integration tutorial: analysis

Consider the integral
int 0ln(4) er sqrt(er) dr
This is a definite integral, so we need to find the area under the integrand f(r), 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(r), 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 substitution to rewrite the integrand in terms of w = er, or can rewrite the integrand: use properties of exponents to combine the numerator and denominator of the fraction to give an expression that is more easily integrable. Once we have the antiderivative, we determine the value of the definite integral by evaluating it at the endpoints (0 and ln(4)) and taking the difference. This is shown below: substitution; rewriting.
Substitution:
Let w = er. Then w' = er, so dw = er dr. The integral can therefore be rewritten as
int er sqrt(er) dr = int sqrt(w) dw = (2/3) w3/2 + C
Thus, substituting back for w,
int er sqrt(er) dr = (2/3) (er)3/2 + C
To evaluate the definite integral, we take this antiderivative, evaluate it at the endpoints of the integral (0 and ln(4)), and take the difference of the values. This gives
[ (2/3) (e(ln(4)))3/2 ] - [ (2/3) (e(0))3/2 ] = 14/3.
Rewriting:
Use properties of exponents to combine the numerator and denominator of the fraction to rewrite the integral:
er sqrt(er) = e(3/2) r.
Thus,
int er sqrt(er) dr =inte(3/2) rdr
which can be evaluated using substitution (of "ax"), to obtain
inte(3/2) rdr = ((2/3) e(3/2) r) + C.

Explanation for rewritten term(s)
Substitution (of "ax")
Let w = 3/2 r. Then w' = 3/2, so (1/(3/2)) dw = dr. The integral can therefore be rewritten as
int e(3/2) r dr = int ew (1/(3/2)) dw = (2/3) ew + C
Thus, substituting back for w,
int e(3/2) r dr = (2/3) e(3/2) r + C

To evaluate the definite integral, we take this antiderivative, evaluate it at the endpoints of the integral (0 and ln(4)), and take the difference of the values. This gives
[ ((2/3) e(3/2) (ln(4))) ] - [ ((2/3) e(3/2) (0)) ] = 14/3.
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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.