Integration by Parts: Graphic Presentation

Discussion in 'Math' started by Mark44, May 5, 2008.

  1. Mark44

    Thread Starter Well-Known Member

    Nov 26, 2007
    All of you studying integration techniques in calculus have probably come across integration by parts, or you will soon.

    The basic formula is this:
    \int u dv = uv - \int v du

    The purpose of this technique is to start from an integral that you can't figure out, and end up with one that you can.

    There's a graphic representation of how this works, and I thought I would share it with any interest forum readers.

    In the attached figure, we see the graph of y = f(x) above the interval from x = a to x = b. The graph I've drawn happens to be such that it has an inverse, and that is crucial to the graphic presentation, but not to the integration by parts technique itself.

    The area under the graph of y = f(x) between x=a and x= b is shown in a lavender color. This area is given by the definite integral \int f(x) dx, with limits of integration x = b and x = a. Here and throughout this explanation I will give the upper limit of integration first, followed by the lower limit of integration. (Unfortunately LATeX doesn't have the ability to write definite integrals, so I'll do the best I can.)

    The graph that is the upper boundary of the lavender region can also be thought of as the graph of x = f^{-1} (y). The green or turquoise-colored region has an area that can be written as the definite integral \int f^{-1} (y) dy, with limits of integration f(b) and f(a).

    It's pretty easy to see that the two colored regions have an area equal to f(b) * b - f(a) * a. In other words, the area of the L-shaped region that is colored equals the area of the large rectangle minus that of the small rectangle.

    OK, are you with me so far?

    What we have is
    \int f(x) dx + \int f^{-1} (y) dy = f(b) * b - f(a) * a

    The integrals above are definite integrals, meaning they have limits of integration. The limits of integration for the first integral are x = b and x = a. The limits of integration for the second integral are y = f(b) and y = f(a).

    So, by subtracting the second integral from both sides, we have
    \int f(x) dx = f(b) * b - f(a) * a - \int f^{-1} (y) dy
    where, again, both integrals are definite integrals, with the limits of integration as already shown.

    How does this fit in with the u and dv, and v and du? Glad you asked.

    Let's say that:
    (*) u = y = f(x), and
    dv = dx

    We know that the other relationship between x and y is
    x = f^{-1} (y) (but NOT y = f^{-1} (x)!!). That will come in handy later on.

    Substituting from the equations marked (*), the integral on the left side of the equation can be rewritten this way:
    \int f(x) dx = \int u dv, with limits of integration still x = b and x = a.

    The integral on the right side of the equation can also be rewritten:

    \int f^{-1} (y) dy = \int x du = \int v du, with limits of integration y = f(b) and y = f(a).

    In the first equality, I replaced f^{-1} (y) with x (we know that x = f^{-1} (y)), and dy with du (since y = u, then dy = du).

    In the next equality, I replaced x with v, since we know that dx = dv (from (*)), therefore x = v (plus possibly a constant, which we can safely ignore since we're dealing with definite integrals).

    We now have:
    \int u dv (with limits of integration x = b to x = a)
    = f(b) * b - f(a) * a - \int v du (from y = f(b) to y = f(a))

    The final step is realizing f(b) * b - f(a) * a is equal to u*v (evaluated at x = b and x = a).

    In summary, instead of our integrals involving f(x) and dx and f^{-1} (y) and dy, we have arrived at this equation:

    \int u dv (x = b to x = a) = uv|^{x=b}_{x=a} - \int v du (y = f(b) to y = f(a))

    If we dispense with limits of integration so that we're talking about indefinite integrals, you see the more familiar form:
    \int u dv = uv - \int v du

  2. Wendy


    Mar 24, 2008
    Calculas was my bane at college. I understood the concept of area under the curve, but couldn't make the jump how they did it in equation form. The infinately smaller slices made perfect sense. The CRC handbook was my salvation, or at least my grade. They still use those?

    Just curious, where are you going with this, working through a concept, or trying to discuss an idea?
  3. Mark44

    Thread Starter Well-Known Member

    Nov 26, 2007
    I still have by CRC Math Tables, 15th Ed., but I didn't use it all that much for calculus, at least not when I was taking the course. I used it some in my teaching, as it provides a nice summary of math ideas from many of the subdisciplines of math.

    As for my motivation in the integration by parts, it was inspired my mentaaal's question yesterday about substitutions, and why they did or didn't work. It got me to thinking of a picture and explanation I saw once about about integration by parts, so I thought I'd share this idea, for whatever it's worth.