integration tutorial: analysis

Consider the integral
int -ln(2)ln(2) e2y sqrt(5 + e2y) dy
This is a definite integral, so we need to find the area under the integrand f(y), 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(y), 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, only one method is appropriate. We use substitution to rewrite the integrand in terms of w = 5 + e2y. This is shown below.
Substitution:
Let w = 5 + e2y. Then w' = 2 e2y, so (1/2) dw = e2y dy. The integral can therefore be rewritten as
int e2y sqrt(5 + e2y) dy = int sqrt(w) (1/2) dw = (1/3) w3/2 + C
Thus, substituting back for w,
int e2y sqrt(5 + e2y) dy = (1/3) (5 + e2y)3/2 + C
To evaluate the definite integral, we take this antiderivative, evaluate it at the endpoints of the integral (-ln(2) and ln(2)), and take the difference of the values. This gives
[ (1/3) (5 + e2(ln(2)))3/2 ] - [ (1/3) (5 + e2(-ln(2)))3/2 ] = 9 - 7sqrt(21)/8.
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integration analysis
Last Modified: Wed Feb 6 13:53:59 EST 2002
Comments to glarose@umich.edu
©2002 Gavin LaRose, UM Math Dept.