Evaluate the integrals in Exercises 77–90.
79. ∫(from -1...

Question

Evaluate the integrals in Exercises 77–90.

79. ∫(from -1 to 0)6dt/√(3-2t-t²)

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. Compute the definite integral. The integral evaluated from -2 to 0 of 5 dt divided by the square root of 4 minus parentheses of t + 1 to the power of 2. OK. So, it appears for this particular prompt, we're simply asked to compute, or rather solve for the definite integral that is provided to us by the prom itself. So with that in mind, our first step that we need to take is we need to recall and use our substitution method in order to simplify this shifted argument, meaning we need to say that you for this particular prompt will be equal to t plus 1, which will mean that duu will be equal to DT. And we need to note that the integration limits will change to as follows. When T is equal to -2, that will mean that U is equal to -1, and when T is equal to 0, that will mean that U is equal to 1. So once again, when T is equal to -2, you will be equal to -1. But when T is equal to 0, that will mean u is equal to 1. Which will mean our integral will become 5 multiplied by the integral evaluated from -1 to 1 of DU divided by the square root of 4 minus u to the power of 2. Fantastic. Our second step now is we need to factor the constant which is inside of our square root, meaning we need to factor 4 from under our radical sign. So we need to go ahead and write that 5 multiplied by the integral evaluated from -1 to 1 of duu divided by the square root of. 4 minus u to the power of 2, which will be equal to 5 multiplied by the interval of -1 to 1. Of duu divided by the square root of 4 multiplied by parentheses of 1 minus u to the 2 divided by 4. Fantastic, which will be equal to 5 multiplied by the integral evaluated from -1 to 1 of DU divided by the square root of 4 multiplied by parentheses of 1 minus s u divided by 2. To the power of 2. Which will be equal to 5 multiplied by the integral evaluated from -1 to 1 of DU divided by the square root of 1 minus. Parentheses of u divided by 2 to the power of 2. Fantastic. So now, at this point, we need to use the substitution. V will be equal to U divided by 2. And that will mean that you will be equal to 2 V. And that will mean that DU will be equal to 2 DV. And the U limits will be U is equal to -1. And positive1 will correspond to V, which V is equal to plus or minus. 1/2. OK. So, using this information, we will obtain that 5 multiplied by the integral evaluated from -1 to 1 of DU divided by 4 minus u 2, which is equal to 5 multiplied by the integral evaluated from -1 to 1 of. DU divided by 2 multiplied by the square root of 1 minus u divided by 2 to the power of 2. Which will be equal to. When we write in our most simplified integral form, I'm gonna skip a few steps. We will get 5 multiplied by the integral of 1/2 to 1/2. DV, so it's gonna be 5 multiplied by the integral evaluated from -1/2 to 1/2 of DV divided by the square root of 1 minus V to the 2. And once again, this is applying all of our information because we know that V is equal to plus or minus 1/22, and we had to use our new substitution variables into our integral. So With that said, Our third step now is we need to recall and use trigontric substitution in order to remove the square root. So to integrate 1 divided by the square root of 1 minus V to the power of 2, we need to use V is equal to drum roll sin of theta. And that will mean that DV will be equal to cosine theta d theta. And we need to note that the square root of 1 minus V 2 will be equal to the square root of 1 minus sin square of theta, which will be equal to the square root of cosine square of theta, which is simply equal to cosine of theta. So, we also need to note that we will need to take. The, so once again, we need to take cosine of theta is greater than or equal to 0 for our range. So this is for the range used, so that will mean that the V limits will map to theta limits, meaning that V will be equal to -1/2 which will approach 0. Which will be equal to inverse sin of 1/2. So in inverse sin of 1/22, which will be equal to pi divided by 6. V when it's equal to 1/22 will approach 0 and it will be equal to inverse sin of 1/2, which will be equal to pi divided by 6. Fantastic, which will mean our integrand will become. 5 multiplied by the interval, evaluated from -1/2 to 1/2 of DV divided by the square root of 1 minus V 2. Which will be when we perform some of our simplification. It'll be ultimately equal to 5 multiplied by the integral evaluated from negative pi. Divided by 62 pi, divided by 6. Of D theta, which will be equal to 5 theta evaluated from negative pi divided by 6 to positive pi divided by 6. Fantastic. And this is when we're plugging in. Our values, our substitution values, because once again we know that DV is equal to cosine of theta d theta. Divided by cosine of theta, which we simplify. So, moving right along here. So, that will mean that our 4th and final step is we need to apply, rather recall and apply the fundamental theorem of calculus. So we need to evaluate our antiderivative at the endpoint. So we need to take 5 theta evaluated from negative pi divided by 6 to positive pi divided by 6. Which will be equal to 5 multiplied by parentheses of pi divided by 6 minus pi divided by 6, which will be equal to ultimately 5 pi divided by 3. And that is our final answer. Hooray, we did it. We solve for our final answer. So going back to the top, we could safely say that our final answer for our evaluated integral is once again we will note that this integral will be equal to drum roll. 5 pi divided by 3, and that's it, that is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

Evaluate the integrals in Exercises 77–90.79. ∫(from -1 to 0)6dt/√(3-2t-t²)

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Evaluate the integrals in Exercises 77–90.
79. ∫(from -1 to 0)6dt/√(3-2t-t²)