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} xf(x) , dx)
a. evaluate the definite integral. (int_{0}^{1} 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 \( \int_{0}^{1} f(x) dx \)

From the property \( \int_{-1}^{1} f(x) dx = 2\int_{0}^{1} f(x) dx \), we can rearrange to get \( \int_{0}^{1} f(x) dx=\frac{1}{2}\int_{-1}^{1} f(x) dx \). Substituting \( \int_{-1}^{1} f(x) dx = 14 \), we have \( \int_{0}^{1} f(x) dx=\frac{14}{2}=7 \).

Step1: Recall property of odd functions

A function \( g(x)=xf(x) \). For an even function \( f(x) \), \( f(-x) = f(x) \). Then \( g(-x)=-x f(-x)=-x f(x)=-g(x) \), so \( g(x) \) is an odd function.

Step2: Integral of odd function over symmetric interval

The integral of an odd function over a symmetric interval \( [-a,a] \) is \( 0 \). So \( \int_{-1}^{1} xf(x) dx = 0 \) since \( xf(x) \) is odd and the interval \( [-1,1] \) is symmetric about \( 0 \).

Answer:

\( 7 \)

Part b