17Calculus Integrals - (First) Fundamental Theorem of Calculus
17Calculus
This theorem allows us to avoid calculating sums and limits in order to find area. It is actually called The Fundamental Theorem of Calculus but there is a second fundamental theorem, so you may also see this referred to as 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
\[ \int_{a}^{b}{f(x)~dx} = \left. F(x) \right|_{a}^{b} = F(b) - F(a) \]
where \(F' = f\).
So what is this theorem saying? In essence, it says we only have to calculate the indefinite integral and plug in the endpoints, subtract them in the right order and the value we get is the area under a curve.
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 (1) [6min-21secs]
video by MIP4U |
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MIP4U - Proof of the Fundamental Theorem of Calculus (2) [6min-26secs]
video by MIP4U |
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Notation
Notice in the equation above that, after we evaluate the integral, we use a vertical bar in order to 'hang' the limits of integration before we do the substitution. This notation is very common. You will also see some instructors use a right bracket instead of a vertical bar. We prefer the vertical bar unless we also use a left bracket to enclose the \(F(x)\) expression. So on this site, you will see either \(\displaystyle{ \left. F(x) \right|_{a}^{b} }\) or \(\displaystyle{ \left[ F(x) \right]_{a}^{b} }\).
Another way to write this is to explicitly write the variable that the limits of integration will be substituted into, like this \(\displaystyle{ \left. F(x) \right|_{x=a}^{x=b} }\). This is a very good way to do write it but is not very common. If your instructor allows it, we recommend it since, when you get to multi-variable calculus you will be required to write it this way. However, as usual check with your instructor to see what they require.
Before we go on, let's work some basic integrals using this theorem. After that, your next logical step is to learn about the Second Fundamental Theorem of Calculus.
Practice
Unless otherwise instructed, evaluate these integrals using the Fundamental Theorem of Calculus.
Basic
\(\displaystyle{ \int_2^8{ 5~dx } }\)
Problem Statement |
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\(\displaystyle{ \int_2^8{ 5~dx } }\)
Final Answer |
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\( 24 \)
Problem Statement
\(\displaystyle{ \int_2^8{ 5~dx } }\)
Solution
The Organic Chemistry Tutor - 3790 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( 24 \)
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\(\displaystyle{ \int_4^{10}{ 7 ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int_4^{10}{ 7 ~ dx } }\)
Final Answer |
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\( 42 \)
Problem Statement
\(\displaystyle{ \int_4^{10}{ 7 ~ dx } }\)
Solution
The Organic Chemistry Tutor - 3795 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( 42 \)
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\(\displaystyle{\int_{1}^{3}{(x+2)~dx}}\)
Problem Statement
\(\displaystyle{\int_{1}^{3}{(x+2)~dx}}\)
Solution
PatrickJMT - 888 video solution
video by PatrickJMT |
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\(\displaystyle{ \int_1^4{ 5x-4 ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int_1^4{ 5x-4 ~ dx } }\)
Final Answer |
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\( 51/2 \)
Problem Statement
\(\displaystyle{ \int_1^4{ 5x-4 ~ dx } }\)
Solution
The Organic Chemistry Tutor - 3791 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( 51/2 \)
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\(\displaystyle{ \int_{1}^e{ \frac{5}{x} ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{1}^e{ \frac{5}{x} ~ dx } }\)
Final Answer |
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\( 5 \)
Problem Statement
\(\displaystyle{ \int_{1}^e{ \frac{5}{x} ~ dx } }\)
Solution
The Organic Chemistry Tutor - 3792 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( 5 \)
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\(\displaystyle{ \int_{1}^{e}{ \frac{1}{x} ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{1}^{e}{ \frac{1}{x} ~ dx } }\)
Final Answer |
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\( 1 \)
Problem Statement
\(\displaystyle{ \int_{1}^{e}{ \frac{1}{x} ~ dx } }\)
Solution
The Organic Chemistry Tutor - 3799 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( 1 \)
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\(\displaystyle{ \int_4^9{ \frac{1}{\sqrt{x}} ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int_4^9{ \frac{1}{\sqrt{x}} ~ dx } }\)
Final Answer |
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\( 2 \)
Problem Statement
\(\displaystyle{ \int_4^9{ \frac{1}{\sqrt{x}} ~ dx } }\)
Solution
The Organic Chemistry Tutor - 3793 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( 2 \)
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\(\displaystyle{ \int_2^4{ x^3 ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int_2^4{ x^3 ~ dx } }\)
Final Answer |
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\( 60 \)
Problem Statement
\(\displaystyle{ \int_2^4{ x^3 ~ dx } }\)
Solution
The Organic Chemistry Tutor - 3794 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( 60 \)
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\(\displaystyle{ \int_1^2{ 3x^2 - 5x + 2 ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int_1^2{ 3x^2 - 5x + 2 ~ dx } }\)
Final Answer |
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\( 3/2 \)
Problem Statement
\(\displaystyle{ \int_1^2{ 3x^2 - 5x + 2 ~ dx } }\)
Solution
The Organic Chemistry Tutor - 3796 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( 3/2 \)
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\(\displaystyle{ \int_{-1}^{3}{ (2x+3)^2 ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{-1}^{3}{ (2x+3)^2 ~ dx } }\)
Final Answer |
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\( 364/3 \)
Problem Statement
\(\displaystyle{ \int_{-1}^{3}{ (2x+3)^2 ~ dx } }\)
Solution
The Organic Chemistry Tutor - 3797 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( 364/3 \)
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\(\displaystyle{ \int_{1/2}^{1}{ \frac{1}{x^2} ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{1/2}^{1}{ \frac{1}{x^2} ~ dx } }\)
Final Answer |
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\( 1 \)
Problem Statement
\(\displaystyle{ \int_{1/2}^{1}{ \frac{1}{x^2} ~ dx } }\)
Solution
The Organic Chemistry Tutor - 3798 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( 1 \)
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\(\displaystyle{ \int_{4}^{9}{ \sqrt{x} ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{4}^{9}{ \sqrt{x} ~ dx } }\)
Final Answer |
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\( 38/3 \)
Problem Statement
\(\displaystyle{ \int_{4}^{9}{ \sqrt{x} ~ dx } }\)
Solution
The Organic Chemistry Tutor - 3800 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( 38/3 \)
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\(\displaystyle{ \int_{\pi/6}^{\pi/3}{ \cos(x) ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{\pi/6}^{\pi/3}{ \cos(x) ~ dx } }\)
Final Answer |
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\( (\sqrt{3}-1)/2 \)
Problem Statement
\(\displaystyle{ \int_{\pi/6}^{\pi/3}{ \cos(x) ~ dx } }\)
Solution
The Organic Chemistry Tutor - 3802 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( (\sqrt{3}-1)/2 \)
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Intermediate
\(\displaystyle{ \int_{2}^{3}{ \frac{x^3-5x^2}{x} ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{2}^{3}{ \frac{x^3-5x^2}{x} ~ dx } }\)
Final Answer |
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\( -37/6 \)
Problem Statement
\(\displaystyle{ \int_{2}^{3}{ \frac{x^3-5x^2}{x} ~ dx } }\)
Solution
The Organic Chemistry Tutor - 3801 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( -37/6 \)
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\(\displaystyle{ \int_{0}^{1}{ x^2(x^3+5)^2 ~ dx } }\)
Problem Statement |
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\(\displaystyle{ \int_{0}^{1}{ x^2(x^3+5)^2 ~ dx } }\)
Hint |
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If you already know integration by substitution, that is the easy way to do this problem. However, if you have not covered integration by substitution yet, just multiply out the squared factor and then multiply through by \(x^2\). The answer will be the same in both cases.
Problem Statement |
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\(\displaystyle{ \int_{0}^{1}{ x^2(x^3+5)^2 ~ dx } }\)
Final Answer |
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\( 91/9 \)
Problem Statement
\(\displaystyle{ \int_{0}^{1}{ x^2(x^3+5)^2 ~ dx } }\)
Hint
If you already know integration by substitution, that is the easy way to do this problem. However, if you have not covered integration by substitution yet, just multiply out the squared factor and then multiply through by \(x^2\). The answer will be the same in both cases.
Solution
The Organic Chemistry Tutor - 3803 video solution
video by The Organic Chemistry Tutor |
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Final Answer
\( 91/9 \)
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Practice Instructions
Unless otherwise instructed, evaluate these integrals using the Fundamental Theorem of Calculus.
