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.
When radicals are involved, substitute the expression inside the radical with w. Then, use the substitution method: find dw by taking the derivative of w with respect to x, substitute the x's in the integrand with the expression for w, and integrate as usual. Replace the w's with the expression for x.
Let’s look at a really tougher example that involves integrating radical functions. This one; the integral of 4x² times the square root of 3 minus 2x. Now again, the advice I’ve been giving you is substitute for the thing inside the radical. So we will do that again in here. Then we’ll get w equals 3 minus 2x. And that means that dw equals -2 dx. And I need the dw I need to replace dx with something so I need my dw term here.
However I don’t have a I have a 4 and not a -2. Let me just solve this for dx by, multiplying both sides by -1/2. That will get rid of this -2. So I get dx is -1/2 dw. I’ll use that in a second. So far I’ve got this guy taking care of and this guy but to this one.
For this for x, I need to solve this equation for x. And the way I would do that is I could add 2x to both sides I get 2x equals 3 minus w, and subtract w from both sides and then divide by 2. So x equals ½ 3 minus w. So let’s do that so taking this back into here, I’m going to have the integral of 4 times x².
When I square this, I’m going to get ¼ times this squared. That’s how I’m going to write it in. So times ¼ times 3 minus w² and then I have the radical w then dx is -1/2 dw. So that’s quite a bit there. The 4 and the ¼ cancel and I can pull this -1/2 out in front. But I still have my dw, my root w and my 3 minus w², let me expand this. That’s going to be 9 minus 6w plus w², and this is just w to the ½.dw. See where everything went? Minus ½ is now here, I’ve expanded this binomial and I converted this into an exponent. Now I need to distribute this over each of these terms. My goal is to get just powers of w, nothing but powers of w.
So minus ½, 9 times w to the ½ minus 6, w times w ½ is w to the 3/2. Plus w² times w to the ½. This is the same as w to the 4/2. When you multiply you add the exponents, this will be w to the 5/2. There we have it. Nothing but powers of w very easy to integrate so let me do that.
I still have the -1/2 out in front, I’m going to have 9 times w to the, remember I add 1 to the exponent. So 9 times w to the 3/2 and I divide by 3/2 which is the same as multiplying by 2/3. You can see it’s really important that you have to put something between the 9 and the 2/3 to make sure it doesn’t look something like a mixed number. 9 times 2/3 not 9 and 2/3. Back here, minus 6 times the anti derivative of w to the 3/2. That’s going to be w to the 5/2 divided by 5/2. So minus 6 times 2/5 w to the 5/2.
And finally, anti-derivative of w to the 5/2, I add 1 I get w to 7/2 divided by 7/2 that’s 2/7. So plus 2/7 w 7/2 plus c.
So all I have to do is multiply all these constants out and then replace w with 3 minus 2x. I’m going to the constants first, because this is pretty complicated. So the 9 and the 2/3, the 9 and the 3 is going to cancel leaving a 3. And that makes 6. 3 times 2, 6 times -1/2 is -3w to the 3/2. Then I have 6 times 2 12 over 5 divided by 2, that’s going to be 6 over 5. The negatives cancel, so positive 6 over 5 w to the 5/2. And then minus ½ times 2/7, the 2’s cancel I get minus 1/7w to the 7/2 and I need a plus c still.
So now it’s time to substitute back in 3 minus 2x for w. So I have minus 3 times 3 minus 2x, to the 3/2 plus 6/5. 3 minus 2x to the 5/2. So 5, minus 1/7, 3 minus 2x to the 7/2 plus c. Really complicated looking answer. But this should be the answer to my original anti-derivative of 4x² times the square root of 3 minus 2x.
Unit
Antiderivatives and Differential Equations