SUNY Geneseo Department of Mathematics

Substitution in Definite Integrals

Wednesday, December 2

Math 221 02
Fall 2020
Prof. Doug Baldwin

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Substitution in Definite Integrals

Based on “Substitution for Definite Integrals” in section 5.5 of our textbook and this discussion of substitution in definite integrals.

Key Idea

To evaluate a definite integral by substitution, substitute as usual, but also apply the substitution to the bounds.

Examples

There’s one decently solved example in the discussion already: integrate 2x √(x²-1) from 1 to 2. Using the substitution u = x²-1 works well with this integral, but remember to apply the substitution to the bounds too:

Substitute u equals x squared minus 1 in integral of 2 x times square root of x squared minus 1 and its bounds

The rest of the integration uses the power rule for antiderivatives and the Evaluation Theorem to use an antiderivative to evaluate a definite integral. Notice that the payoff for applying the substitution to the bounds comes at the end, when you just plug the new bounds into an antiderivative expressed in terms of u instead of having to undo the substitution:

Integrate to the 1 half from 0 to 1

Simplifying this answer further involves a nice way to think about fractional exponents: they’re (integer) powers of (integer) roots, so if you can factor the “integer power” part into a factor that cancels the “integer root” part and a factor that doesn’t, you can often get a simpler final expression:

2 thirds times 3 to the 3 halves is 2 thirds times square root of 3 squared times 3 or 2 root 3

Integrate cos(2Θ) from 0 to π.

The substitution u = 2Θ is a natural one to try here, and it works well. Once the substitution has been applied, area-under-the-curve considerations suggest that the integral should be 0, and indeed it is:

Integrate cosine 2 Theta with substitution u equals 2 Theta

Integrate sin²(4x) cos(4x) from 0 to π/8.

One effective substitution for this is u = sin(4x):

Integrate sine squared of 4 x times cosine 4 x via substitution u equals sine 4 x

But another way to do this is via 2 smaller substitutions. This leads to the same result as the first substitution, but the simpler substitutions may be easier to spot and apply than one more complicated one:

Integrate sine squared 4 x times cosine 4 x via double substitution u equals 4 x and v equals sine u

The moral is that you can do multiple substitutions on a single integral if that’s easier.

Applications of integrals to more general area problems than just the area between a curve and the x axis: area between 2 curves.

Read “Area of a Region between Two Curves” in section 6.1 of the textbook by class time Friday.

Please also contribute to this discussion of areas between curves.

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