Calculus

Integration by Parts

The product rule, run backwards and read as a picture. Plot one factor against the other and the curve cuts a rectangle into two pieces: the area under it and the area beside it. One of those is the integral you were asked for and the other is the one you are left with, and they have to add up to the corners. Choosing u is choosing which of the two you would rather do.

a few wide onestoo fine to see

Every number is integrated numerically from the two functions themselves, so the identity is checked on screen rather than assumed.

What to observe

  1. The curve is one factor plotted against the other as x runs from a to b. Nothing else is plotted, and that single curve does all the work: it splits the corner of the plane into the area underneath it andthe area beside it. Those two are exactly ∫v du and ∫u dv.
  2. Watch what the two pieces have to add up to. The big rectangle isu(b)·v(b), the small one tucked in the corner is u(a)·v(a), and everything between them is covered by the two shaded regions with nothing left over and no overlap. That is the whole theorem:∫u dv + ∫v du = [uv], which is just the product rule with an integral sign in front of it.
  3. Drag strips. The vertical ones stack along the u-axis and measure ∫v du; the horizontal ones stack up the v-axis and measure ∫u dv. Same curve, same region, sliced two different ways, and the residual in the panel falls towards zero as they get thinner.
  4. Now the part that matters in practice. Switch which factor you call u and the curve is replaced by a completely different one, but the two areas still tile the same kind of corner. What changes is theleftover: with u = x the integral you are handed back is ∫eˣ dx, and with u = eˣ it is ∫(x²/2)eˣ dx, which is worse than what you started with. That is the whole of the LIATE rule of thumb, and you can see it rather than memorise it.
  5. Pick ∫ ln x dx, where there is apparently only one factor. Taking u = ln x and dv = dx makes the leftover ∫x·(1/x) dx, which is just ∫1 dx. The trick is not that a second factor appeared, it is that dv = dxwas always available.
  6. Pick ∫ x² eˣ dx. The leftover is ∫2x eˣ dx: not finished, but one power of x lighter than before. Apply the same move again and the x is gone entirely. Integration by parts is often not a solution but astaircase, and each step is one of these pictures.
  7. One quiet freedom: v is any antiderivative of dv, so the constant is yours to choose. For ∫x sin x dx this page uses v = 1 − cos x instead of the usual −cos x, purely so the curve starts on the axis and the region stays in view. The identity never noticed.

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