Area between Curves

0:00 / 0:00

Area Between Curves
The area between a curve f(x)f\left(x\right)f(x) and the x-axis on the interval [a,b]\left[a,b\right][a,b] can be represented by 
A=∫abf(x)dx  or  lim⁡n→∞∑i=1nf(xi ∗)Δx\displaystyle A=\int_a^bf\left(x\right)dx\ \ or\ \ \lim_{n\to\infty}\sum_{i=1}^nf\left(x_i^{\ \ast}\right)\Delta xA=∫ab​f(x)dx  or  n→∞lim​i=1∑n​f(xi ∗​)Δx

Steps for finding area between 2 curves
1. Sketch the graph
2. Rearrange the curve equations
If one curve is consistantly the "top curve" and the other curve is consistantly the "bottom curve": we need to set up the equations as y=...y=...y=...
If one curve is consistantly the "right curve" and the other curve is consistantly the "left curve": we need to set up the equations as x=...x=...x=...
3. Find the bounds of your integral
This could be given by the question
Otherwise set the two curve equations equal one another and solve for the point(s) of intersection
Wize Tip
If you end up with more than 2 points of intersection, that means that we have multiple bounded regions  →  need to set up an integral for each region.

4. Set up the integral for each region
If the curve equations are y=...y=...y=...: A=∫x=ax=b(top curve−bottom curve)dx\displaystyle A=\int_{x=a}^{x=b}\left(\text{top curve}-\text{bottom curve}\right)dxA=∫x=ax=b​(top curve−bottom curve)dx
If the curve equations are x=...x=...x=...: A=∫y=cy=d(right curve−left curve)dy\displaystyle A=\int_{y=c}^{y=d}\left(\text{right curve}-\text{left curve}\right)dyA=∫y=cy=d​(right curve−left curve)dy

Find the area enclosed by the curves y=x−4,  y=1,  x=13y=\sqrt{x-4},\ \ y=1,\ \ x=13y=x−4​,  y=1,  x=13.

Extra Practice:
Try doing this using 2 different methods (one with the equations set up as y=...y=...y=... and the other with the equations set up as x=...x=...x=...)

Practice: Area between Curves
Find the integral that represents the area of the region bounded by the curevs x=y2−4y+2x=y^2-4y+2x=y2−4y+2 and 5−x=y2−y5-x=y^2-y5−x=y2−y.

A=∫−1/23[−2y2+5y+3]dy\displaystyle A=\int_{-1/2}^3[-2y^2+5y+3]dyA=∫−1/23​[−2y2+5y+3]dy

A=∫25[−2y2+5y+3]dy\displaystyle A=\int_{2}^5[-2y^2+5y+3]dyA=∫25​[−2y2+5y+3]dy

A=∫−52[2y2−5y−3]dy\displaystyle A=\int_{-5}^2[2y^2-5y-3]dyA=∫−52​[2y2−5y−3]dy

A=∫05[2y2−5y+2]dy\displaystyle A=\int_{0}^5[2y^2-5y+2]dyA=∫05​[2y2−5y+2]dy

A=∫01/2[−2y2+5y+2]dy\displaystyle A=\int_{0}^{1/2}[-2y^2+5y+2]dyA=∫01/2​[−2y2+5y+2]dy

I don't know

Wize University Calculus 2 Textbook > Applications of Integrals

Popular Courses

Area Between Curves

Practice: Area between Curves

Extra Practice

Area Between Curves

Area Between Curves

Area Between Curves

Area Between Curves

Area Between Curves