Examples 7.1.20(b):
Find the value of the following integrals:
This time we will use the Integral Evaluation Shortcut, or First
Fundamental Theorem of Calculus, which requires us to find the
antiderivative for each of the functions.
- x5 - 4 x2 dx on the interval [0, 2].
- 1/x2 + cos(x) dx on the interval [1, 4].
- (1 + x2)-1 dx on the interval [-1, 1].
1. Here is a numerical approximation of x5 - 4 x2 dx on the interval [0, 2].
P(x) = 1/6 x6 - 4/3 x3 + CThen we clearly have:
P'(x) = x5 - 4 x2so that P is an antiderivate of the integrand. Therefore:
x5 - 4 x2 dx = P(b) - P(a)With our choices of a and b we can evaluate the integral to
P(2) - P(0) = 1/6 26 - 4/3 23 = 1/6 64 - 4/3 * 8 = 0
2. Here is a numerical approximation of 1/x2 + cos(x) dx on the interval [1, 4].
P(4) - P(1) = -1/4 + sin(4) - (-1/1 + sin(1)) = -.8482734801
3. Here is a numerical approximation of (1 + x2)-1 dx on the interval [-1, 1].
arctan(x) = (1 + x2)-1But then the integral evaluates to
arctan(1) - arctan(-1) = 2 /4 = /2 = 1.570796327