Evaluate the definite integral. ∫ 0 ln 2 e 5 y 3 + e 5 y d y \int_0^{\ln2}\frac{e^{5y}}{3+e^{5y}}dy

this problem, we're asked to evaluate the definite integral from zero to the natural log of two of e to the power of five y over three plus e to the power of y five y d y. Now remember that whenever we have to use substitution to evaluate a definite integral, there are two different ways that we could go about this. Now depending on your textbook or your professor, you may have learned one method but not the other, or you may just have a preference for with which method you like to use. Now I'm going to go through both possible ways that you could solve this here, and of course you should get the same answer regardless of which method. Feel free to skip around according to which method you like to use. Here, we're going to start with what we're referring to as method one. We're going to first treat this as an indefinite integral, put in that original function, and evaluate at our original bounds at the end. So with this first method, method one, let's get started with our first step, choosing u and finding du. Now in this particular problem, if I choose u as this bottom function, that will make du appear nicely in my integral. So if u is equal to three plus e to the five y, that then makes du equal to five e to the five y, applying the chain rule to take the derivative of this here. Because the derivative of three is zero, the integral or the derivative of e to the five y is five e to the five y. This is five e to the five y dy. Now we move on to step two here, which is rewriting our integral only in terms of this u and du. So if this is u, then du is five e to the five y. We can see that we're missing that five there, so we need to go ahead and multiply by five and its reciprocal in order to get our substitution to work. So if I multiply by five here, that means I need to multiply by one fifth as well. So I'm not actually changing the value of my integral. So this allows me to rewrite my integral as the integral of one over u du with that one fifth on the outside there because that five e to the five y dy is my du, and I have one over u and with that one fifth on the outside. So now we can move on to step three here, integrating with respect to u. Now the integral of one over u is just the natural log of u, so this is one fifth times the natural log of u and then plus my constant of integration c since we're currently treating this as an indefinite integral. With step three done, we move on to step four, replacing u with our original function g of x, in this case, g of y since that's what we're integrating with respect to. So this is one fifth times the natural log of three plus e to the five y. And now since we've replaced that with our original function, we've brought that back in. We, of course, wanna end here by evaluating our antiderivative at those original bounds. So in this case, zero and the natural log of two. Those are original bounds here. So that's what we're gonna plug in here using our fundamental theorem of calculus. This is one fifth times the natural log of three plus e to the power of five natural log of two. So that's that upper bound and then subtracting off the natural log of three plus e to the five times zero. Now e to the zero is just one. So that makes this one. And then simplifying here, one fifth times the natural log of three plus plus e to the five natural log of two. If I move that five up into that exponent here, so this is the natural log of two to the power of five using our log rules here, that e and that natural log are going to cancel out. So this is gonna leave me here with the natural log of three plus two to the power of five and then subtracting off here the natural log of four because three plus one is four having canceled or not canceled, but e to the power of zero is just one. So we have the natural log of three plus two to the fifth minus a natural log of four. Now we can simplify this a little bit more. One fifth times the natural log, three plus two to the fifth, that's 35. So natural log of 35 minus natural log of four. Now since we are working with natural logs here, we can simplify this further. So in its most simplified form, but not making this a decimal, this would be one fifth times the natural log of 35 over four. Or we can go ahead and put this into our calculator and get a decimal here. And this decimal will end up being 0.434 for the final answer to this definite integral. Now that was our first method, so we're done here. We've evaluated this at those original bounds. In our second method, we're going to start out largely in the same way, but as we are making our substitution, we're also going to transform our bounds along with the rest of our integral. So for our second method here, method two, we have our integral from zero to the natural log of two of e to the five y over three plus e to the five y d y. So starting out the same way, choosing u, finding du, we're gonna make the same exact choice that we did up above. If we choose u to be that three plus e to the power of five y, that makes du equal to the derivative of that. So five e to the power of five y d y. So we know what u and du are. We move on to step two, rewriting our integral only in terms of u and du including our bounds. Now we have to do the same thing that we did up above again here. We need to multiply by a constant and its reciprocal in order to make this substitution work because I can see here that this is u, but du is five times e to the five y d y. So if I go ahead and multiply it by that five and one fifth, that will make my substitution work out. So I have u and I have du, but we also need to go ahead and transform form those bounds by plugging into u equals g of x. In this case, g of y. Here, three plus e to the five y. So g of zero and g of the natural log of two. So those are our original bounds. We're plugging them into our choice for you to get our new bounds here. So g of zero here, this is three plus e to the five times zero. This makes this e to the power of zero, which is just one. So this is three plus one, which is four. Then g times the g of the natural log of two, this is three plus e to the five natural log of two. Remember, using our log rules here, I can move that five into the exponent of my two. So that's natural log of two to the power of five. That causes e and natural log to cancel out. So this is just three plus two to the power of five, which is 35. So these are my new bounds here four and thirty five and I can fully rewrite my integral here. So this is one fifth times the integral from four to 35 those are new bounds of one over u du. Now we can go ahead and integrate here with respect to u. So continuing on here finding our antiderivative, that's one fifth and then the integral of one over u is the natural log of u, so this is natural log of u, bounded from four to 35. So this is then going to be applying our final bounds here at those new bounds that we found four and thirty five. Onefive times the natural log of my upper bound 35 minus the natural log of my lower bound four. Remember, just like above, we can simplify this using our log rules into one fifth times the natural log of 35 over four or simply as a decimal if we go ahead and plug this into our calculator we will end up getting 0.434 for that final answer. Of course this is the same answer that we got up above because no matter which method you use to solve this you should get the same answer. So feel free to let us know if you have any questions here, and I'll see you in the next one.

8. Definite Integrals

Substitution for Definite Integrals

Business Calculus

8. Definite Integrals

Substitution for Definite Integrals

Struggling with Business Calculus?

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Multiple Choice

Evaluate the definite integral.
$\int_{0}^{\ln 2} \frac{e^{5 y}}{3 + e^{5 y}} d y$

0.139

0.434

-0.139

-0.675

Verified step by step guidance

Rewrite the integral in its given form: $ \int_0^{\ln 2} \frac{e^{5y}}{3 + e^{5y}} \, dy $. This is a definite integral with limits of integration from 0 to $ \ln 2 $.

Perform a substitution to simplify the integral. Let $ u = 3 + e^{5y} $, so $ du = 5e^{5y} \, dy $. This substitution will help simplify the denominator.

Adjust the differential $ dy $ using the substitution: $ dy = \frac{du}{5e^{5y}} $. Notice that $ e^{5y} $ is part of the numerator, which will cancel out with the substitution.

Change the limits of integration to match the substitution. When $ y = 0 $, $ u = 3 + e^{5(0)} = 4 $. When $ y = \ln 2 $, $ u = 3 + e^{5(\ln 2)} = 3 + 2^5 = 35 $. The new integral becomes $ \int_4^{35} \frac{1}{5u} \, du $.

Integrate the simplified expression. The integral of $ \frac{1}{u} $ is $ \ln |u| $. After integrating, evaluate the result at the new limits $ u = 4 $ and $ u = 35 $, and multiply by the constant $ \frac{1}{5} $.