9–40. Integration by parts Evaluate the following integrals us...

Question

9–40. Integration by parts Evaluate the following integrals using integration by parts.

14. ∫ s · e⁻²ˢ ds

Accepted Answer

Welcome back, everyone. Evaluate the integral of X multiplied by the negative XDX using integration by parts. For this problem, let's recall the integration by parts formula. Integral of UDV is equal to U minus the integral of VDU. In this problem, we have a product of X and E to the negative X. We're going to set U equals X, because whenever we have a polynomial and an exponential function, U prefers to have polynomial against exponential functions according to the LIPET rule, right? So we're setting U equals X, which means that the remaining part, which is DV becomes E raises the power of negative XD X. Now, if U is equal to X, then D U is equal to D X. And in terms of V, we can identify it by integrating E, raises the power of negative XD X. So V would be eats the negative X divided by the slope of the linear term in the exponent, so we're dividing by -1. We can basically add the negative sign in front, so we can show that V equals negative, eats the negative X, right? And now we are ready to apply the formula. The integral of X multiplied by E to the negative XDX is U multiplied by V. Which is X multiplied by negative E to the negative X, or simply negative X to the negative X, we can seek out the negative sign and write it in front. And we're subtracting the integral of VDU. So we're subtracting the integral of negative E to the negative X, and the U is DX, right? Now we can take out the negative sign again, so we're basically adding the integral. Of each to the negative XD X. And we already know that the integral of eats the negative XD X is. Negative. E to the negative X, right? From the previous part. So now let's go ahead and calculate the result. We get negative X E to the negative X. Plus Negative E to the negative X, and we're going to add a constant of integration C for an indefinite integral. Notice that our terms have negative signs, so we can factor out negative. E To the negative X because E to the negative X is also a repeating term. And then inresies we're going to have eggs. Plus 1, so now we have our factored form and now let's add a constant of integration C. And that would be our final answer for this problem, negative E to the negative X multiplied by X + 1 plus C. Thank you for watching.

9–40. Integration by parts Evaluate the following integrals using integration by parts.14. ∫ s · e⁻²ˢ ds

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9–40. Integration by parts Evaluate the following integrals using integration by parts.
14. ∫ s · e⁻²ˢ ds