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U Substitution Example 4 Video
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 U Substitution Example 4 Video
This video from IntegralCALC shows you how to solve the U Substitution Example 4 Math problem.
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U Substitution Example 4 Hey guys, here’s another U substitution problem. This one is X2 × v2X34dx. You might not know that this immediately is U substitution problem but maybe the best way even to make a guess is you know that it’s not integration by parts. If you can kind of rule things out, you know it’s not integration by parts, you know it’s integration by partial fraction, it’s too complicated to do in a basic way and if you start converting the problem by taking out the square root etcetera, you might sort to get a better feel for it. So let’s kind of talk through it as we go if we change this X2 as a rule and you can always change something in a square root by reading that same thing to the one half 2X3  41/2 is the same as the square root of 2X3  4. So we change that dx. So you know you can’t do this easily. One good way to tell is, you know that if you use U substitution, and you’re going to use U for this part right here, you’re going to end up taking the derivative which is going to turn this X3 into an X2 and your X2 might have a chance to cancel with this X2. so if you ever identify that whenever you’re going to use as your U in a potential U substitution problem has something wit an exponent that’s one higher than what’s out here since this is X3 and this is X2, this being one higher than this, good chance that it might be a U substitution problem. So let’s try it. We’ll say, U is 2X3  4, then taking the derivative of U here, du is 6X2, multiplying 3 by 2 to get 6 and then subtracting 1 from the exponent. 6X2 and then of course the derivative of 4 or a constant is zero, so here’s a derivative and then always add dx so that we can solve for dx. So now we’re going to solve for dx by dividing both sides by 6X2. So it will end up being du/6X2 = dx. Now that we have that we can plug U in for 2X3  4 and we can plug du/ 6X2 in for dx. So let’s go ahead and do that, we’re going to have the integral of X2 × U, don’t miss that 1/2 right there, you don’t want to have. And then for dx, we have du/6X2. Okay so this is great, it worked out how we thought it was going to. We have an X2 in the numerator and X2 in the denominator, so they both cancel so X2/X2. Now what we need to do is simplify this so that we can take the integral. I’m going to go ahead and erase everything up here and start working from the top again, so put a lot of space, I’m actually going to leave the U in so we can use it for reference later, the U was 2X3 4. Okay so, we’re going to circle back up here, first we’re going to take the 6 out and the 6 is in the denominator, so it’s going to come out in front of the integral sign as 1/6. So we’re going to have 1/6 times the integral of—and all that’s left is U1/2du. So the way we’re going to integrate that is have the 1/6 out here then the parenthesis U, of course we always ad one to the exponent so 1/2 + 1 is 1 1/2 or 3/2 and then of course the coefficient which here is implied as a one, 1 divided by the new exponent 3/2 of course is 2/3. And that’s right, so that’s the integral there and then of course, we always add c to the end so now all we have to do is plug back in for U and simplify. So I’m going to go ahead so that we don’t get confused here after I kind of small it up. Multiply these 2 so we’ve got 2/18 looking like 2/18 which I’m just going ahead and simplify that right now to 1/9, 1/9 times plug back in for U, 2X3  4. And then 23/2 and that’s everything plus C, and sorry for the things close, but that’s our answer right there.