Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
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Cornell University
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
To solve an integral with the substitution method, look for a composite function. In most cases, the function containing the other function will be what you substitute (with u or w, whichever variable is preferred. Here, we will use w). When using the substitution method, you must turn all x's into w's. So, when you define w, solve for x; then, you will be able to replace x's in any other part of the integrand with an expression containing w's. For example, if w = 2x + 3, then x = (w - 3)/2. In addition, the dx must be converted to dw. To do this, simply take the derivative of w with respect to x. Continuing with the previous example, if w = 2x + 3, dw/dx is 2. So dw = 2dx, and dx = ½dw.
After substituting all of the x's in the integrand with the appropriate expressions with w, then integrate as usual. Remember to replace the w's in the result with the expression of x that was substituted.
I have done the integral, I’m going to solve by the method of substitution. Here I’m asked to perform the antidifferentiation and then check by differentiating my answer. So in the end, we’ll check our answer.
First of all, the thing I look for, when I’m looking to use the method of substitution, is a composite function, somewhere in the integrand. And here I see a composite function. The outside is the square root function and inside I have x² minus 9. I’m going to substitute for that inside part, x² minus 9. I have to take the derivative of this with respect to x. And the reason for this is, I need to convert this dx into a dw.
Everything in this integral is going to be converted from xs to wx. That’s the way the method of substitution has to work. And you have to have complete conversion. If you don’t actually get rid of all the xs, then the substitution hasn’t worked. So hopefully this will work.
Dw/dx is 2x. And so my dw is going to be 2xdx. And I have an x here and a dx, so that looks like a pretty good substitution. The only thing that’s not quite right, is I don’t have a 2 outside. So I can fix that by multiplying both sides by ½. 1/2dw is xdx. So this x corresponds to this x and this dx with this dx. And so these two become my 1/2dw.
And this square root of x² minus 9, is the square root of w. So I can pull a constant out in front; the ½. I can change this to an exponent. W to the 1/2dw and use the power rule for antiderivatives on this guy. You get ½w and I raise the exponent by 1. ½ plus 1 is 2/3 and I divide by 3/2 plus c.
Now, dividing by 3/2 is the same as multiplying by 2/3, and so I’m going to get ½ times 2/3 which is 1/3. So this will be 1/3w to the 3/2 plus c. The last thing left to do is to re-substitute for w. w is x² minus 9. So I’m going to have 1/3x² minus 9 to the 3/2 plus c.
So this is my set of anti derivatives for this function. Now I need to check my answer by differentiating. Let’s take this up here and differentiate it. The derivative with respect to x of 1/3, I have x² minus 9 to the 3/2 plus c. First, the 1/3 multiple can pull right out of the derivative and then I get, the 3/2 come down in front, x² minus 9 to the 3/2 minus 1. 3/2 minus 1 is ½, times the derivative of the inside, that’s the chain rule. So times 2x and the derivative of the plus c is zero. 1/3 and 3/2, the 3's cancel, I get ½ times 2x, the 2's cancel. I get x.
Let me just erase that and put an x in here. And this is the same as the square root of x² minus 9. Just going back to check. x² minus 9, root (x² minus 9) times x, that’s exactly what I started out with so that means that this checks. This is the correct answer; 1/3 x² minus 9 to the 3/2 plus c.
Unit
Antiderivatives and Differential Equations