Day 30, Mat227

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Announcements:
- Your 6.6 homework is returned: graded
  - #9, p. 459
  - #12
  - #47
- I just have to share one story from my Peace Corps days:
  
  No student has ever worked harder than Kpandja.
Today we begin section 7.1: Integration by parts.
- First of all, the big picture: integration by parts is just the product rule backwards. This integration technique, like all integration techniques, is really just a differentiation technique in reverse.
  Consider the method of u-substitution, for example. That's just the chain rule in reverse.
- Derivation of the integration by parts formula:
  We start with the product rule:
  
  ${\left[u(x)v(x)\right]'=u(x)v'(x)+v(x)u'(x)}$
  Now integrate both sides:
  
  ${\int_{}^{}\left[u(x)v(x)\right]'dx=\int_{}^{}\left[{u(x)v'(x)+v(x)u'(x)}\right]dx}$
  
  ${\int_{}^{}u(x)v'(x)dx=u(x)v(x)-\int_{}^{}v(x)u'(x)dx}$
  
  ${\int_{}^{}udv=uv-\int_{}^{}vdu}$
- We'll motivate this technique by considering an important example: ${\int x\ln(x)dx}$
- Integration by parts may need to be carried out multiple times: sometimes the idea is to simplify the integral each time, until a really simple one arises allowing us to calculate the final solution (e.g. Example 6, p. 491). Sometimes it's something of a trick: we compute the integral multiple times in order to return to the original integral, allowing us to solve an equation for the original integral (e.g. Example 4, p. 490).
- Suggested strategy:
  - Choose u so that u' is simpler than u itself.
  - Choose v' so that $v=\int_{}^{}v'dx$ can be evaluated.
  - Sometimes v'=1 is a good choice.
- Of course, all this works with definite integrals:
- Examples:
  1. #9, p. 492 (sometimes the function to integrate is 1!)
  2. #10 (ditto)
  3. #26
  4. #39
  5. #40
  6. #51
  7. #66,

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