Evaluate the definite integral. ∫ 0 1 t t 2 + 1 d t \int_0^1\frac{t}{\sqrt{t^2+1}}dt

this problem, we're asked to evaluate the definite integral from zero to one of t over the square root of t squared plus one d t. Now because this is a definite integral and we're going to have to evaluate here using substitution, there are two different ways that we can go about this. Now, depending on your textbook or your professor, you may have learned only one of these methods and not the other, or you may just have a preference for one over the other. Here, I'm going to go through both different ways that you could have gone about solving this integral, so feel free to skip around accordingly. Now we're going to start with our first method in which we're going to treat this as an indefinite integral from the beginning and then apply our original bounds at the end. So getting started here with our first step here in method one, the first thing that we need to do is choose u and find du. So looking at this particular problem, ignoring those bounds for now, because I have this t squared plus one under my radical, that's a good clue that that's what I should choose to be u. So starting with our first method here, method one, we're going to choose u as that t squared plus one. And what does that then make du? This makes du equal to two t dt. That's the derivative of that t squared plus one multiplied by dt. So with this in mind, we can go ahead and move on to step two and rewrite this integral only in terms of this u and du. Now I can see that I have that u clearly in my integral and I have t dt, but I need this to be two t dt. So I wanna go ahead and make this work by multiplying here by a constant and its reciprocal. So I wanna go ahead and multiply by that two. If I'm multiplying by two, I also need to multiply it by the reciprocal of two, which is one half, and I'm gonna go ahead and just stick that on the outside of my integral. So now with this two t dt having multiplied by that constant, this gives me du. So I can rewrite this integral here as the integral one half times the integral of one over the square root of u du. So we're ignoring those bounds for now. We've completed step two here. Now we wanna go ahead and integrate this with respect to u. So this is one over the square root of u. We can think about this as a power as u to the power of negative one half so that we can easily apply the power rule here. So going ahead and finding my antiderivative, I'm keeping that one half that's that's on the outside of my integral and then multiplying this by the antiderivative of that u to the power of negative one half. So we wanna add one to that exponent, negative one half plus one gives us one positive one half. Then we wanna divide by that positive one half here as well. So dividing by positive one half is the same thing as multiplying by two. So this is two u to the power of positive one half, that new power from our power rule. Now here, we wanna consider our constant of integration, although we are going to end up not needing that later, so it's up to you. At this point, since we're treating this as an indefinite integral, I'm gonna keep that there for now. So looking at this having integrated with respect to u, this one half and this two will cancel out. So really, I'm just left with u to the power of one half plus c. Then we move on to step four, which is replacing u with our original function, g of x in our steps here, but we know that we're working with t. So we want our original function of t, which looking up here, we know that we chose that to be t squared plus one. So we wanna go ahead and plug that back in. So this is t squared plus one to the power of one half. We can also just think of that as being a square root if we wanna write it that way. So the square root of t squared plus one and then plus my constant of integration c. Now at this point, since we have replaced u with g of x, we're moving on to our final step here, which is evaluating our antiderivative at those original bounds. So we don't really have to worry about that constant of integration, and we can go ahead and bound this from zero to one. Those are the original bounds of our definite integral. So now evaluating that, applying our fundamental theorem of calculus here, this is going to give us the square root of one squared plus one minus the square root of zero squared plus one. Now if I go ahead and work this out, this is gonna be root two minus the square root of one, one which is just one. So root two minus one, this is an acceptable answer, but if you want this as a decimal this ends up being about 0.414. So that was our first method here, but remember we could choose to evaluate this in another way where we actually change our bounds alongside our substitution. So let's take a look at our second method that we could choose to solve this integral. So remember, this is the integral from zero to one of t over the square root of t squared plus one d t. So we're looking at this second method here. We're gonna end up doing a lot of the same steps. So remember starting from the beginning here, choosing u and finding du, this is gonna end up being the exact same thing that we saw in that first method. So if we choose u to be that t squared plus one, that makes du equal to two t dt. This means that we need to go ahead and multiply it by two and also one half in order to get du in this integral. So we have du and we have a u. Now in this second step here, as we are rewriting our integral only in terms of u and du, I already did a little bit of this, I multiplied it by that constant there in order to make this substitution work. But in this second part of step two of method two, we need to go ahead and transform our bounds along with the rest of our substitution that we're doing. So we wanna go ahead and plug our bounds zero and one into our choice for u. So this means that we're looking here at g of one and g of zero because those are our original bounds, zero and one, and we're plugging them into our choice for u, which was t squared plus one. So plugging that in here, that's one squared plus one, which gives us two, and then zero squared plus one, which gives us one. So our new bounds are going to be from one to two. So as we are rewriting our integral, this is going to be the integral from one to two. And looking at everything that we did to do our substitution, we have that one half, we have u, and we have du. So this is going to be one half times the integral from one to two of one over root u du. So we can see that everything about this integral is the same that we saw in method one, except here we went ahead and transformed our bounds along with the rest of our integral. So that's where these differentiate from each other. So now that we have rewritten this integral including our bounds in terms of u and du, we can move on to step three, integrating with respect to u. So going ahead and doing that here, I'm keeping that one half, that constant as usual, and then applying the power rule here. Remember, one over root u is the same thing as writing this as u to the power of a negative one half. Going ahead and adding one to that power, that's gonna give us positive one half. If I divide by one half, I'm really multiplying by two. So this is two u to the power of positive one half, and I don't have to worry about that constant of integration c because we know that this is bounded from one to two since we transformed those bounds. Alright. Now we see this one half and this two are going to cancel out here. So we have u to the power of one half, which remember we can also think of as being the square root of u bounded from one to two. Then we just have to go ahead and plug in those bounds. That's a root two minus root one, which is just one. So this is really root two minus one or as a decimal we could also write this as 0.414. And of course, we got the same exact answer that we did up above because no matter which method you should use you should get the same answer. Now 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

Calculus

8. Definite Integrals

Substitution

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Multiple Choice

Evaluate the definite integral.
$\int_{0}^{1} \frac{t}{\sqrt{t^{2} + 1}} d t$

1.414

0.414

0.707

Verified step by step guidance

Rewrite the integral in a more convenient form: $ \int_0^1 \frac{t}{\sqrt{t^2+1}} \, dt $. Notice that the numerator $ t $ is the derivative of the expression inside the square root, $ t^2 + 1 $. This suggests a substitution method.

Let $ u = t^2 + 1 $. Then, $ du = 2t \, dt $, or equivalently, $ \frac{1}{2} du = t \, dt $. Update the limits of integration: when $ t = 0 $, $ u = 1 $; and when $ t = 1 $, $ u = 2 $.

Substitute $ u $ and $ du $ into the integral: $ \int_0^1 \frac{t}{\sqrt{t^2+1}} \, dt = \int_1^2 \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \, du $.

Simplify the integral: $ \int_1^2 \frac{1}{\sqrt{u}} \cdot \frac{1}{2} \, du = \frac{1}{2} \int_1^2 u^{-1/2} \, du $.

Evaluate the integral: Use the power rule for integration, $ \int u^n \, du = \frac{u^{n+1}}{n+1} + C $, where $ n \neq -1 $. Apply this to $ u^{-1/2} $, and then substitute the limits of integration to find the result.