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:

Step1: Recall property of even functions

For an even function \( f(x) \), we know that \( f(-x)=f(x) \). The integral of an even function over a symmetric interval \([-a,a]\) can be split as \( \int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx \).
Here, \( a = 1 \), and we are given \( \int_{-1}^{1}f(x)dx=14 \).

Step2: Apply the property to find the integral from 0 to 1

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

Answer:

7