## 7.1. Riemann Integral

### Examples 7.1.21(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.
*x*on the interval^{5}- 4 x^{2}dx*[0, 2]*.*1/x*on the interval^{2}+ cos(x) dx*[1, 4]*.*(1 + x*on the interval^{2})^{-1}dx*[-1, 1]*.

**1.**
Here is a numerical approximation of
x^{5} - 4 x^{2} dx
on the interval *[0, 2]*.

Then we clearly have:P(x) = 1/6 x^{6}- 4/3 x^{3}+ C

so thatP'(x) = x^{5}- 4 x^{2}

*P*is an antiderivate of the integrand. Therefore:

With our choices ofx^{5}- 4 x^{2}dx = P(b) - P(a)

*a*and

*b*we can evaluate the integral to

P(2) - P(0) = 1/6 2^{6}- 4/3 2^{3}= 1/6 64 - 4/3 * 8 = 0

**2.**
Here is a numerical approximation of
* 1/x ^{2} + cos(x) dx*
on the interval

*[1, 4]*.

*1/x*is continuous over the interval

^{2}+ cos(x)*[-1, 4]*so that the function is Riemann integrable. If we define

*P(x) = -1/x + sin(x) + C*then

*P'(x) = 1/x*so that again we found an antiderivative of the original integrand. Using the Integral Evaluation Shortcut over

^{2}+ cos(x)*[-1, 4]*we can compute the value of the integral to

P(4) - P(1) = -1/4 + sin(4) - (-1/1 + sin(1)) = -.8482734801

**3.**
Here is a numerical approximation of
*(1 + x ^{2})^{-1} dx*
on the interval

*[-1, 1]*.

*(1 + x*is less obvious than in the previous cases until we remember that

^{2})^{-1}But then the integral evaluates toarctan(x) = (1 + x^{2})^{-1}

arctan(1) - arctan(-1) = 2 /4 = /2 = 1.570796327