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6.2 Integration by Substitution M.L.King Jr. Birthplace, Atlanta, GA

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Presentation on theme: "6.2 Integration by Substitution M.L.King Jr. Birthplace, Atlanta, GA"— Presentation transcript:

1 6.2 Integration by Substitution M.L.King Jr. Birthplace, Atlanta, GA
Greg Kelly Hanford High School Richland, Washington Photo by Vickie Kelly, 2002

2 How did I know that? Because I know that by the Chain Rule… Which implies that…

3 The chain rule allows us to differentiate a wide variety of functions, but we are able to find antiderivatives for only a limited range of functions. We can sometimes use substitution to rewrite functions in a form that we can integrate.

4 Now we substitute x2 back in for u so that our answer is…
We can’t integrate this until we make this all one variable NO!!! Let u = x2 But we also know that… Which implies that… So now we can say that… Now we substitute x2 back in for u so that our answer is…

5 The variable of integration must match the variable in the expression.
Don’t forget to substitute the value for u back into the problem!

6 Note that this only worked because of the 2x in the original.
Example (Exploration 1 in the book) One of the clues that we look for is if we can find a function and its derivative in the integrand. The derivative of is Note that this only worked because of the 2x in the original. Many integrals can not be done by substitution.

7 Solve for dx.

8

9 Example: (Not in book) We solve for because we can find it in the integrand.

10

11 The technique is a little different for definite integrals.
We can find new limits, and then we don’t have to substitute back. new limit new limit We could have substituted back and used the original limits. =

12 Wrong! The limits don’t match! Using the original limits:
Leave the limits out until you substitute back. Wrong! The limits don’t match! This is usually more work than finding new limits

13 The technique is a little different for definite integrals.
new limit new limit Changing the limits as we did above is the preferred method of substitution on the AP Exam.

14 Example: (Exploration 2 in the book)
Don’t forget to use the new limits.

15 Separable differential equation
Combined constants of integration

16 We now have y as an implicit function of x.
We can find y as an explicit function of x by taking the tangent of both sides. Notice that we can not factor out the constant C, because the distributive property does not work with tangent.

17 In another generation or so, we might be able to use the calculator to find all integrals.
Until then, remember that only part of the AP exam and half the nation’s college professors do not allow calculators. You must practice finding integrals by hand until you are good at it! p


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