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Section 8.2 – Integration by Parts

Section 8.2 – Integration by Parts. Find the Error. The following is an example of a student response. How can you tell the final answer is incorrect? Where did the student make an error?. The integral of a product is not equal to product of the integrals. Evaluate: .

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Section 8.2 – Integration by Parts

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  1. Section 8.2 – Integration by Parts

  2. Find the Error The following is an example of a student response. How can you tell the final answer is incorrect? Where did the student make an error? The integral of a product is not equal to product of the integrals. Evaluate: This should remind us of the Product Rule. Is there a way to use the Product Rule to investigate the antiderivative of a product? This is not the antiderivativeof since

  3. Integration by Parts: An Explanation When u and v are differentiable functions of x: The Product Rule tells us… If we integrate both sides… If we simplify the integrals… If we solve for one of the integrals…

  4. Integration by Parts When evaluating the integral and f (x)dx=u dv (with u and vbeing differentiable functions of x), then, the following holds: Rewrite the function into the product of u and dv. The integral equals… The integral of the product of the antiderivative of dv and the derivative of u. u times the antiderivative of dv.

  5. Integration by Parts: The Process Since f (x)dx=u dv, success in using this important technique depends on being able to separate a given integral into parts u and dv so that… • dv can be integrated. • is no more difficult to calculate than the original integral. The following does NOT always hold, but is very helpful: Frequently, the derivative of u, or any higher order derivative, will be zero.

  6. Example 1 Evaluate: Pick the u and dv. Find du and v. Differentiate. Integrate. Apply the formula.

  7. Example 2 Evaluate: Pick the u and dv. Find du and v. Differentiate. Integrate. Apply the formula. Pick the u and dv. You may need to apply Integration by Parts Again. Find du and v. Apply the formula.

  8. White Board Challenge Evaluate:

  9. Example 3 Notice the cubic function will go to zero. So it is a good choice for u. Evaluate: Since, multiple Integration by Parts are needed, a Tabular Method is a convenient method for organizing repeated Integration by parts. Integrate the dv. Differentiate the u. Start with + + Alternate – + Connect the diagonals. – Must get 0. Find the sum of the products of each diagonal:

  10. Example 4 Evaluate: Pick the u and dv. Find du and v. Differentiate. Integrate. This was a bad choice for u and dv.

  11. Example 4: Second Try Try the opposite this time. Evaluate: Pick the u and dv. Find du and v. Differentiate. Integrate. Apply the formula.

  12. Example 5 Evaluate: If there is only one function, rewrite the integral so there is two. Pick the u and dv. Find du and v. Differentiate. Integrate. Apply the formula.

  13. Example 6 Evaluate: Pick the u and dv. Find du and v. Apply the formula. You may need to apply Integration by Parts Again. Pick the u and dv. Find du and v. Apply the formula. If you see the integral you are trying to find, solve for it.

  14. Example 7 Evaluate: If there is only one function, rewrite the integral so there is two. Pick the u and dv. Find du and v. Apply the formula.

  15. Integration by Parts: Helpful Acronym When deciding which product to make u, choose the function whose category occurs earlier in the list below. Then take dv to be the rest of the integrand. L I A T E ogarithmic nverse trigonometric lgebraic rigonometric xponential

  16. White Board Challenge Evaluate:

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