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# The curve with equation $y^2 = x^2 (x + 3)$ is called Tschirnhausen's cubic. If you graph this curve you will see that part of the curve forms a loop. Find the area enclosed by the loop.

## $\frac{24}{5} \sqrt{3}$

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Applications of Integration

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### Video Transcript

in this problem, we have to use integration, define the area under a curve. And so we have got to talk about how we do that for the function that were given. We're given. Our function is why squared equals X squared Times X plus three. So we can simplify that a little bit and say that why is equivalent to plus or minus the square root of X square Times X plus three. So when we graft that this is a very rough drawing. But this is what the graph would mostly look like. We have an intersection at 00 and if you drew this better than I did, you would see that this is symmetric and so we can do a little bit more simplification with our function. We can say that why equals the absolute value of X Times Square root of X plus three. But working with absolute values and intervals is a little bit more challenging than what we need here. So we can say when X is less than zero. Why equals negative X time to square root of X plus three And now remember when we look at this graph, or maybe you put it into a graphing, um, utility to look make it look better than my drawing you find in its symmetric. So what you could basically do. I find the area of the top portion or the bottom portion off our function and multiply it by two. And that's what we're going to do. We'll gather. The area is equivalent to two times the integral of negative 3 to 0 of negative X times the square root of X plus three with respect to X. So now we can do some U substitution. We can say let you be X plus three. So U minus three is equal to X and d use equal to D X, and now we can change. The limits of integration are a value which was negative. Three. We can add three and we'll get zero for a B value. Well, we had zero before. We'll add three and we'll get three. So our new integral now using new substitution will look like this. Are area will be equivalent to negative two times the integral from 0 to 3 of U minus three times the square root of you in do you and now we can simplify that and say that our area equals negative. Two times the integral from 0 to 3 of you raised the three halves minus three times you raise to the half power and do you? So now we can take the anti derivatives will get negative two times two or five. You raised a five over to minus three times. Pardon me, That should know. Sorry to Over three. You raised a three over to when will evaluate this from 0 to 3. Sorry for the confusion. I just cause they're hopefully it becomes clear in this next step. So now we'll basically plug in. Our limits of integration will get negative two times. Two over. Five times three raised to four over. Two times three raised to one half, minus two times three raised to two over. Two times three. Race to the one half. Now I know that these could be simplified, which we're going to do when we simplify. This whole expression will get negative two times 2/5 times nine times a spirit of three minus two times three squared of three. Now we can put all of this in a common denominator and we'll get negative 36 plus 60/5 times square root of three. And finally, what gets the area of our curve is 24/5 times the square root of three. So I hope that this shows you how we can use integration to find the area under a curve. Specifically, we looked at a symmetric function and how we can go through those steps to find the area using you substitution as well.

University of Denver

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Applications of Integration

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