## 17Calculus - Definite Integrals Using Substitution

On this page, we assume you have studied both integration by substitution and definite integrals. When you combine the two techniques to evaluate integrals, there are some special techniques that will help you work them correctly. Those techniques are what we cover on this page.

When you have a definite integral, you need to do additional work to correctly handle the limits of integration. Additionally, there is a mistake that many students make that could not only cause you to get an incorrect answer but also, if your instructor is paying attention, could cost you extra points. First, let's talk about the correct way to handle the limits of integration and then show you how NOT to do it.

There are basically two techniques, both of them correct but pay attention to what your instructor requires, since they may want you to use one of them and not the other.

First Way: Change The Limits of Integration

The first and best (in our opinion) way is to use the expression for u to change the limits of integration. Let's do an example to see how this works.

Evaluate $$\displaystyle{ \int_{-2}^{1}{ x \sqrt{x+2} ~dx } }$$.

$$-2\sqrt{3}/5$$

Evaluate $$\displaystyle{ \int_{-2}^{1}{ x \sqrt{x+2} ~dx } }$$.

Solution

Problem Statement: Evaluate $$\displaystyle{ \int_{-2}^{1}{ x \sqrt{x+2} ~dx } }$$.
Solution: Let $$u=x+2 \to du=1~dx$$ and $$x=u-2$$.
We need change the limits of integration. The lower limit is $$x=-2$$, so we use the equation for u to get the lower limit in terms of u, i.e. $$x=-2 \to u=-2+2=0$$. We do the same thing with the upper limit of integration. $$x=1 \to u=1+2=3$$. So our integral is now
$$\displaystyle{ \int_{0}^{3}{ (u-2)\sqrt{u} ~du } = }$$ $$\displaystyle{ \int_{0}^{3}{u^{3/2}-2u^{1/2}} = }$$ $$\displaystyle{ \left[ (2/5)u^{5/2}-(4/3)u^{3/2} \right]_{0}^{3} = }$$ $$\displaystyle{ (2/5)3^{5/2} -(4/3)3^{3/2} = }$$ $$\displaystyle{ 2(3^{3/2})(3/5-2/3) = }$$ $$\displaystyle{ 6\sqrt{3}(-1/15) = -2\sqrt{3}/5 }$$

$$-2\sqrt{3}/5$$

Notice in that example, we changed the limits of integration from what they were in terms of x to their values when the variable to integration is changed to u. The limits of integration are tied to the variable of integration. So we could write the integral as $$\displaystyle{ \int_{x=-2}^{x=1}{ x \sqrt{x+2} ~dx } }$$

Okay, so that is the first way and, in our opinion, the best way to handle definite integrals when we use integration by substitution.

Second Way: Drop The Limits and Go Back

The other way is to drop the limits of integration, evaluate the integral, go back to x's in the integral and then substitute the original limits of integration. This next example, shows how to do that.

Evaluate the same integral as the last example, but do not change the limits of integration and use correct notation.

Evaluate the same integral as the last example, but do not change the limits of integration and use correct notation.

Solution

Let $$u=x+2 \to du=1~dx$$ and $$x=u-2$$.
$$\displaystyle{ \int{ (u-2)\sqrt{u} ~du } = }$$ $$\displaystyle{ \int{u^{3/2}-2u^{1/2}} = }$$ $$\displaystyle{ (2/5)u^{5/2}-(4/3)u^{3/2} }$$
Now we convert back to x to get
$$\displaystyle{ \left[ (2/5)(x+2)^{5/2}-(4/3)(x+2)^{3/2} \right]_{-2}^{1} = -2\sqrt{3}/5 }$$

How NOT To Do It

Notice in the solution to the last example, we dropped the limits of integration when we wrote the integral in terms of u. It would be incorrect to write $$\displaystyle{ \int_{-2}^{1}{ (u-2)\sqrt{u} ~du} }$$ since the limits of integration are in terms of x but the integral is in terms of u. Now, theoretically it MIGHT be okay to write $$\displaystyle{ \int_{x=-2}^{x=1}{ (u-2)\sqrt{u} ~du } }$$ but no one does that (but, of course, check with your instructor to see what they expect). So not only is the notation $$\displaystyle{ \int_{-2}^{1}{ (u-2)\sqrt{u} ~du } }$$ incorrect, but after you have completed the integration it is extremely easy to forget to go back to x's in the integral before substituting the limits of integration.

As an instructor, I ALWAYS take off points for writing the integral this way. So be extra careful. We recommend that you always change the limits of integration like we did in the first example. Here is a video that explains very well when to change the limits of integration, why instructors are so picky about it and possible mistakes that could trip you up. We highly recommend this video.

### PatrickJMT - U-Substitution: When Do I Have to Change the Limits of Integration? [8min-14secs]

video by PatrickJMT

Practice

Unless otherwise instructed, evaluate these integrals using correct notation. Give your answers in exact, simplified form.

$$\displaystyle{ \int_{1}^{2}{\frac{2x}{\sqrt{x^2+1}}~dx} }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{1}^{2}{\frac{2x}{\sqrt{x^2+1}}~dx} }$$. Give your answer in exact, simplified and factored form.

Solution

### 887 video

video by PatrickJMT

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Show that $$\displaystyle{ \int_{a}^{b}{f(x)~dx} = }$$ $$\displaystyle{ \int_{a}^{b}{f(a+b-x)~dx} }$$

Problem Statement

Show that $$\displaystyle{ \int_{a}^{b}{f(x)~dx} = }$$ $$\displaystyle{ \int_{a}^{b}{f(a+b-x)~dx} }$$

Solution

Use the substitution $$u=a+b-x \to du=-dx$$
$$x=a \to u=b$$
$$x=b \to u=a$$
$$\displaystyle{ \int_{a}^{b}{f(a+b-x)~dx} = \int_{b}^{a}{f(u)(-du)} = }$$ $$\displaystyle{ \int_{a}^{b}{f(u)~du} }$$
Since we can use any variable of integration, $$\displaystyle{ \int_{a}^{b}{f(u)~du} = \int_{a}^{b}{f(x)~dx} }$$
Therefore $$\displaystyle{ \int_{a}^{b}{f(x)~dx} = \int_{a}^{b}{f(a+b-x)~dx} }$$   [qed]

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$$\displaystyle{ \int_{0}^{4}{x\sqrt{x^2+9}~dx} }$$

Problem Statement

Evaluate $$\displaystyle{ \int_{0}^{4}{x\sqrt{x^2+9}~dx} }$$. Give your answer in exact, simplified, factored form.

Solution

Be careful that you don't do what she does in this video with her notation. She leaves her limits of integration in terms of x after she does the substitution with u. If you watched the last video in the tutorial, you know that doing this would result in loss of points with many instructors since it is wrong.

### 889 video

video by Krista King Math

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You CAN Ace Calculus

 basics of integrals substitution definite integrals Some practice problems on this page require the use of integration by parts, which is a more advanced technique usually introduced in second semester calculus. If you haven't studied integration by parts yet, no worries. You can skip those practice problems and come back later, once you have covered integration by parts.

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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