7–84. Evaluate the following integrals.
9. ∫ from 4 to 6...

Question

7–84. Evaluate the following integrals.

9. ∫ from 4 to 6 [1 / √(8x – x²)] dx

Accepted Answer

Hello. In this video, we are going to be evaluating the following integral. We are given the integral from 3 to 7 of 1 divided by the square root of 20 X minus X2. Now, since we have a squared X inside of the square root function, this is an indication that we should approach this integral by using some type of inverse trigonometric function. But currently, the integral that is given to us does not allow us to use inverse trigonometric functions because they are not in the form of sine inverse, cosine inverse, or tangent inverse or so on. So instead, what we need to do is we need to rewrite this integral in order to apply our inverse trigonometric functions. Before we look at the integral, let's just take a look at what's inside the square root function. We are given 20 X minus X squared. Now, we can rewrite this into a form of a squared constant minus the squared variable in order to apply the inverse trig functions, but we'll need to complete the square first. Let's go ahead and reorder these terms in order from highest power to lowest power. That is going to give us negative X2 + 20 X. And if we were to factor out a -1, we have negative X2 minus 20 X. If we were to complete the square on this quantity, we end up with negative X2 minus 20 X plus 100, but in order to balance the equation, we have to subtract a 100 from this. But keep in mind that we originally factor out a negative one from this from this trio of terms, so we have to include that negative 1 into our balancing equation. So we have negative X2 minus 20 X plus 100 + 100. And this trinomial is going to simplify to be negative x minus 10 2 + 100. And if we were to reorder these terms, we are left with 100 minus X minus 10 squared. So, we can rewrite the given integral to be the integral from 3 to 7. Of one divided by the square root. Of 100 minus X minus 10 squared. DX. Let's go ahead and use a U substitution on X minus 10. If we allow U to equal to X minus 10, then D U is equal to DX. So we end up with the integral. Of one divided. By the square root of 100 minus U 2 DU. Now, since we are using a U substitution over a defined integral, we have to rewrite our bounds from terms of X to terms of U. We will take our bounds and plug them into the U substitution, and that is going to give us the bounds from -7 to -3. Now, notice that this integral is in the form where we can apply the sine inverse function. Recall that if we have the integral. Of one divided by the square root of a squared constant term minus a squared variable in this case you squared, then this integral is equivalent to sine inverse of u. Divided by a plus C. In this case here, our squared constant A2d is going to be 100, which means that A is equal to 10. And so if we were to use this definition on the current integral, The antiderivative is going to be ine inverse. Of you divided by 10. And this will be evaluated from -7 to -3. And so now what we need to do is we just need to evaluate the following limits. So by plugging in our bounds, we end up with sine inverse of -3 divided by 10 minus sign inverse of -7 divided by 10. Now recall that sign is an odd function. Because sine is an odd function, sine of negative theta is equal to negative sine theta. And so that's going to allow to allow us to rewrite this antiderivative as negative sign inverse of 3 divided by 10, plus sign inverse of 7 divided by 10. And by reordering the terms with the positive leading coefficient, we will have a final value of sine inverse of 7 divided by 10 minus sign inverse of 3 divided by 10, and this is going to be the final answer to the given definite integral. And with that being said, the overall solution is going to be B. So, I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

7–84. Evaluate the following integrals.
9. ∫ from 4 to 6 [1 / √(8x – x²)] dx

Key Concepts

Definite Integrals

Integration Techniques

Improper Integrals

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