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Lesson 53 – Integration by Parts

Lesson 53 – Integration by Parts. Calculus - Santowski. Lesson Objectives. Use the method of integration by parts to integrate simple power, exponential, and trigonometric functions both in a mathematical context and in a real world problem context. (A) Product Rule.

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Lesson 53 – Integration by Parts

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  1. Lesson 53 – Integration by Parts Calculus - Santowski Calculus - Santowski

  2. Lesson Objectives • Use the method of integration by parts to integrate simple power, exponential, and trigonometric functions both in a mathematical context and in a real world problem context Calculus - Santowski

  3. (A) Product Rule • Recall that we can take the derivative of a product of functions using the product rule: • So we are now going to integrate this equation and see what emerges Calculus - Santowski

  4. (B) Product Rule in Integral Form • We have the product rule as • And now we will integrate both sides: Calculus - Santowski

  5. (B) Product Rule in Integral Form • We will make some substitutions to simplify this equation: Calculus - Santowski

  6. (C) Integration by Parts Formula • So we have the formula • So what does it mean? • It seems that if we are trying to solve one integral • and we create a second integral !!!! • Our HOPE is that the second integral is easier to solve than the original integral Calculus - Santowski

  7. (D) Examples • Integrate the following functions: Calculus - Santowski

  8. (D) Examples • The easiest way to master the method is by practicing, so determine • We need to select a u and a dv • So we have a choice ….. • Comment: Is the second integral any easier than the original??? Calculus - Santowski

  9. (D) Examples • So let’s make the other choice as we determine • Checkpoint: Is our second integral any “easier” than our first one??? • Now verify by differentiating the answer Calculus - Santowski

  10. Example 4: LIPET This is still a product, so we need to use integration by parts again.

  11. Example: LIPET logarithmic factor

  12. (D) More Examples • Integrate the following functions: Calculus - Santowski

  13. Example 1: LIPET polynomial factor

  14. Example 5: LIPET This is the expression we started with!

  15. Example 6: LIPET

  16. This is called “solving for the unknown integral.” It works when both factors integrate and differentiate forever. Example 6:

  17. (E) Further Examples • For the following question, do it using the method requested. Reconcile your solution(s) Calculus - Santowski

  18. (E) Further Examples • Definite Integrals  Evaluate: Calculus - Santowski

  19. Calculate Let Then,

  20. To evaluate this integral, we use the substitution t = 1 + x2 (since u has another meaning in this example). Then, dt = 2x dx. So, x dx =½dt.

  21. When x = 0, t = 1, and when x = 1, t = 2. Hence,

  22. Therefore,

  23. As tan-1x≥ for x≥ 0 , the integral in the example can be interpreted as the area of the region shown here.

  24. Further Examples

  25. Further Examples

  26. Test Qs Calculus - Santowski

  27. INTEGRATION BY PARTS • Evaluate ∫ exsinx dx • ex does not become simpler when differentiated. • Neither does sin x become simpler.

  28. INTEGRATION BY PARTS • Nevertheless, we try choosing u = exand dv = sin x • Then, du = ex dx and v = – cos x.

  29. INTEGRATION BY PARTS • So, integration by parts gives: • The integral we have obtained, ∫excos xdx,is no simpler than the original one. • At least, it’s no more difficult. • Having had success in the preceding example integrating by parts twice, we do it again.

  30. INTEGRATION BY PARTS • This time, we use u = exand dv = cos x dx • Then, du = ex dx, v = sin x, and

  31. INTEGRATION BY PARTS • At first glance, it appears as if we have accomplished nothing. • We have arrived at ∫ exsin x dx, which is where we started.

  32. INTEGRATION BY PARTS • However, if we put the expression for ∫ excos x dx from Equation 5 into Equation 4, we get: • This can be regarded as an equation to be solved for the unknown integral.

  33. INTEGRATION BY PARTS • Adding to both sides ∫ exsin x dx, we obtain:

  34. INTEGRATION BY PARTS • Dividing by 2 and adding the constant of integration, we get:

  35. INTEGRATION BY PARTS • The figure illustrates the example by showing the graphs of f(x) = ex sin x and F(x) = ½ ex(sin x – cos x). • As a visual check on our work, notice that f(x) = 0when F has a maximum or minimum.

  36. (F) Internet Links • Integration by Parts from Paul Dawkins, Lamar University • Integration by Parts from Visual Calculus Calculus - Santowski

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