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

17Calculus Integrals - Area Under Curves

Limits

Using Limits

Limits FAQs

Derivatives

Graphing

Applications

Derivatives FAQs

Integrals

Trig Integrals

Area/Volume

Applications

Infinite Series

Applications

Tools

Parametrics

Conics

Polar Coordinates

Laplace Transforms

Tools

Calculus Tools

Additional Tools

Articles

Area Under A Curve

To find the area under a curve that is above the x-axis, you just need to integrate the curve between two specific points. This is a natural idea from definite integrals. Here are a couple of videos explaining this.

rootmath - Intro to Area Under a Curve [4min-28secs]

video by rootmath

rootmath - Exact Area Under A Curve [9min-55secs]

video by rootmath

Okay, to really understand this idea, let's work these practice problems.
Unless otherwise instructed, calculate the area under these curves, between the two points, if given.

\(y=x^2 ~~ [0,4]\)

Problem Statement

\(y=x^2 ~~ [0,4]\)

Solution

2114 video

video by Michel vanBiezen

close solution
\(y=x^3 ~~ [1,3]\)

Problem Statement

\(y=x^3 ~~ [1,3]\)

Solution

2115 video

video by Michel vanBiezen

close solution
\( y=1/x^2 ~~ [1,4] \)

Problem Statement

\( y=1/x^2 ~~ [1,4] \)

Solution

2117 video

video by Michel vanBiezen

close solution
\( y=\sin x ~~ [0,\pi] \)

Problem Statement

\( y=\sin x ~~ [0,\pi] \)

Solution

2118 video

video by Michel vanBiezen

close solution
\(y=x^2-2x+8 ~~ [1,2]\)

Problem Statement

\(y=x^2-2x+8 ~~ [1,2]\)

Solution

2116 video

video by Michel vanBiezen

close solution
\( y=\sin x \cos x \) \( [\pi/4, \pi/2] \)

Problem Statement

\( y=\sin x \cos x \) \( [\pi/4, \pi/2] \)

Solution

2119 video

video by Michel vanBiezen

close solution
\(\displaystyle{ y = \frac{x^2}{\sqrt{x^3+9}} ~~ [-1,1] }\)

Problem Statement

\(\displaystyle{ y = \frac{x^2}{\sqrt{x^3+9}} ~~ [-1,1] }\)

Solution

2120 video

video by Michel vanBiezen

close solution

When Area Under The Curve is Zero or Negative

Logically, area should always be positive. However, there will be times when you work the problem correctly and you end up with a negative number. What's the deal?

That's easy. When the area is below the x-axis, the area will turn out to be negative. So it is not that we have negative area. The sign tells us where MOST of the area is, above the x-axis for positive area or below the x-axis for negative area.

It is also possible to get zero for the area. This occurs when the area above the x-axis has exactly the same area as the area below the x-axis.

Here is a good video explaining in more detail why the area is negative. He uses an example and explains this very well.

Section 5.3 - Integrals and Negative Area [5min-21secs]

From this discussion, you now understand why knowing what the graphs look like is so important. Let's work these practice problems. Some of these will have negative or zero area. Make sure you understand why.

Calculate the area under the curve \( y = 60x - 6x^2 \) on \( 0 \leq x \leq 15 \).

Problem Statement

Calculate the area under the curve \( y = 60x - 6x^2 \) on \( 0 \leq x \leq 15 \).

Final Answer

0

Problem Statement

Calculate the area under the curve \( y = 60x - 6x^2 \) on \( 0 \leq x \leq 15 \).

Solution

2433 video

video by Michel vanBiezen

Final Answer

0

close solution
Calculate the area under the curve \(y=x^2-6x\) between 1 and 3.

Problem Statement

Calculate the area under the curve \(y=x^2-6x\) between 1 and 3.

Final Answer

\(-46/3\)

Problem Statement

Calculate the area under the curve \(y=x^2-6x\) between 1 and 3.

Solution

2434 video

Final Answer

\(-46/3\)

close solution

You CAN Ace Calculus

Topics You Need To Understand For This Page

Related Topics and Links

related topics on other pages

For discussion of area under a parametric curve, see the parametric calculus page.

For discussion of area under a polar curve, see the polar calculus page.

external links you may find helpful

area under curves youtube playlist

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

Calculus Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations Topics Listed Alphabetically

Precalculus Topics Listed Alphabetically

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.

calculus motivation - music and learning

Get great tutoring at an affordable price with Chegg. Subscribe today and get your 1st 30 minutes Free!

The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free.

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 © 2010-2020 17Calculus, All Rights Reserved     [Privacy Policy]     [Support]     [About]

mathjax.org
Real Time Web Analytics
17Calculus
We use cookies to ensure that we give you the best experience on our website. By using this site, you agree to our Website Privacy Policy.