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

7–64. Integration review Evaluate the following integrals.

44. ∫ from 0 to √3 of (6x³) / √(x² + 1) dx

Accepted Answer

Welcome back, everyone. Calculate the definite integral. Integral from 1 to 2 of 10 X cub to divided by square root of X2 + 2DX. For this problem, we're going to begin with U substitution. Let's suppose that our function under the radical is U, so U equals X2 + 2. Our next step is to identify the derivative DU divided by D X. That's the derivative of X2 + 2, which is 2 X. Now what we can do is simply express DX from this form, and to identify DX we basically have to divide the U by 2 X. So D X equals DU divided by 2 X. Now, what we want to do is simply focus on the integrant, and essentially what we get is 10 X cubed multiplied by the X, right? So we're multiplying by DU, which is divided by 2 X, and then our denominator becomes square root of U. Now notice that 10 X cubed divided by 2 X is equal to 5 x 2, so we get 5 X 2 DU in the numerator, and this is divided by square root of U. And now what we can do is simply express x2d. X2 is equal to U minus 2, we can subtract 2 from our U substitution, so X2 is equal to U minus 2. And now we have shown that our integran is 5 multiplied by U minus 2 DU divided by square root of U. And now we can split it into. Two parts, right? So first of all, we have 5 EU divided by square root of you. Minus 1 Divided by square root of you and all of that is multiplied by DU, right? Now, of course, we can simplify you divided by square root of you, which is square root of you. So the first integran will involve 5 square root of you. And for the second one, let's leave it as it is for now. What we're going to do next is simply change the limits of integration. We know that X1 is equal to 1, which means that you 1. The lower limit of integration is going to be 12 + 2, and that's 3. And our upper limit of integration is 2. So, in terms of you, that would be 2 squad plus 2, which is 6. Meaning our integral in terms of you becomes. Integral from 3 to 6. Of 5 square root of u minus 10 divided by square root of u. The And then we can evaluate to integrals, right? So we can rewrite this as The integral from 3 to 6 off. Now 5 can be factored out. And then we end up with you to the power of 12, right? We can rewrite square roots of U as U to the power of 1/2, followed by the U. And then we're subtracting the integral from 3 to 6. We can factor out 10. And then we get use the power of -15 because We end up with one divided by square root of you, right? The EU And from here we can just apply the power rules. So for the first integral we get 5 multiplied by you the power of 3 halves. Divided by 3 halves. Now 5 divided by 3 halves says, 5 multiplied by 2 divided by 3. So that'd be 10 divided by 3, we can just write it as our constant. 0 divided by 3. And then for the 2nd integral, we are subtracting 10. Applying the power rule, we get you the power of -12 plus 1 is 1/2, and we're dividing by 1/2. 10 divided by 12 is 10 multiplied by 2. So that's 20, and that's our next constant. 20 multiplied by you to the power of 1/2. And now we are evaluating the results from 3 to 6, right? So now let's go ahead and simplify. First of all, we can factor out you, right? And essentially we can show that this is equal to. If we completely factor it out, well, we can begin with. 10/3 Followed by you to the power of one half which is square root of you. In parentis, we get you. Right? Because you multiplied by square root of you is used to the power of three halves. And then we are subtracting. Since the original term was 20 to the power of 1/2, we first will want to consider our constant. 20 divided by 10/3 would be 20 multiplied by 3 divided by 10, and that's 6, right? And then that's it, right, because we have used the power of one half and we have factored out square root of you. And now let's evaluate it from 3 to 6. So first of all, we substitute the upper limit of integration. That'd be 10/3 square root of 6 multiplied by 6 minus 6. And then we are subtracting the value of the function at the lower limit of integration. Notice that we have also factored out the constants and 3s, right? So we take minus. Square root of 3. Multiplied by 3 minus 6. And now let's valuate the results, so that'd be 10/3. We noticed that our first term is 0 because we have multiplication by 0. And then We are multiplying 10/3 by square root of 3 multiplied by -3. So we're multiplying by 3 square root of 3 is positive because negative multiplied by negative is positive. And now we can cancel out the 3. And showed that the final answer is sun square root of 3. Well then, we have got our final answer and thank you for watching.

7–64. Integration review Evaluate the following integrals.
44. ∫ from 0 to √3 of (6x³) / √(x² + 1) dx

Key Concepts

Definite Integrals

Integration Techniques

Fundamental Theorem of Calculus

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