integration tutorial: analysis

Consider the integral
int -22 3s5 - 4s3 ds
This is a definite integral, so we need to find the area under the integrand f(s), which we do by using the Fundamental Theorem of Calculus: we find an antiderivative, evaluate it at the endpoints of the integral, and take the difference of the values.
To find an antiderivative of f(s), we go through our list of integration methods:
  1. Recognize elementary antiderivatives
  2. Rewrite the integrand to make it easier
  3. Use substitution to reverse the chain rule or simplify the integrand
  4. Use integration by parts
  5. Use inspection to see the value of a definite integral
to find one that works for this integrand. In this case, there are several methods that work. We can use basic antidifferentiation to determine the antiderivative, or use inspection to determine the value of the integral. This is shown below: basic antidifferentiation; inspection.
Basic antidifferentiation:
In this case, by using rules for basic antiderivatives we can just write down the antiderivative of the integrand:
int 3s5 - 4s3 ds = (1/2)s6 - s4 + C
To evaluate the definite integral, we take this antiderivative, evaluate it at the endpoints of the integral (-2 and 2), and take the difference of the values. This gives
[ (1/2)(2)6 - (2)4 ] - [ (1/2)(-2)6 - (-2)4 ] = 0.
Inspection:
Note that we are integrating on odd function over an interval symmetric about the y-axis. Because the function is odd, the area to the left of the axis will be the negative of the area to the right, and the value of the integral must be 0.
[ ]
integration analysis
Last Modified: Wed Feb 6 13:53:59 EST 2002
Comments to glarose@umich.edu
©2002 Gavin LaRose, UM Math Dept.