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You CAN Ace Calculus

17calculus > integrals > definite integrals

 basics of integrals substitution Definite integrals are usually introduced early in the study of integration after covering the basics and integration by substitution. However, 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 filter out those practice problems and come back later, once you have covered integration by parts.

### Calculus Main Topics

Integrals

Integral Applications

Single Variable Calculus

Multi-Variable Calculus

### Tools

math tools

general learning tools

WikiBooks: Infinite Sums

Wolfram Alpha: The Second Fundamental Theorem of Calculus

Wikipedia: Fundamental Theorem of Calculus (It is interesting that in wikipedia, the first and second fundamental theorems are switched, which is different than is sometimes taught in first semester calculus and discussed in Larson Calculus.)

Definite Integrals

Since the definite integral is so closely tied to summations, we start with a discussion of summations and the associated notation. Then we cover the fundamental theorems of calculus.

Summation Notation

The first thing you need to be comfortable with in order to understand definite integrals is summation notation. This notation is just a compact way to write a list of numbers or terms that are added together. The notation looks like this.
$$\displaystyle{ \sum_{i=1}^{n}{a_n} }$$

If you haven't seen this before or you need a refresher on what all this means, here is a good video explaining it. This is a good overview of summation notation. In this video, he starts by explaining the general notation and then he works several examples.

 PatrickJMT - Summation Notation

Summation and the Definite Integral

The general sum $$\displaystyle{ \sum_{i=1}^{n}{[f(x_i) \cdot \Delta x_i]} }$$ is an approximation to the area under the curve $$f(x)$$. Let's look more closely at this sum and get an idea of what is going on.

For each $$i$$, we calculate the area of a rectangle on the interval $$[a,b]$$. The height of the rectangle is $$f(x_i)$$ and the width is $$\Delta x_i$$. If the same area is broken into more and more rectangles, we can get a better and better approximation of the area.

To get the actual area, we can take the limit of the above sum as follows.
$$\displaystyle{ \lim_{n \to \infty}{ \left[ \sum_{i=1}^{n}{(f(x_i) \cdot \Delta x_i)} \right] } = \int_{a}^{b}{f(x)~dx} }$$
Let's break down the equation. When taking the limit we get

$$\Delta x_i ~~~ \to ~~~ dx$$

$$\displaystyle{ \lim_{n \to \infty}\sum_{i=1}^{n}{} ~~~ \to ~~~ \int_{a}^{b}{} }$$

The idea is that the width of each interval goes to $$dx$$. The limit of the sum goes to $$\int_{a}^{b}{}$$. You can think of the integrand, $$f(x)$$, being swept from $$a$$ to $$b$$.

Okay, time for a video. Here is a good video on the definition of the definite integral. It will give you an intuitive understanding what it means and how the notation works.

 PatrickJMT - The Definite Integral - Understanding the Definition

Before we actually evaluate the Reimann sum with limits, let's see how to approximate the value of a definite integral. This is a great video with explanation of an example.

 PatrickJMT - Approximating a Definite Integral Using Rectangles

From the first video in this section, you know that the limit given above is the definition of the definite integral using Reimann Sums. Here is an example showing how to calculate a definite integral using this definition. The example is worked in two consecutive videos.

 PatrickJMT - Calculating a Definite Integral Using Riemann Sums

Before we go on, let's work some practice to make sure we understand these ideas.

Practice 1

$$\displaystyle{\sum_{k=1}^{6}{\frac{1}{k^2}}}$$

solution

Practice 2

Use summation notation to rewrite the finite sum $$\displaystyle{ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \frac{1}{64} }$$

solution

Practice 3

$$\displaystyle{\sum_{r=1}^{8}{(r-1)(r+2)}}$$

solution

Practice 4

$$\displaystyle{\sum_{i=1}^{4}{(-1)^i i^2}}$$

solution

The (First) Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus allows us to integrate a function between two points by finding the indefinite integral and evaluating it at the endpoints. The equation is
$$\displaystyle{\int_{a}^{b}{f(x)~dx} = \left. F(x) \right|_{a}^{b} = F(b) - F(a)}$$ where $$F' = f$$.

Okay, let's take a few minutes and watch a couple of videos explaining the proof of this theorem.

 MIP4U - Proof of the Fundamental Theorem of Calculus

Before we go on, let's work some basic integrals using this theorem.

Practice 5

$$\displaystyle{\int_{1}^{3}{x+2~dx}}$$

solution

The Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus

For a continuous function $$f$$ on an open interval $$I$$ containing the point $$a$$,
then the following equation holds for each point in $$I$$ $f(x) = \frac{d}{dx} \left[ \int_{a}^{x}{f(t)~dt} \right]$

The Second Fundamental Theorem of Calculus shows that integration can be reversed by differentiation. There are several key things to notice in this integral.
- The integral has a variable as an upper limit rather than a constant.
- The variable is an upper limit (not a lower limit) and the lower limit is still a constant.
- The upper limit, $$x$$, matches exactly the derivative variable, i.e. $$dx$$.
The theorem itself is simple and seems easy to apply. But you need to be careful how you use it. If one of the above keys is violated, you need to make some adjustments. Here are some variations that you may encounter.

If the variable is in the lower limit instead of the upper limit, the change is easy. Just use this result.
$$\displaystyle{ \int_{a}^{b}{f(t)dt} = -\int_{b}^{a}{f(t)dt} }$$

If the upper limit does not match the derivative variable exactly, use the chain rule as follows.
Given $$\displaystyle{\frac{d}{dx} \left[ \int_{a}^{g(x)}{f(t)dt} \right]}$$
Letting $$u = g(x)$$, the integral becomes $$\displaystyle{\frac{d}{du} \left[ \int_{a}^{u}{f(t)dt} \right] \frac{du}{dx}}$$

Finally, another situation that may arise is when the lower limit is not a constant. One way to handle this is to break the integral into two integrals and use a constant $$a$$ in the two integrals, For example,
$$\displaystyle{\int_{g(x)}^{h(x)}{f(t)dt} = \int_{g(x)}^{a}{f(t)dt} + \int_{a}^{h(x)}{f(t)dt}}$$
Then evaluate each integral separately and combine the result.

Okay, so let's watch a video clip explaining this idea in more detail. Warning: Do not make this any harder than it appears. As this video explains, this is very easy and there is no trick involved as long as you follow the rules given above.

 PatrickJMT - Fundamental Theorem of Calculus Part 1

Even though this appears really easy, it is easy to get tripped up. So make sure you work these practice problems.

 Basic Problems

Practice 6

For $$\displaystyle{g(x)=\int_{1}^{x}{(t^2-1)^{20}~dt}}$$, find $$g'(x)$$

solution

Practice 7

For $$\displaystyle{h(x)=\int_{x}^{2}{[\cos(t^2)+t]~dt}}$$, find $$h'(x)$$

solution

 Intermediate Problems

Practice 8

For $$\displaystyle{g(x)=\int_{1}^{\sqrt{x}}{\frac{s^2}{s^2+1}~ds}}$$, find $$g'(x)$$

solution

Practice 9

For $$\displaystyle{g(x)=\int_{\tan(x)}^{x^2}{\frac{1}{\sqrt{2+t^4}}~dt}}$$, find $$g'(x)$$

solution

U-Substitution and Definite Integrals

Note: This section requires understanding how to do integration by substitution.

When you have a definite integral, you need to do additional work to 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. 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 $$\displaystyle{ \int_{-2}^{1}{ x \sqrt{x+2} ~dx } }$$.

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. 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.

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?
 Basic Problems

Practice 10

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

solution

Practice 11

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

solution

 Intermediate Problems

Practice 12

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

solution

### Trapezoidal and Simpson's Rules

At this time, we do not cover the Trapezoidal Rule and Simpson's Rule on this site. However, here are some videos that might help you if you are studying these topics.

 title: Approximating Integrals: Simpsons Rule
 title: Approximating Integrals: Simpsons Rule Error Bound
 title: Approximating Integrals: Trapezoid Rule

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