Area Between Curves

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 6.1 – Area Between Curves

Expected Educational Results

• Objective 10–1: I can set up the integral to compute the area of the region bounded by two or more curves.
• Objective 10–2: I can evaluate the integral to compute the area of the region bounded by two or more curves using FTC-II.
• Objective 10–3: I can use technology to estimate the area of the region bounded by two or more curves using FTC-II.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Area Between Curves

Definitions

Area Between Two Curves – 1:

The area $A$$A$ of the region bounded by the curves $y=f(x)$$y=f(x)$ and $y=g(x)$$y=g(x)$ where $f$$f$ and $g$$g$ are continuous and $f(x)\ge g(x)$$f(x)\geq g(x)$ for all $x$$x$ in $[a,b]$$[a,b]$ is $A={\int }_a^b(f(x)-g(x))dx$$A=\int_a^b{\Big(f(x)-g(x)\Big)\,dx}$

Area Between Two Curves – 2

The area $A$$A$ of the region bounded by the curves $y=f(x)$$y=f(x)$ and $y=g(x)$$y=g(x)$ where $f$$f$ and $g$$g$ are continuous for all $x$$x$ in $[a,b]$$[a,b]$ but $f(x)\ngeqq g(x)$$f(x)\ngeq g(x)$ or $g(x)\ngeqq f(x)$$g(x)\ngeq f(x)$ for all $x$$x$ on $[a,b]$$[a,b]$ is $A={\int }_a^b|f(x)-g(x)|dx$$A=\int_a^b{\Big|f(x)-g(x)\Big|\,dx}$

Area Between Two Curves – 3

The area $A$$A$ of the region bounded by the curves $x=f(y)$$x=f(y)$ and $x=g(y)$$x=g(y)$ where $f$$f$ and $g$$g$ are continuous and $f(y)\ge g(y)$$f(y)\geq g(y)$ for all $c\le y\le d$$c\leq y\leq d$ is $A={\int }_c^d(f(y)-g(y))dy$$A=\int_c^d{\Big(f(y)-g(y)\Big)\,dy}$

Area Between Two Curves – 4

The area $A$$A$ of the region bounded by the curves $x=f(y)$$x=f(y)$ and $x=g(y)$$x=g(y)$ where $f$$f$ and $g$$g$ are continuous for all $c\le y\le d$$c\leq y\leq d$ but and $f(y)\ngeqq g(y)$$f(y)\ngeq g(y)$ or $g(y)\ngeqq f(y)$$g(y)\ngeq f(y)$ for all $c\le y\le d$$c\leq y\leq d$ is $A={\int }_c^d|f(y)-g(y)|dy$$A=\int_c^d{\Big|f(y)-g(y)\Big|\,dy}$

Procedure

1. Sketch the curves

2. Identify the region enclosed by the curves

3. Sketch one (1) typical rectangle within the region where every potential rectangle starts and ends at different curves. There are two potential rectangles to draw:

   1. Vertical rectangle [the integral will be in terms of the $x$$x$-variable only]
   2. Horizontal rectangle [the integral will be in terms of the $y$$y$-variable only]

4. Find intersection(s) of curves [only the $x$$x$- or $y$$y$-coordinate of the intersections is needed, depending on the typical rectangle]

5. Set up integral in one of the following ways:

   1. ${\displaystyle {\int }_a^b(f(x)-g(x))dx}$$\displaystyle\int_a^b{\Big(f(x)-g(x)\Big)}dx$, where $f(x)\ge g(x)$$f(x)\geq g(x)$, $x=a$$x=a$ is the smaller $x$$x$-coordinate of the intersection or smaller given value, and $x=b$$x=b$ is the larger $x$$x$-coordinate of the intersection or larger given value;
   2. ${\displaystyle {\int }_a^b(f(y)-g(y))dy}$$\displaystyle\int_a^b{\Big(f(y)-g(y)\Big)}dy$, where $f(y)\ge g(y)$$f(y)\geq g(y)$ [i.e., $f(y)$$f(y)$ is to the right of $g(y)$$g(y)$], $y=a$$y=a$ is the smaller $y$$y$-coordinate of the intersection or smaller given value, and $y=b$$y=b$ is the larger $y$$y$-coordinate of the intersection or larger given value;

6. If there are more than two intersections, then you will need to set up more than one integral and add the areas.

7. Evaluate the integral.

Question 04

For each of the following, use the embedded DESMOS graph to graph both functions and the $a$$a$- and $b$$b$-values. Identify the enclosed region, set-up the integral to compute the area, then evaluate the integral.

1. Find the area between $y=\sin (x)$$y=\sin{(x)}$ and $y=x^2+2$$y=x^2+2$ from $a=0$$a=0$ to $b=\frac{\pi }{2}$$b=\frac{\pi}{2}$.
2. Find the area between $y=2\cos (x)$$y=2\cos{(x)}$ and $y=-2{\sec }^2(x)$$y=-2\sec^2{(x)}$ from $a=0$$a=0$ to $b=\frac{\pi }{4}$$b=\frac{\pi}{4}$.
3. Find the area between $y=e^{2x}$$y=e^{2x}$ and $y=3-2x$$y=3-2x$ from $a=-1$$a=-1$ to $b=0$$b=0$.

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Last Modified: Thursday, 17 September 2020 2:48 EDT