In Exercises 27–40, use a substitution to change the integral ...

Question

In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.

∫ x^2 √(2x - x^2) dx

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Evaluate the interval, the interval of x 2 multiplied by the square root of 4 x minus x 2 dx. By setting u is equal to x minus 2. The integral can be rewritten as the interval of x 2 multiplied by the square root of 4 x minus x 2 dx is equal to the integral of u2 + 4 u + 4 multiplied by the square root of 4 minus u to the 2 duu, which is equal to i1 plus i2 + i3. Where i1 is equal to the integral of u to the 2 multiplied by the square root of 4 minus u to the 2 duu, i2 is equal to 4 multiplied by the integral of u multiplied by the square root of 4 minus u to the power of 2 duu. And i3 is equal to 4 multiplied by the integral of the square root of 4 minus u to the 2 d u. In addition, by using a standard interval formula, i1 will be equal to 2 multiplied by inverse sin of u divided by 2. Minus u multiplied by the square root of 4 minus u 2 multiplied by parentheses of 2 minus u 2 all divided by 4. Plus c1 and by substitution i2 is equal to -4 divided by 3 multiplied by parentheses of 4 minus u to the 2, all to the power of 3 divided by 2 plus c2. OK, so it appears for this particular prompt we're asked to take all the information that is provided to us and we're simply asked to evaluate the integral, which once again is the integral of x 2 multiplied by the square root of 4 x minus x 2 dx. And we need to be able to recognize that the problem itself gives us the first few steps that we need to use in order to solve this problem. So a lot of the solving has already been done for us, which is a bonus. So that means as a result, since a lot of steps have been solved for us, our first step is we need to consider. I subscript 3, I3 for short, I3 is equal to 4 multiplied by the interval of the square root of 4 minus u to the 2 duu. And we need to note that we could recall and use the following formula which states that the interval of the square root of a 2 minus x 2 dx will be equal to x divided by 2 multiplied by the square root of a 2 minus x 2. Plus a to the power of 2 divided by 2 multiplied by inverse sign of. Drum roll x divided by A plus C. Fantastic. So with that said, we need to note that for this case, X will be equal to you. A will be equal to 2. And we need to note that A 2 will be equal to 4. Therefore, we could go ahead and write that i3 will be equal to 4 multiplied by parentheses of u divided by 2 multiplied by the square root of 4 minus u to the 2 plus 2 multiplied by inverse sign of u divided by 2. And you could write that in parentheses or you could write it without parentheses. And don't forget parenthesis plus C3, where C3 represents some constant variable, and this will simplify to be equal to 2 multiplied by u multiplied by the square root of 4 minus u to the power of 2 + 8 multiplied by inverse sin of u divided by 2 plus C3. Fantastic. Our 2nd step now, so this is step 2. So for step two, we need to combine our results now. So we need to take the integral of x 2 multiplied by the square root of 4 x minus x 2 dx, which will be equal to i1 + i2 + i3. Which will be all equal to. 2 multiplied by inverse sign of U divided by 2. Minus u multiplied by the square root of 4 minus u to the power of 2 multiplied by parentheses of 2 minus u to the 2 divided by 4 minus 4 divided by 3 multiplied by 4 minus u to the 2 in parentheses to the power of 3 divided by 2. Plus 2 multiplied by u multiplied by the square root of 4 minus u to the 2 plus 8 multiplied by inverse sign of u divided by 2 plus c1 + C2 + C3. And when we combine all the inverse sign terms, we will be able to write 2 multiplied by inverse sign of u divided by 2 plus 8 multiplied by inverse sin of u divided by 2, which will be equal to 10 multiplied by inverse sin of u divided by 2. So next we need to combine the terms that involve u multiplied by the square root of 4 minus u to the 2, and when we do that, we will get negative u multiplied by the square root of 4 minus u to the 2 multiplied by parentheses of 2 minus u 2 divided by 4. Plus 2 multiplied by u multiplied by the square root of 4 minus u to the power of 2. Which will be equal to, so ultimately we need to write this expression in its most simple form, which in an effort to save time, I'm going to skip a few steps, but when you write this result in its most simplified form, meaning we can't simplify it any further, we will find that it's equal to u multiplied by the square root of 4 minus u to the power of 2. Divided by 4, multiplied by parentheses of 6 plus u to the 2. And we got this simplified result by using a little bit of algebra to manipulate our expression to make sure all of the terms cancel out and we're just left with our most simplified expression as a result. So, once again, you're just manipulating your expression until you can't manipulate it anymore. So we now at this point need to rewrite the entire expression as the interval of x 2 multiplied by the square root of 4 x minus x 2. The x is equal to 10 multiplied by inverse sin of u divided by 2 plus u multiplied by the square root of 4 minus u to the power of 2. Divided by 4 multiplied by parentheses of 6 plus u to the 2. -4 divided by 3 multiplied by parentheses of 4 minus u to the 2 to the power of 3 divided by 2 plus c and let us note that simply that c will be equal to c1 plus c2. Plus C3. So we're just gonna call C1 plus C2 plus C3 simply equal to C. So now what we need to do is we need to replace you by x minus 2, and we need to recall a note using blue again, we need to recall a note that the square root of 4 minus parentheses of x minus 2 to 2 will be equal to the square root of 4 x minus x 2. So that means we can go ahead and write that 10 multiplied by inverse sin of x minus 2 divided by 2 plus parentheses of x minus 2 multiplied by the square root of 4 x minus x to the 2 divided by 4 multiplied by parentheses of 6 plus parentheses of x minus 2 to the power of 2. -4 divided by 3 multiplied by 4. X minus x to the power of 2. All to the power of 3 divided by 2. Plus C. OK. And we need to note that since 6 plus the parentheses of X minus 2 to 2. Is going to, so let's make a note once again in blue. That for our expression 6 + parentheses of x minus 22 will be ultimately equal to x 2 minus 4 x + 10. That will mean that since this is the case we can go ahead and write that the interval of x 2 multiplied by the square root of 4 x minus x 2 dx will be simply equal to 10 multiplied by inverse sin of parentheses of x minus 2 divided by 2. Plus x minus 2 in parentheses multiplied by the square root of 4 x minus x to the power of 2 divided by 4. Multiplied by parentheses of x 2 minus 4 x + 10 minus 4 divided by 3 multiplied by parentheses of. 4 x minus x to the 2, all to the power of 3 divided by 2 plus c, and this is our correct and final answer. Hooray, we did it. We saw for our final answer. So going back to the top to recap what we've learned, we could safely say without a doubt that our final answer. Without a doubt, once again, we'll be drum roll. So let's put our final answer here. It just shrink it down a little bit so we can fit it into one spot. So our final answer once again without a doubt will be 10 multiplied by inverse sin of parentheses of x minus 2 divided by 2 plus parentheses of x minus 2 in parentheses multiplied by the square root of 4 x minus x 2 divided by 4 multiplied by parentheses of x 2 minus 4 x plus 10. -4 divided by 3 multiplied by parentheses of 4 x minus x 2 to the power of 3 divided by 2 plus c. And that's it. That is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.∫ x^2 √(2x - x^2) dx

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In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.
∫ x^2 √(2x - x^2) dx