AP®MathsCollege BoardCalculus BCStudy GuidesUnit 8: Applications of IntegrationAreas & Arc LengthsArea Between a Curve & x-Axis

Area Between a Curve & x-Axis (College Board AP® Calculus BC): Study Guide

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on 18 May 2026

Area between a curve & x-axis

How do I find an area between a curve and the x-axis?

The value found when calculating the definite integral of a function $y = f (x)$ with respect to $x$ between $x = a$ and $x = b$ , $\int_{a}^{b} f (x) d x$
- as long as $f (x) \geq 0$ on the interval $[a, b]$
- is equal to the area between the curve and the $x$ -axis between $x = a$ and $x = b$

Graph of y = f(x) showing the area under the curve between x = a and x = b, shaded in purple and labeled R; integrals are used to find the area.

Consider finding the area between the graph of $y = 5 + 2 x - x^{2}$ and the $x$ -axis, between $x = 1$ and $x = 3$

Graph of the function y = 5 + 2x - x^2 with a shaded region between x = 1 and x = 3, area calculated as 28/3 square units using integration.

This method of finding areas uses the idea of a definite integral as calculating an accumulation of change
- $f (x) \cdot ∆ x$ is the area of a rectangle with height $f (x)$ and width $∆ x$
- $f (x) d x$ is the limit of this area element as $∆ x \to 0$
- The integral $\int_{a}^{b} f (x) d x$ sums up all these infinitesimal area elements between $x = a$ and $x = b$

What if I am not told the limits of integration?

If limits are not provided they will often be the $x$ -axis intercepts
- Set $y = 0$ and solve the equation to find the $x$ -axis intercepts first

Graph showing the shaded area R under the curve y = x(5 - x).
Finding the x-intercepts first, which are 0 and 5.
Calculation shows area of R = 125/6 square units.

Remember that the $y$ -axis (i.e. $x = 0$ ) may also be one of the limits

Worked Example

The area bounded by the curve with equation $y = 9 - x^{2}$ , the $x$ -axis and the vertical lines with equations $x = 1$ and $x = 2$ is shaded below.

Graph of y = 9 - x² with a shaded area between x = 1 and x = 2 and the x axis, bounded by dashed lines.

Find the area of the shaded region.

Answer:

Find the definite integral from $x = 1$ to $x = 2$

$\begin{array}{rcl} \int_{1}^{2} (9 - x^{2}) d x & = & {[9 x - \frac{1}{3} x^{3}]}_{1}^{2} \\ = & (9 \cdot 2 - \frac{1}{3} \cdot 2^{3}) - (9 \cdot 1 - \frac{1}{3} \cdot 1^{3}) \\ = & (18 - \frac{8}{3}) - (9 - \frac{1}{3}) \\ = & \frac{46}{3} - \frac{26}{3} \\ = & \frac{20}{3} \end{array}$

$\frac{20}{3}$ units squared

When is the area integral negative?

If the area lies underneath the $x$ -axis the value of the definite integral will be negative
- However, an area cannot be negative
- The area is equal to the modulus (absolute value) of the definite integral
If the area has some parts which are above the $x$ -axis, and some which are below the $x$ -axis
- then see the method outlined in the 'Multiple Areas' study guide

Examiner Tips and Tricks

Always check whether you need to find the value of an integral, or an area.

When areas below the $x$ -axis are involved, these will be two different values.

Graph of y = x^2 - 6x + 5 with shaded region R between x = 2 and x = 4. Integral calculation shows integral is -22/3, so the area is +22/3

Examiner Tips and Tricks

In an FRQ, make sure you clearly state the integrand and the limits of integration.

Worked Example

The finite region $R$ , shown in the figure below, is bounded by the curve with equation $y = x^{2} - 4 x + 3$ and the $x$ -axis.

Find the exact area of $R$ .

Graph of y = x² − 4x + 3 with a shaded region R under the curve between x = 1 and x = 3 below the x-axis on standard coordinate axes

Answer:

Find the definite integral from $x = 1$ to $x = 3$

$\begin{array}{rcl} \int_{1}^{3} (x^{2} - 4 x + 3) d x & = & {[\frac{1}{3} x^{3} - 2 x^{2} + 3 x]}_{1}^{3} \\ = & (\frac{1}{3} \cdot 3^{3} - 2 \cdot 3^{2} + 3 \cdot 3) - (\frac{1}{3} \cdot 1^{3} - 2 \cdot 1^{2} + 3 \cdot 1) \\ = & (9 - 18 + 9) - (\frac{1}{3} - 2 + 3) \\ = & 0 - \frac{4}{3} \\ = & - \frac{4}{3} \end{array}$

Take the absolute value to find the area

$\frac{4}{3}$ units squared

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Area Between a Curve & x-Axis (College Board AP® Calculus BC): Study Guide

Area between a curve & x-axis

How do I find an area between a curve and the x-axis?

What if I am not told the limits of integration?

Worked Example

When is the area integral negative?

Examiner Tips and Tricks

Examiner Tips and Tricks

Worked Example

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

Author: Jamie Wood

Reviewer: Dan Finlay