17Calculus Integrals - Second Fundamental Theorem of Calculus
17Calculus
The Second Fundamental Theorem of Calculus |
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For a continuous function \(f\) on an open interval \(I\) containing the point \( a\), then the following equation holds for each point in \(I\) \[f(x) = \frac{d}{dx} \left[ \int_{a}^{x}{f(t)~dt} \right]\] |
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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 [9min-25secs]
video by PatrickJMT |
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Even though this appears really easy, it is easy to get tripped up. So make sure you work the practice problems.
Practice
Basic |
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For \(\displaystyle{g(x)=\int_{1}^{x}{(t^2-1)^{20}~dt}}\), find \(g'(x)\).
Problem Statement
For \(\displaystyle{g(x)=\int_{1}^{x}{(t^2-1)^{20}~dt}}\), find \(g'(x)\).
Solution
PatrickJMT - 883 video solution
video by PatrickJMT |
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For \(\displaystyle{h(x)=\int_{x}^{2}{[\cos(t^2)+t]~dt}}\), find \(h'(x)\).
Problem Statement
For \(\displaystyle{h(x)=\int_{x}^{2}{[\cos(t^2)+t]~dt}}\), find \(h'(x)\).
Solution
PatrickJMT - 884 video solution
video by PatrickJMT |
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Intermediate |
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For \(\displaystyle{g(x)=\int_{1}^{\sqrt{x}}{\frac{s^2}{s^2+1}~ds}}\), find \(g'(x)\).
Problem Statement
For \(\displaystyle{g(x)=\int_{1}^{\sqrt{x}}{\frac{s^2}{s^2+1}~ds}}\), find \(g'(x)\).
Solution
PatrickJMT - 885 video solution
video by PatrickJMT |
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For \(\displaystyle{g(x)=\int_{\tan(x)}^{x^2}{\frac{1}{\sqrt{2+t^4}}~dt}}\), find \(g'(x)\).
Problem Statement
For \(\displaystyle{g(x)=\int_{\tan(x)}^{x^2}{\frac{1}{\sqrt{2+t^4}}~dt}}\), find \(g'(x)\).
Solution
PatrickJMT - 886 video solution
video by PatrickJMT |
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\(\displaystyle{ \int_{-3}^{5}{ \sqrt{\abs{x}^3} ~dx } }\)
Problem Statement |
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Evaluate \(\displaystyle{ \int_{-3}^{5}{ \sqrt{\abs{x}^3} ~dx } }\)
Final Answer |
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\(\displaystyle{ \int_{-3}^{5}{ \sqrt{\abs{x}^3} ~dx } }\) \( = \dfrac{18}{5}\sqrt{3} + 10\sqrt{5}\)
Problem Statement
Evaluate \(\displaystyle{ \int_{-3}^{5}{ \sqrt{\abs{x}^3} ~dx } }\)
Solution
Since we have an absolute value, we need to handle the negative \(x\)-values separate from the positive. So we will break the integral at \(x=0\) into two integrals. |
\(\displaystyle{ \int_{-3}^{5}{ \sqrt{\abs{x}^3} ~dx } }\) |
\(\displaystyle{ \int_{-3}^{0}{ \sqrt{(-x)^3} ~dx } }\) \(\displaystyle{ + \int_{0}^{5}{ \sqrt{x^3} ~dx } }\) |
Let's work with the first integral. Let \(u=-x \to du = -dx\) |
\(x=-3 \to u = 3\) and \( x=0 \to u=0\) |
\(\displaystyle{ \int_{-3}^{0}{ \sqrt{(-x)^3} ~dx } }\) |
\(\displaystyle{ \int_{3}^{0}{ -\sqrt{u^3} ~du } }\) |
Swap the limits of integration and absorb the negative sign. |
\(\displaystyle{ \int_{0}^{3}{ \sqrt{u^3} ~du } }\) |
\(\displaystyle{ \int_{0}^{3}{ u^{3/2} ~du } }\) |
Integrating yields \( \dfrac{3^{5/2}}{5/2} = \dfrac{2}{5}(9\sqrt{3}) = \dfrac{18}{5}\sqrt{3} \) |
We can evaluate the second integral directly. |
\(\displaystyle{ \int_{0}^{5}{ \sqrt{x^3} ~dx } }\) |
\(\displaystyle{ \int_{0}^{5}{ x^{3/2} ~dx } }\) |
\(\displaystyle{ \left[ \frac{x^{5/2}}{5/2} \right]_0^5 }\) |
\(\displaystyle{ \frac{2}{5}5^{5/2} = \frac{2}{5}25\sqrt{5} = 10\sqrt{5}}\) |
Combining the solutions gives us our final answer. |
Final Answer
\(\displaystyle{ \int_{-3}^{5}{ \sqrt{\abs{x}^3} ~dx } }\) \( = \dfrac{18}{5}\sqrt{3} + 10\sqrt{5}\)
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Advanced |
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Evaluate \(\displaystyle{\int_0^1{ \frac{t^7-1}{\ln t}~dt }}\)
Problem Statement |
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Evaluate \(\displaystyle{\int_0^1{ \frac{t^7-1}{\ln t}~dt }}\)
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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Evaluate \(\displaystyle{\int_0^1{ \frac{t^7-1}{\ln t}~dt }}\)
Final Answer |
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\(3\ln 2\)
Problem Statement
Evaluate \(\displaystyle{\int_0^1{ \frac{t^7-1}{\ln t}~dt }}\)
Hint
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
World Wide Center of Mathematics - 2215 video solution
video by World Wide Center of Mathematics |
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Final Answer
\(3\ln 2\)
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