Integration: The definite integral
Area of a surface between curves
The area of a surface #\orange S# enclosed by the lines #x=a# and #x=b# and the graphs of the functions #\blue f# and #\green g# with #\blue f(x) \geq \green g(x)# on the interval #\ivcc{a}{b}# is equal to
\[\begin{array}{c} \displaystyle \int_a^b \blue f(x)- \green g(x) \; \dd x \end{array} \]
Just as when calculating the area between graphs and the #x#-axis it can differ which graph is above the other one. Therefore we once again use the minus sign to correct for that.
Procedure |
Example |
|
Determine the area of a surface enclosed by the graph #\blue f# and the graph #\green g#, the #x#-axis and the lines #x=a# and #x=b#. |
The area enclosed by #\blue f(x)=x^3+x#, #\green g(x)=x+1# and #x=0# and #x=2# |
|
Step 1 |
Determine the #x#-values of the intersection points of the graph #\blue f# and #\green g# by solving the equation #\blue f(x)=\green g(x)#. We call these intersection points #x_1#, #x_2#, #\ldots#, #x_n# if there are #n# intersection points. |
#x_1=1# |
Step 2 | For each interval #\ivco{a}{x_1}#, #\ivoo{x_1}{x_2}#, #\ldots#, #\ivoc{x_n}{ b}# determine if the #y#-values of #\blue f# are greater than or less than #\green g#. |
\[\blue f(x)\begin{cases}\lt \green g(x)&\text{if } x\in \ivco{0}{1}\\ |
Step 3 |
The area of the surface is equal to: \[\pm \int_a^{x_1} \blue f - \green g\; \dd x \pm \int_{x_1}^{x_2} \blue f - \green g \; \dd x \pm \ldots \pm \int_{x_n}^{b} \blue f - \green g\; \dd x \] Here, there's a plus sign in front of the integral if #\blue f \gt \green g# on that interval, and a minus sign if #\blue f \lt \green g#. |
\[\begin{array}{c}-\int_{0}^{1} x^3+x-(x+1)\; \dd x \\ + \int_{1}^2 x^3+x-(x+1)\; \dd x \end{array}\] |
Step 4 |
Calculate the definite integrals and determine the area. |
#\frac{7}{2}# |
We first determine the surface for which we have to calculate the area. It is enclosed by the two intersection points of #f(x)# and #g(x)#. With some simple calculating we find that #x=0# and #x=1#. Within the interval the value of #f(x) # is greater than or equal to the value of #g(x)#. We calculate the area in the following manner.
\[\begin{array}{rcl}\displaystyle
\int_{0}^{1} \left(f(x)-g(x)\right){\dd}x&=&\displaystyle \int_{0}^{1} \left(\sqrt{x}-x^2\right){\dd}x\\
&& \displaystyle\phantom{xxx}\blue{f(x) \text{ and } g(x) \text{ entered}}\\
&=&\displaystyle \left[{{2\cdot x\cdot \sqrt{x}-x^3}\over{3}}\right]_{0}^{1}\\
&&\phantom{xxx}\blue{\text{definition definite integral}}\\
&=&\displaystyle {{2\cdot 1\cdot \sqrt{1}- 1^3}\over{3}}-{{2\cdot 0\cdot \sqrt{0}- 0^3}\over{3}}\\
&&\phantom{xxx}\blue{\text{boundaries entered}}\\
&=& \displaystyle {{1}\over{3}}\\
&&\phantom{xxx}\blue{\text{simplified}}
\end{array}\]
Or visit omptest.org if jou are taking an OMPT exam.