The integrals in Exercises 1–34 converge. Evaluate the integra...

Question

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.

∫₀² (s + 1) / √(4 − s²) ds

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. Evaluate the improper integral. The integral evaluated from 0 to 1 of 2 x + 3, all divided by the square root of 1 minus x 2 dx. OK. So it appears for this particular prompt we're asked to evaluate this improper integral that is provided to us by the prompt itself. So now that we know that we're ultimately trying to evaluate this improper integral, our first step that we need to take based on our prior knowledge is we need to rewrite our integral with a limit. So that means we need to take the integral evaluated from 0 to 1 of 2 X. Plus 3. Divided by. The square root of 1 minus x 2. DX and this will be equal to the limit as B approaches 1 of the integral evaluated from 0 to B of 2x plus 3 divided by the square root of 1 minus x 2 dx. Our second step now is we need to find the antiderivative of the integral of 2 x + 3 divided by the square root of 1 minus x 2 dx. And in order to solve for this antiderivative, as we should recall, is we need to first split our integral, and when we do that, we'll be able to write the integral of 2 x + 3 divided by the square root of 1 minus x 2. Dx will be equal to the integral of 2 x divided by the square root of 1 minus x to the power of 2 dx plus the integral of 3 divided by the square root of 1 minus x to the power of 2 dx. So Let us take a moment here to focus our attention on the integral of 2 x divided by the square root of 1 minus X2. Let us use the substitution method to to evaluate this specific integral by itself. So when we use the substitution method, and let's call this integral one. For now, so we don't get confused, let us state that you will be equal to 1 minus x 2. So, that will mean that DU by DX will be equal to -2X, which will mean that DU will be equal to 2XDX. Fantastic. So, when we evaluate this particular integral, or rather when we solve for this integral, we will find that the integral of 2 x divided by the square root of 1 minus. X to the power of 2. And don't forget your DX will be equal to drum roll. Negative, the integral of 1 divided by the square root of u d u, which will be equal to negative, the integral of u to the power of 1/2 d u. Which will be equal to u to the power of 1/2 + 1 divided by u to the power, so it's going to be no u, so it's gonna be equal to u to the power of 1/2 + 1 divided by 1/2 + 1. Plusc subscript 1 where c denotes some constant variable and this will be equal to u to the power of 1/2 divided by 1/22 plus c subscript 1 which will be equal to -2 multiplied by the square root of u plus c subscript 1. Which will be equal to -2 multiplied by the square root of 1 minus x 2 plus C subscript 1. Hooray, we are on a roll. So now let's focus our attention on our second integral, so we'll call this number 2 integral, which is the integral. Of 3 divided by the square root of 1 minus x to the power of 2, the x and using our substitution, we can't use substitution. Method on this particular integral. So we need to recall and use the following formula, which states that the integral of DX divided by the square root of 1 minus X2. Is equal to inverse sin of x plus C. So that will mean that the integral of 3 divided by the square root of 1 minus X 2 DX will be equal to 3 multiplied by inverse sin of X plus C subscript 2. Fantastic. Moving right along here, we now need to combine both of our integrals, so when we do, we will find that the integral of 2. X + 3. Divided by the square root of 1 minus x 2. D X Will be equal to the integral of 2 x divided by the square root of 1 minus x 2. DX plus the integral of 3 divided by the square root of 1 minus x 2. The X Which will be equal to -2 multiplied by the square root of 1 minus X2 + 3 multiplied by inverse sin of X + C. Where, let's make a note, where C is equal to C1 plus C2. Let's make a note of that. So that will mean moving right along here to step number 3. We now need to evaluate our definite integral. So when we do that, we will find that the integral evaluated from 0 to B of 2 x + 3 divided by the square root of 1 minus x to the 2. The x will be equal to parentheses of -2 multiplied by the square root of 1 minus x 2. Plus 3 multiplied by inverse sine of x. Evaluated from 0 to B. Will be equal to parentheses of 2 multiplied by the square root of 1 minus b to the power of 2 + 3 multiplied by inverse sin of b minus so it's gonna be minus negative or rather minus parentheses of 2 multiplied by the square root of 1 minus 0 to 2. Plus 3 multiplied by inverse sin of 0, fantastic. Which will be equal to -2 multiplied by the square root of 1 minus b 2 + 3 multiplied by inverse sin of B minus -2. Plus 3 multiplied by 0. And when we simplify this expression, we will find that it's simply equal to -2 multiplied by the square root of 1 minus b 2 + 3 multiplied by inverse sin of b. + 2. Fantastic. Moving right along here. So our 4th step is we need to take the limit as B approaches 1, so we need to take the limit as B approaches 1. Of parentheses of -2 multiplied by the square root of 1 minus b 2 + 3 multiplied by inverse sin of b + 2. Which will be equal to -2 multiplied by the square root of 1 minus 12 + 3 multiplied by inverse sin of 1 + 2. Which will be equal to -2 multiplied by 0 + 3 multiplied by pi divided by 2 + 2, which when we simplify this expression and we write it in its most simplified form, we will find that it's simply equal to 3 pi plus 4 divided by 2 and this is our correct and final answer. Hooray, we did it. So going back to look at the prom itself, we can safely say, based on what we've learned together today, that our final answer without a doubt for this improper integral once again is 3 pi plus 4 divided by 2. And that's it. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.∫₀² (s + 1) / √(4 − s²) ds

Watch next

The integrals in Exercises 1–34 converge. Evaluate the integrals without using tables.
∫₀² (s + 1) / √(4 − s²) ds