17Calculus - Integrals and Integration
The idea of integration is that integration 'undoes' the derivative, i.e. integration is the 'inverse' of differentiation. I used quotes around the word 'inverse' because in actuality differentiation and integration are mostly inverses but not completely. Let's start with what an integral is and some notation first.
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What Is An Integral?
Before we get to the details of integration, let's watch a quick video on exactly what integrals are. The material in this video is covered on several 17calculus pages but the video gives you a good overview of integration, what it means and some of the notation.
Physical Chemistry - What is an integral? [7min-21secs]
video by Physical Chemistry |
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As you know, if you have a function, \(g(x)\), the derivative of \(g(x)\) is written
\(\displaystyle{
\frac{d}{dx}[g(x)] = G(x)
}\)
In this equation, we are using \(G(x)\) to represent the new function we get after taking the derivative. When integrating, the notation looks like this
\(\int{F(x)~dx} = f(x) + C \)
In this equation, \(F(x)\) is the function we are integrating and \(f(x)+C\) is the result. The curved vertical line \(\int{}\) and the \(dx\) are both necessary in this notation. They act like brackets to indicate what is being integrated, which we call the integrand.
Okay, so what are integrals and how do we go about calculating them? Here is a great introduction video to integrals, what they represent and the notation. This is one of the best introductory videos you will find anywhere on any topic. So your time will be well spent watching this.
Krista King Math - Integrals - Calculus [ excellent video ] [5min-55secs]
video by Krista King Math |
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You will notice in the equation \(f(x)+C\), we have \(+C\) at the end. We need this because the derivative of a constant is zero. When we go in reverse, it is not possible to recover constants without more information. So, we use the \(+C\) as a placeholder for an unknown constant. Here is a video that explains this very well. If you are already confident with integrals, you can easily skip this video without losing anything. But, if you need a bit more explanation with some examples, this video may help.
Khan Academy - The Indefinite Integral or Anti-derivative [9min-28secs]
In this video he starts with the derivative \( \frac{d}{dx}[x^2] = 2x \).
Then explains how to work this backwards, i.e. given \(2x\) how do you get \(x^2\). Then he shows that the derivative of \( x^2+1\) is also \(2x\) and the derivative of \( x^2+2 \) is also \(2x\). So that the derivative of \( x^2 + any~constant\) is \( 2x\), usually written \(x^2+C\) where \(C\) just represents a constant.
Then he looks at the derivative of \(y=Ax^n\) which is \( (A \cdot n) x^{n-1} \) to try to figure out what \( y=\int{x^3 ~ dx} \) is.
This is a neat way of learning integration. He then generalizes the power rule for integration from this example.
Another example: \( \int{5x^7 dx} \) He is correct about not forgetting the \(+C\). Many instructors will take off points if you leave it off.
To sum up, although his explanation is a little choppy, this is a pretty good introduction video to integration.
video by Khan Academy |
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Some Basic Formulas
Here are a few basic formulas that you will need for upcoming pages. You already know the derivatives of these, so the anti-derivatives should contain no surprises.
\(\displaystyle{ \int{ k~dx } = kx + C }\) |
\(\displaystyle{ \int{ x^r~dx } = \frac{x^{r+1}}{r+1} + C }\) | |
\(\displaystyle{ \int{ e^x~dx } = e^x + C }\) |
\(\displaystyle{ \int{ \frac{1}{x} dx } = \ln(x) + C }\) | |
\(\displaystyle{ \int{ \sin(x) ~dx} = -\cos(x) + C }\) |
\(\displaystyle{ \int{ \cos(x) ~dx } = \sin(x) + C }\) |
Notes
1. In the above list, \(k\) is a constant and \(r\) is a rational number.
2. For the natural logarithm, be careful that you understand that \( \int{ \ln(x)~ dx } \neq 1/x + C \). This is a common mistake when students are first learning integration.
3. Watch the negative on the integral for sine.
4. There are comparable anti-derivatives for the other trig functions as well. You can find those details on the trig integration page.
Okay, time for some practice problems.
Practice
Unless otherwise instructed, evaluate these integrals, giving your answers in exact form.
Basic |
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\(\displaystyle{\int{3x^2+2x+1~dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{3x^2+2x+1~dx}}\), giving your answer in exact form.
Solution |
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996 video
video by Krista King Math |
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close solution
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\(\displaystyle{\int{3x^4+5x-6~dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{3x^4+5x-6~dx}}\), giving your answer in exact form.
Solution |
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997 video
video by Krista King Math |
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close solution
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\(\displaystyle{\int{1-2x^2+3x^3~dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{1-2x^2+3x^3~dx}}\), giving your answer in exact form.
Solution |
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998 video
video by Krista King Math |
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\(\displaystyle{\int{\frac{-1}{x^2}dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{\frac{-1}{x^2}dx}}\), giving your answer in exact form.
Solution |
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999 video
video by Krista King Math |
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\(\displaystyle{\int{\frac{3}{x^3}+2x^{3/2}-1~dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{\frac{3}{x^3}+2x^{3/2}-1~dx}}\), giving your answer in exact form.
Solution |
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1000 video
video by Krista King Math |
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\(\displaystyle{\int{x^{5/2}-\frac{5}{x^4}-\sqrt{x}~dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{x^{5/2}-\frac{5}{x^4}-\sqrt{x}~dx}}\), giving your answer in exact form.
Solution |
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1001 video
video by Krista King Math |
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\(\displaystyle{\int{\frac{3}{2}x^{1/2}+7~dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{\frac{3}{2}x^{1/2}+7~dx}}\), giving your answer in exact form.
Solution |
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1002 video
video by Krista King Math |
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close solution
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\(\displaystyle{\int{1+2x-4x^3~dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{1+2x-4x^3~dx}}\), giving your answer in exact form.
Solution |
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1005 video
video by PatrickJMT |
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\(\displaystyle{\int{(x+1)(x^2+3)~dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{(x+1)(x^2+3)~dx}}\), giving your answer in exact form.
Solution |
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1007 video
video by PatrickJMT |
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\(\displaystyle{\int{-5e^x+7\sin(x)~dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{-5e^x+7\sin(x)~dx}}\), giving your answer in exact form.
Solution |
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1315 video
video by PatrickJMT |
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\(\displaystyle{\int{2e^x-1+\sin(x)\csc(x)~dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{2e^x-1+\sin(x)\csc(x)~dx}}\), giving your answer in exact form.
Solution |
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1316 video
video by PatrickJMT |
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Intermediate |
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\(\displaystyle{\int{\frac{2}{x^{3/4}}-\frac{3}{x^{2/3}}~dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{\frac{2}{x^{3/4}}-\frac{3}{x^{2/3}}~dx}}\), giving your answer in exact form.
Solution |
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1003 video
video by Krista King Math |
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\(\displaystyle{\int{2x\sqrt{x}-\frac{1}{\sqrt{x}}dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{2x\sqrt{x}-\frac{1}{\sqrt{x}}dx}}\), giving your answer in exact form.
Solution |
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1004 video
video by Krista King Math |
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\(\displaystyle{\int{\frac{3x-2}{\sqrt{x}}dx}}\)
Problem Statement |
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Evaluate \(\displaystyle{\int{\frac{3x-2}{\sqrt{x}}dx}}\), giving your answer in exact form.
Solution |
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1006 video
video by PatrickJMT |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
Related Topics and Links
external links you may find helpful |
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Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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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Practice Instructions
Unless otherwise instructed, evaluate these integrals, giving your answers in exact form.
