 2:** By changing variables and using the property of even functions, we showed the negative half of the integral is equal to the positive half, so the total is twice the integral over \([0, a]\). --- ### **Final Answer:** \[ \boxed{ \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx } \]