\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\units}[1]{\,\text{#1}} \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Integrals - Second Fundamental Theorem of Calculus

17Calculus
Single Variable Calculus
Derivatives
Integrals
Multi-Variable Calculus
Precalculus
Functions

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 [9min-25secs]

video by PatrickJMT

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

Math Word Problems Demystified

Practice

Basic

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

Log in to rate this practice problem and to see it's current rating.

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

Log in to rate this practice problem and to see it's current rating.

Intermediate

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

Log in to rate this practice problem and to see it's current rating.

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

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{ \int_{-3}^{5}{ \sqrt{\abs{x}^3} ~dx } }\)

Problem Statement

Evaluate \(\displaystyle{ \int_{-3}^{5}{ \sqrt{\abs{x}^3} ~dx } }\)

Final Answer

\(\displaystyle{ \int_{-3}^{5}{ \sqrt{\abs{x}^3} ~dx } }\) \( = \dfrac{18}{5}\sqrt{3} + 10\sqrt{5}\)

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

\(\displaystyle{ \int_{-3}^{5}{ \sqrt{\abs{x}^3} ~dx } }\) \( = \dfrac{18}{5}\sqrt{3} + 10\sqrt{5}\)

Log in to rate this practice problem and to see it's current rating.

Advanced

Evaluate \(\displaystyle{\int_0^1{ \frac{t^7-1}{\ln t}~dt }}\)

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.

Problem Statement

Evaluate \(\displaystyle{\int_0^1{ \frac{t^7-1}{\ln t}~dt }}\)

Final Answer

\(3\ln 2\)

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

Final Answer

\(3\ln 2\)

Log in to rate this practice problem and to see it's current rating.

Really UNDERSTAND Calculus

Log in to rate this page and to see it's current rating.

Topics You Need To Understand For This Page

Related Topics and Links

external links you may find helpful

To bookmark this page and practice problems, log in to your account or set up a free account.

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

effective study techniques

Shop Amazon - Rent Textbooks - Save up to 80%

As an Amazon Associate I earn from qualifying purchases.

I recently started a Patreon account to help defray the expenses associated with this site. To keep this site free, please consider supporting me.

Support 17Calculus on Patreon

Practice Search

Do NOT follow this link or you will be banned from the site!

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

Links and banners on this page are affiliate links. We carefully choose only the affiliates that we think will help you learn. Clicking on them and making purchases help you support 17Calculus at no extra charge to you. However, only you can decide what will actually help you learn. So think carefully about what you need and purchase only what you think will help you.

We use cookies on this site to enhance your learning experience.

17calculus

Copyright © 2010-2022 17Calculus, All Rights Reserved     [Privacy Policy]     [Support]     [About]

mathjax.org
Real Time Web Analytics