Question

suppose f is an even function and (int_{-1}^{1} f(x) , dx = 14). a. evaluate (int_{0}^{1} f(x) , dx) b. evaluate (int_{-1}^{1} x f(x) , dx) a. evaluate the definite integral. (int_{0}^{1} f(x) , dx = 7) (simplify your answer.) b. evaluate the definite integral. (int_{-1}^{1} x f(x) , dx = square) (simplify your answer.)

Explanation:

Part a

Step1: Recall property of even functions

For an even function \( f(x) \), we know that \( \int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx \). Here, \( a = 1 \) and \( \int_{-1}^{1} f(x) dx = 14 \).

Step2: Solve for \( \int_{0}^{1} f(x) dx \)

Using the property, we have \( 14 = 2\int_{0}^{1} f(x) dx \). Divide both sides by 2: \( \int_{0}^{1} f(x) dx=\frac{14}{2} = 7 \).

Step1: Analyze the function \( g(x)=xf(x) \)

Check if \( g(x) \) is odd or even. \( g(-x)=(-x)f(-x) \). Since \( f \) is even, \( f(-x)=f(x) \), so \( g(-x)= -xf(x)= -g(x) \). Thus, \( g(x) \) is an odd function.

Step2: Use property of odd functions

For an odd function \( g(x) \), \( \int_{-a}^{a} g(x) dx = 0 \). Here, \( a = 1 \) and \( g(x)=xf(x) \), so \( \int_{-1}^{1} xf(x) dx = 0 \).

Answer:

\( 7 \)

Part b