7–64. Integration review Evaluate the following integrals.
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Question

7–64. Integration review Evaluate the following integrals.

18. ∫ from 3 to 7 of (t - 6) * √(t - 3) dt

Accepted Answer

Hello. In this video, we are going to be evaluating the following integral. We are given the integral from 4 to 9 of the quantity Y minus 8 multiplied by the square root of Y minus 4. Now, since we are given a composite function, the square root of Y minus 4, this is an indication that we should approach this integral by using a U substitution. The U substitution that we can use is we're going to pick Y minus 4, the innermost function to be substituted. So we are going to let U equal to Y minus 4. And now we need to come up with a substitution of DY in terms of you. In order to do this, we will take the derivative of our U substitution with respect to Y, and that is going to give us DU is equal to D Y. Now, notice that we have a substitution for Y minus 4, but we don't have a substitution for Y minus 8. In fact, when we took the derivative of U equal to Y minus 4, we are just left with DU as equal to D Y. There is a way to rewrite this Y term into terms of U. If we were to solve for Y in our original U substitution, we will end up with Y is equal to U + 4. This is going to be a substitution for the Y term in the original integral with respect to you, and that is going to allow us to rewrite the integral as the integral of you. Plus 4 minus 8 multiplied by the square root of u which can also be rewritten as U to the half power d u. Now, recall that we are doing a use substitution over a defined integral. Now what we need to do is we need to rewrite our terms, our bounds from terms of Y into terms of you. And in order to do this, we are going to take each of our bounds and plug them into our original U substitution. So when we plug in 4 into our U substitution, 4 minus 4 gives us a lower bound of 0. And when we plug in 99 minus 4 gives us an upper bound of 5. So now what we need to do is we need to simplify the integral. U + 4 minus 8 is going to give us U minus 4, so we have the integral from 0 to 5 of U minus 4 multiplied by U2 power, DU. Next, we are going to substitute or distribute to the power into the quantity U minus 4, and that's going to allow us to rewrite the integral as the integral from 0 to 5 of U to the three halves power. Minus 4 multiplied by u to the power dU and now what we need to do is we need to take the antiderivative of each of our terms. That is going to leave us with u to the three halves plus 1 divided by 3 halves plus 1. -4 multiplied by you the power plus 1. Divided by 1/2 + 1, and this will be evaluated from 0 to 5. The antiderivative is going to simplify. To be 2/5 or Yes, 2/5. Multiplied by U to the 5 halves power, minus 4 multiplied by 2/3. Multiplied by U3 half power, evaluated from 0 to 5. Multiplying -4 into 2/3 is going to give us negative 8/3, so we can rewrite the second term to be negative 8/3 multiplied by U to the 3 houses power. Now what we need to do is we need to evaluate this function of U from 0 to 5. Now, since we have you as a multiple in each of our terms, if we were to plug in 0, we end up with 0. So we really need to evaluate the upper bound, which is going to be 5. So when we plug in 5 into our antiderivative, we get 2/5s multiplied by 5, raised to the 5 halves power, minus 8/3. Multiplied by 5, raised to the 3 halves power, then we will subtract the value of the lower bound which we discussed is to be zero, because plugging in zero for any value view is going to give us 0. Now we just need to simplify. 55 halves power simplifies to be 25 multiplied by the square root of 5. So we have 25ths multiplied by 25 multiplied by the square root of 5 minus 8/3 multiplied by 5 rates to 3 halves power, which simplifies to 5 multiplied by the square root of 5. And by further simplifying these terms, we will end up with a final value of -10 divided by 3, multiplied by the square root of 5, and this is going to be the value of the definite integral that we were working with. And with that being said, the overall solution to this problem 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.

7–64. Integration review Evaluate the following integrals.
18. ∫ from 3 to 7 of (t - 6) * √(t - 3) dt

Key Concepts

Definite Integrals

Integration Techniques

Fundamental Theorem of Calculus

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