Editing Integration by parts (section)

==Visualization==
[[Image:Integration by parts v2.svg|thumb|280px |Graphical interpretation of the theorem. The pictured curve is parametrized by the variable t.]]
Consider a parametric curve <math>(x, y) = (f(t), g(t))</math>. Assuming that the curve is locally [[Injective function|one-to-one]] and [[Locally integrable function|integrable]], we can define
<math display="block">\begin{align}
  x(y) &= f(g^{-1}(y)) \\
  y(x) &= g(f^{-1}(x))
\end{align}</math>

The area of the blue region is

<math display="block">A_1=\int_{y_1}^{y_2}x(y) \, dy</math>

Similarly, the area of the red region is
<math display="block">A_2=\int_{x_1}^{x_2}y(x)\,dx</math>

The total area ''A''<sub>1</sub> + ''A''<sub>2</sub> is equal to the area of the bigger rectangle, ''x''<sub>2</sub>''y''<sub>2</sub>, minus the area of the smaller one, ''x''<sub>1</sub>''y''<sub>1</sub>:

<math display="block">\overbrace{\int_{y_1}^{y_2}x(y) \, dy}^{A_1}+\overbrace{\int_{x_1}^{x_2}y(x) \, dx}^{A_2}\ =\ \biggl.x \cdot y(x)\biggl|_{x_1}^{x_2} \ =\ \biggl.y \cdot x(y)\biggl|_{y_1}^{y_2}</math>
Or, in terms of ''t'',
<math display="block">\int_{t_1}^{t_2}x(t) \, dy(t) + \int_{t_1}^{t_2}y(t) \, dx(t) \ =\ \biggl. x(t)y(t) \biggl|_{t_1}^{t_2}</math>
Or, in terms of indefinite integrals, this can be written as
<math display="block">\int x\,dy + \int y \,dx \ =\ xy</math>
Rearranging:
<math display="block">\int x\,dy \ =\ xy - \int y \,dx</math>
Thus integration by parts may be thought of as deriving the area of the blue region from the area of rectangles and that of the red region.

This visualization also explains why integration by parts may help find the integral of an inverse function ''f''<sup>−1</sup>(''x'') when the integral of the function ''f''(''x'') is known. Indeed, the functions ''x''(''y'') and ''y''(''x'') are inverses, and the integral ∫ ''x''&nbsp;''dy'' may be calculated as above from knowing the integral ∫ ''y''&nbsp;''dx''. In particular, this explains use of integration by parts to integrate [[logarithm]] and [[inverse trigonometric function]]s. In fact, if <math>f</math> is a differentiable one-to-one function on an interval, then integration by parts can be used to derive a formula for the integral of <math>f^{-1}</math> in terms of the integral of <math>f</math>. This is demonstrated in the article, [[Integral of inverse functions]].