Evaluate the integrals in Exercises 1–24 using integration by ...

Question

Evaluate the integrals in Exercises 1–24 using integration by parts.

∫ (r² + r + 1) e^r dr

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 integral using integration by parts. The integral of x 2 + 2 x + 2 in parentheses multiplied by e to the power of x dx. OK. So it appears for this particular prompt, we're asked to evaluate the integral that is provided to us by the prompt itself by using integration by parts. So now that we know that we're ultimately trying to solve for this integral by using integration by parts, our first step that we need to take based on our prior knowledge. is we need to identify the parts of the integram in order to assign as You, so once again we need to sign our integrand parts, or once again, parts of the integram using. You and DV. So, we need to find what we're gonna denote as you and DV. So with that said, Our next step after that, once we've assigned our parts of the integrand as U and DB, we will need to now, at that point, compute the corresponding derivatives and integrals for each part, and after that, we will then need to recall and apply the integration by parts formula to our integral. And afterwards, we will need to integrate the resulting simple interval that we produced. And finally, lastly, we will need to substitute back in our variables in order to obtain the final solution, and we will need to repeat this process if it's required or if needed. So with that said, Let us note, so let us state that you will be equal to x 2 + 2 x + 2, and that DV will be equal to EXDX. So now, we need to find the derivative of U is equal to x 2 + 2 x + 2. In order to get DU by DX is equal to 2 x + 2. And then we will need to multiply both sides by DX to get that DU is equal to parentheses 2 x + 2. DX. So we now need to find the integral of DV is equal to e to the power of xDX. So that will mean that the integral of DV is equal to the integral of e to the power of xDX, which will mean that V is equal to e to the power of x. So now, at this stage, we need to apply the integration by parts formula, which as we should recall, states that the integral of U multiplied by DV. Is going to be equal to drum roll U multiplied by V minus the integral of VDU. OK. Moving right along here. So, we could go ahead and use our integration by parts formula to write that the integral of parentheses of x to the 2. Plus 2 X. + 2 Multiplied by e to the power of x. D X is equal to parentheses of x 2 + 2 x + 2. is going to be multiplied by, so once again, in parentheses x 2 + 2 x + 2 in parentheses multiplied by e to the power of x minus the integral of e to the power of x multiplied by parentheses of 2 x + 2DX. And now Moving right along here, we need to find the interval of the integral of E to the power of x multiplied by parentheses of 2 x + 2 DX. And we need to let You will be equal to 2 x + 2. And DV be equal to e to the power of xDX. And we will now need to find the derivative of U is equal to 2 x + 2. To get DU by DX is equal to 2. And we will need to multiply both sides by DX to get. DU is equal to 2DX. So now we need to find the integral of DV is equal to e to the power of x, DX. So we need to take the integral of DV is equal to the integral of E the power of xDX, which will mean that V is equal to e to the power of x. And again, we need to apply the integration by parts formula, and when we do, we will be able to write the integral of e to the power of x multiplied by parentheses of. 2 X + 2. DX will be equal to parentheses 2X. It's gonna be parentheses of 2 X. + 2. Multiplied by, so once again in parentheses 2 x + 2 multiplied by e to the power of x. Minus minus the integral. Of e to the power of x multiplied by 2 EX. Fantastic. And this will be equal to parentheses of 2 x plus 2 in parentheses multiplied by e to the power of x minus 2 multiplied by the integral of e to the power of x dx. Which will be equal to drum roll 2 x + 2 multiplied by E X minus 2 multiplied by e to the power of x, which will be equal to 2 x + 2 minus 2 in parentheses multiplied by e to the power of x. Which will be simply equal to 2 multiplied by x multiplied by e to the power of X. Fantastic. So now we need to substitute back in, and when we do, when we perform substitution, we will find that we could write the integral of e to the power of x multiplied by parentheses of 2 x + 2 dx will be equal to 2 multiplied by x multiplied by e to the power of x, and we're substituting this into. The integral of parentheses x 2 + 2 x + 2 multiplied by e to the power of xDX. And this will be equal to parentheses x 2 + 2 x + 2 in parentheses multiplied by e x minus the integral e x multiplied by parentheses of 2 x + 2DX, and we also need to add the integral constant c, so we need to add c. And when we do that, we will be able to write that the integral of parentheses of x 2 + 2 x + 2 in parentheses multiplied by E X D X. Will be equal to x 2 + 2 x + 2 in parentheses multiplied by e to the power of x. -2 multiplied by x multiplied by e to the power of x + C. And this will be equal to parentheses of x 2 + 2 X, and this will be + 2. -2 x parentheses multiplied by e to the power of x + C. Which will be equal to x 2 + 2 in parentheses multiplied by e x + C. So that will mean, without a doubt the integral of parentheses of x 2 + 2 x + 2 multiplied by e to the power of xDX will be simply equal to. E to the power of x multiplied by parentheses of x 2 + 2 + c and that is our correct and final answer. Hooray, we did it. So going back to the top to review what we've learned, we can safely say that our final answer for our evaluated integral once again without a doubt will be. E to the power of X multiplied by parentheses of x 2 + 2 + C. 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 1–24 using integration by parts.∫ (r² + r + 1) e^r dr

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Evaluate the integrals in Exercises 1–24 using integration by parts.
∫ (r² + r + 1) e^r dr