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Use the Table of Integrals on Reference Pages 6-10 to evaluate the integral.

$ \displaystyle \int_0^2 x^3 \sqrt{4x^2 - x^4}\ dx $

2$\pi$

Integration Techniques

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Campbell University

Baylor University

University of Michigan - Ann Arbor

Boston College

Okay, This question wants us to evaluate this and enroll with help from our table. So as is common with these radical questions, we like some sort of use squared in there. So let's try the substitution of you equals X squared. And that would mean that do you equals two x d x or d'you over too equals x t X. So now from there, we can start substituting after we change our limits. So you of two is two squared ages four and you have zero is still zero. So now we get the integral from 0 to 4 of x times x squared. We're just splitting up this inner girl here of for you minus you squared d x. So we see we have an X t X and we can convert everything to a do you now. So now we're left with this integral as our leftover expression, which is a much more digestible form. Then what we're given so no, If we look at our table, we can evaluate this and we get that this is equal to us. Follow us. Keep the 1/2 in front and then we get a four sign in verse of X squared minus two over too, plus X times square root of four minus X squared times extra. The fourth, minus X squared minus six in that term is all divided by three. And we're going from zero 24 We're sorry, 0 to 2 because we went back into the ex world. So now our final anti derivative, before we start plugging things in is two times the sign in verse of X squared minus two over too, plus X over three square toe, four minus X squared times extra fourth of minus X squared minus six. And again, this came directly from the table. So now it's plug in and see what our final answer is. Well, if we plug in to are right, term vanishes end. If we plug in zero, that right term also vanishes. So all we have to deal with is the sine inverse terms. So it's equal to two times sine inverse of Force four minus two over too minus sign in verse of negative to over two, which is equal to two times the sine inverse of one minus sign in verse of negative one, which is equal to two times. Well, remember the domain that sign in verse is defined on we get a pie over too minus a minus pi over too, which gives us two times to pi over too, or to pie. So a very simple answer for a very complex anti derivative.

University of Michigan - Ann Arbor

Integration Techniques