What is the largest value that
∫ from a to b x√(2x - x²)...

Question

What is the largest value that

∫ from a to b x√(2x - x²) dx

can have for any a and b? Give reasons for your answer.

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. What is the largest possible value of the integral evaluated from A to B of 2x multiplied by the square root of 4 x minus x 2 dx for any real A and B? OK. So it appears for this particular problem that we're asked to determine what is the largest possible value of this specific integral that is given to us for any real A and B. So now that we know that we're ultimately trying to solve for the largest possible value of this specific integral, our first step that we need to take. We need to note that the integrand is real only when 4 x minus x 2 is greater than or equal to 0. So that means that we can go ahead and write that x multiplied by parentheses of 4 minus x is greater than or equal to 0. Which will mean that x is greater than or equal to 0 and that x is less than or equal to 4. That will mean that x is an element of square brackets of 0.4, where square brackets indicate a closed interval, and on the closed interval square brackets of 0.4, we need to note that the integrand. F of X will be equal to 2 multiplied by X, so 2 x multiplied by the square root of 4 x minus x to the power of 2 will be non-negative. So for any real A, B, with the integral being defined, the integral will be maximized by integrating over the whole support square brackets of 0.4. Fantastic. So, keeping that in mind, thus we can go ahead and write that the maximum value will be the max of A, B, so this will be the max of A, B of the interval, evaluate from A to B of 2X multiplied by the square root of 4 X minus X to the power of 2. DX will be equal to the integral evaluated from 0 to 4 of 2 x multiplied by the square root of 4 x minus x 2 dx. And we now need to recall and use the complete the square method for the erratican. So we need to take 4 x minus x 2 will be equal to parentheses of x2 minus 4x, which will be equal to ses of x 2 minus 4x plus 4. Plus 4. Which will be all equal to 4 minus parentheses of x minus 2 to 2. Fantastic. So, let us note that we will let X minus 2 be equal to 2 multiplied by sin of T. So that x will be equal to 2 + 2 multiplied by sin of t. Then that will mean then that DX, so once again, that will mean then, so then DX by DT, so DX by DT. Will be equal to cosine of T. Which will mean that DX will be equal to, so once again, DX is going to be equal to drum roll, 2 multiplied by cosine of TDT. So, therefore, with that in mind, we can go ahead and write that 4 minus parentheses of X minus 2, so 4 minus parentheses of X minus 2 to 2 will be equal to 4 minus 4 multiplied by sin square of T, which will be equal to 4 multiplied by cosine square of T. Which means we could write that the square root of 4 x minus X 2 will be equal to the square root of 4 multiplied by cosine square of T, which will be equal to the absolute value of 2 multiplied by cosine of T. So now we need to rewrite our limits, so we need to note that T will be equal to inverse sin of X minus 2 divided by 2. And the principal range of the arc sign function will be square brackets of pi divided by 2, pi divided by 2. So that will mean that when X is equal to 0, that will mean that T will be equal to inverse of parentheses of 0 minus 2 divided by 2. Which will be equal to when we perform that evaluation, it will be equal to negative pi divided by 2, and when x is equal to 4, that will mean that T will be equal to inverse sin of 4 minus 2 divided by 2, which will be equal to positive pi divided by 2. So that will mean that 2 multiplied by cosine of T will be greater than or equal to 0 on the closed interval square brackets of pi divided by 2. So it's gonna be square brackets of negative pi divided by 2, positive pi divided by 2. Once again, it's a closed interval. Thus we can go ahead and write that the interval evaluated from 0 to 4 of 2 x multiplied by the square root of 4 x minus x to the 2 dx will be equal to the integral evaluated from negative pi divided by 2 to positive pi divided by 2 of 2 multiplied by parentheses of 2 plus. 2 multiplied by sin of t multiplied by parentheses of 2 multiplied by cosine of t, multiplied by 2 multiplied by cosine of t. DT, which will be equal to the integral evaluated from negative pi divided by 2 to pi divided by 2 of parentheses of 16 multiplied by cosine squared of t. Plus 16 multiplied by cosine squared oft multiplied by sin of t. DT means we can't forget our DT. And this will be. Equal to drum roll, 16 multiplied by square brackets of the integral evaluated from negative pi divided by 2, to pi divided by 2 of cosine squared of TDT plus the integral evaluated from negative pi to positive pi divided by 2. Of cosine squared T multiplied by sin of TDT, and don't forget your square bracket. Fantastic. And we need to note that cosine squat is even. So let's make a note here. So, cosine squat is even, so I'll put an E in quotation marks, and sin of T is odd, so I'll put an O in quotation marks. Thus, that will mean as a result that Cosine squared of t multiplied by sin of t will be odd. So, therefore, we can go ahead and write that 16 multiplied by square brackets of the interval evaluated from pi divided by 2 to pi divided by 2 of cosine squared of T. And once again We need to make sure we put DT plus the integral evaluated from negative pi divided by 2 to pi divided by 2 of cosine squared of T multiplied by sin of T DT, which will be when we write our final answer in its most simplified. Simplified expression, which I'm gonna skip a step, but we'll find that this will be simply equal to 32 multiplied by the integral evaluated from 0 to pi divided by 2. Of cosine square of TDT. So, at this point, we need to recall that cosine square of T is equal to 1/2 multiplied by 1 + cosine of 2T. Thus, we can go ahead and write that 32 multiplied by the interval evaluated from 0 to pi divided by 2 of cosine squared of TDT. Will be equal to. 32 multiplied by 1/2 multiplied by the integral evaluate from 0 to pi divided by 2. And we will find that this will be of 1 plus, so 1 plus cosine of 2 t. DET. Fantastic, which you can write two T in parentheses or you could just write it without parentheses. And we will find that this will be. Equal to, when we have our answer in its most simplified form right before we perform the evaluation, you will find that it's equal to 16 multiplied by square brackets of T plus 1/2 multiplied by sin of 2t, and it'll be evaluated from 0 to pi divided by 2. And we need to note that at E is equal to pi divided by 2, that will mean that pi divided by 2 plus 1/2 multiplied by sin of drum roll, sin of 2 multiplied by pi divided by 2, which will be equal to when we plug this expression into our calculator, will be simply equal to pi divided by 2. Fantastic. So that will mean that. We can go ahead and start to solve for at T equals 0. So when T is equal to 0, you'll have 0 + 1/2 multiplied by sin of 2 multiplied by 0, which will be simply equal to 0. Thus, as a result, we could write that 16 multiplied by square brackets of t plus 1/2 multiplied by sin of 2 t evaluated from 0 to pi divided by 2 will be equal to 16 multiplied by square brackets of pi. Divided by 2 minus 0, which will give us an answer of 8 pi. And 8 pi is our correct and final answer. Hooray, we did it. So therefore, the maximum possible value is 8 pi. So going back to the top to recap what we've learned, we can safely say once again without a doubt that our Final answer for the largest possible value of this specific integral once again will be 8 pi. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

What is the largest value that∫ from a to b x√(2x - x²) dxcan have for any a and b? Give reasons for your answer.

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What is the largest value that
∫ from a to b x√(2x - x²) dx
can have for any a and b? Give reasons for your answer.