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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 skip those practice problems and come back later, once you have covered integration by parts. |
external links you may find helpful |
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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.) |
Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |
Arc Length (Vector Functions) |
Arc Length Function |
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Cross Product |
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Cylindrical Coordinates |
Lagrange Multipliers |
Line Integrals |
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Partial Integrals |
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Principal Unit Normal Vector |
Differential Equations |
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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.
If you want a complete lecture on this topic, we recommend this video.
video by Prof Leonard
Summation Notation |
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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.
video by PatrickJMT
Summation and the Definite Integral |
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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.
video by PatrickJMT
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.
video by PatrickJMT
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.
video by PatrickJMT
video by PatrickJMT
Before we go on, let's work some practice problems to make sure we understand these ideas.
Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems |
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Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new. |
Definite Integrals - Practice Problems Conversion |
[1-879] - [2-880] - [3-881] - [4-882] - [5-888] - [6-883] - [7-884] - [8-885] - [9-886] |
[10-887] - [11-1746] - [12-889] |
Please update your notes to this new numbering system. The display of this conversion information is temporary. |
\(\displaystyle{\sum_{k=1}^{6}{\frac{1}{k^2}}}\)
Problem Statement |
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\(\displaystyle{\sum_{k=1}^{6}{\frac{1}{k^2}}}\)
Solution |
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video by Krista King Math
close solution |
Write\(\displaystyle{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+}\) \(\displaystyle{\frac{1}{16}+\frac{1}{32}+\frac{1}{64}}\) in summation notation.
Problem Statement |
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Write\(\displaystyle{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+}\) \(\displaystyle{\frac{1}{16}+\frac{1}{32}+\frac{1}{64}}\) in summation notation.
Solution |
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video by Krista King Math
close solution |
\(\displaystyle{\sum_{r=1}^{8}{(r-1)(r+2)}}\)
Problem Statement |
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\(\displaystyle{\sum_{r=1}^{8}{(r-1)(r+2)}}\)
Solution |
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video by Krista King Math
close solution |
\(\displaystyle{\sum_{i=1}^{4}{(-1)^i i^2}}\)
Problem Statement |
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\(\displaystyle{\sum_{i=1}^{4}{(-1)^i i^2}}\)
Solution |
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video by PatrickJMT
close solution |
The (First) Fundamental Theorem of Calculus |
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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.
video by MIP4U
video by MIP4U
Before we go on, let's work some basic integrals using this theorem.
\(\displaystyle{\int_{1}^{3}{(x+2)~dx}}\)
Problem Statement |
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\(\displaystyle{\int_{1}^{3}{(x+2)~dx}}\)
Solution |
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video by PatrickJMT
close solution |
The Second Fundamental Theorem of Calculus |
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The Second Fundamental Theorem of Calculus |
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For a continuous function \(f\) on an open interval \(I\) containing the point \( a\), |
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.
video by PatrickJMT
Even though this appears really easy, it is easy to get tripped up. So make sure you work these practice problems.
Basic Problems |
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For \(\displaystyle{g(x)=\int_{1}^{x}{(t^2-1)^{20}~dt}}\), find \(g'(x)\).
Problem Statement |
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For \(\displaystyle{g(x)=\int_{1}^{x}{(t^2-1)^{20}~dt}}\), find \(g'(x)\).
Solution |
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video by PatrickJMT
close solution |
For \(\displaystyle{h(x)=\int_{x}^{2}{[\cos(t^2)+t]~dt}}\), find \(h'(x)\).
Problem Statement |
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For \(\displaystyle{h(x)=\int_{x}^{2}{[\cos(t^2)+t]~dt}}\), find \(h'(x)\).
Solution |
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video by PatrickJMT
close solution |
Intermediate Problems |
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For \(\displaystyle{g(x)=\int_{1}^{\sqrt{x}}{\frac{s^2}{s^2+1}~ds}}\), find \(g'(x)\).
Problem Statement |
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For \(\displaystyle{g(x)=\int_{1}^{\sqrt{x}}{\frac{s^2}{s^2+1}~ds}}\), find \(g'(x)\).
Solution |
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video by PatrickJMT
close solution |
For \(\displaystyle{g(x)=\int_{\tan(x)}^{x^2}{\frac{1}{\sqrt{2+t^4}}~dt}}\), find \(g'(x)\).
Problem Statement |
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For \(\displaystyle{g(x)=\int_{\tan(x)}^{x^2}{\frac{1}{\sqrt{2+t^4}}~dt}}\), find \(g'(x)\).
Solution |
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video by PatrickJMT
close solution |
Advanced Problems |
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\(\displaystyle{\int_0^1{ \frac{t^7-1}{\ln t}~dt }}\)
Problem Statement |
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\(\displaystyle{\int_0^1{ \frac{t^7-1}{\ln t}~dt }}\)
Hint |
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Let \(\displaystyle{g(x) = \int_0^1{ \frac{t^x-1}{\ln t}~dt }}\) and notice that our integral is \(g(7)\). Calculate \(g'(x)\). Integrate the result to get \(g(x)\) and then find \(g(7)\).
Note: This is a very unusual procedure that you will probably not see in your class or textbook.
Problem Statement |
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\(\displaystyle{\int_0^1{ \frac{t^7-1}{\ln t}~dt }}\)
Final Answer |
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\(3\ln 2\) |
Problem Statement |
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\(\displaystyle{\int_0^1{ \frac{t^7-1}{\ln t}~dt }}\)
Hint |
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Let \(\displaystyle{g(x) = \int_0^1{ \frac{t^x-1}{\ln t}~dt }}\) and notice that our integral is \(g(7)\). Calculate \(g'(x)\). Integrate the result to get \(g(x)\) and then find \(g(7)\).
Note: This is a very unusual procedure that you will probably not see in your class or textbook.
Solution |
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video by World Wide Center of Mathematics
Final Answer |
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\(3\ln 2\) |
close solution |
Trapezoidal and Simpson's Rules |
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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 if you are studying these topics.
video by PatrickJMT
video by PatrickJMT
video by PatrickJMT