More on the Area Problem.


We have seen that the area of the region under the graph of a positive function f (x) on [a, b] is given by the definite integral $\displaystyle \int_{a}^{b}$f (x) dx. The above properties may help extend this result. Indeed, consider the region $ \Omega$ bounded above by the graph of f (x) and below by the graph of g(x), and bounded on the sides by the vertical lines x = a and x = b, with a $ \leq$ b. Of course, we assume that g(x) $ \leq$ f (x) for x $ \in$ [a, b]. Then we have

Area($\displaystyle \Omega$) = $\displaystyle \int_{a}^{b}$$\displaystyle \Big($f (x) - g(x)$\displaystyle \Big)$dx  .

Note that here we do not assume both functions are positive since f (x) - g(x) is a positive function.

Remark. Sometimes you may be asked to find the area of the region bounded by the graph of two or more functions. The idea is to split the region into subregions of the type described above and then use the formula for the subregions.

Example. Let us find the area of the region $ \Omega$ bounded by the graphs of

f (x) = x2    and    g(x) = 8$\displaystyle \sqrt{x}$  .

First let us graph these functions.

Notice that the region is bounded above by g(x) and below by f (x). To find the boundary points which will give the vertical side lines we need to solve

f (x) = g(x)    or      x2 = 8$\displaystyle \sqrt{x}$  .

Easily we get x = 0 and x = 4. So the answer is

Area($\displaystyle \Omega$) = $\displaystyle \int_{0}^{4}$$\displaystyle \Big($8$\displaystyle \sqrt{x}$ - x2$\displaystyle \Big)$dx  .

We have

$\displaystyle \int_{0}^{4}$$\displaystyle \Big($8$\displaystyle \sqrt{x}$ - x2$\displaystyle \Big)$dx = $\displaystyle \left[\vphantom{\frac{16}{3}x^{3/2} - \frac{1}{3} x^3}\right.$$\displaystyle {\textstyle\frac{16}{3}}$x3/2 - $\displaystyle {\textstyle\frac{1}{3}}$x3$\displaystyle \left.\vphantom{\frac{16}{3}x^{3/2} - \frac{1}{3} x^3}\right]_{0}^{4}$ = $\displaystyle {\textstyle\frac{64}{3}}$   .

Example. Find the area of the region $ \Omega$ bounded by the three lines

x + y = 2,    x - y = - 1,    and    x + 2y = 2  .

First we graph the three lines to see the region.

Before we proceed with any integration we need to find the points of intersection of the three lines represented by the functions

f (x) = 2 - x,    g(x) = x + 1,    and    h(x) = - $\displaystyle {\textstyle\frac{1}{2}}$x + 1  .

The three points of intersection are (0,1), (2,0), and (1/2, 3/2).

The region $ \Omega$ may be subdivided into two regions $ \Omega_{1}^{}$ bounded above by the graph of g(x) and below by h(x) with 0 $ \leq$ x $ \leq$ $\displaystyle {\textstyle\frac{1}{2}}$, and $ \Omega_{2}^{}$ bounded above by the graph of f (x) and below by h(x) with $\displaystyle {\textstyle\frac{1}{2}}$ $\displaystyle \leq$ x $\displaystyle \leq$ 2.

So

Area($\displaystyle \Omega_{1}^{}$) = $\displaystyle \int_{0}^{1/2}$$\displaystyle \Big($g(x) - h(x)$\displaystyle \Big)$dx = $\displaystyle \int_{0}^{1/2}$$\displaystyle {\textstyle\frac{3}{2}}$x dx = $\displaystyle {\textstyle\frac{3}{16}}$  ,

and

Area($\displaystyle \Omega_{2}^{}$) = $\displaystyle \int_{1/2}^{2}$$\displaystyle \Big($f (x) - h(x)$\displaystyle \Big)$dx = $\displaystyle \int_{0}^{1/2}$$\displaystyle \Big($1 - $\displaystyle {\frac{x}{2}}$$\displaystyle \Big)$dx = $\displaystyle {\textstyle\frac{9}{16}}$  ,

which implies

Area($\displaystyle \Omega$) = Area($\displaystyle \Omega_{1}^{}$) + Area($\displaystyle \Omega_{2}^{}$) = $\displaystyle {\textstyle\frac{3}{16}}$ + $\displaystyle {\textstyle\frac{9}{16}}$ = $\displaystyle {\textstyle\frac{3}{4}}$   .


[Back] [Next]
[Trigonometry] [Calculus]
[Geometry] [Algebra] [Differential Equations]
[Complex Variables] [Matrix Algebra]

S.O.S MATHematics home page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.


Mohamed A. Khamsi
Helmut Knaust

Copyright 1999-2017 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour