Integration: The definite integral
Area
The area of the surface #\orange S# above the #x#-axis and bound by the graph of #\blue{f}#, the lines #x=a# and #x=b# is equal to
\[\int_a^b \blue f(x) \; \dd x\]
We have now seen how to calculate the area of a surface above the #x#-axis, but in the same manner we can calculate a surface below the #x#-axis.
The area of the surface #\orange S# that lies below the #x#-axis and is enclosed by the graph of #\blue{f}#, the lines #x=a# and #x=b# is equal to:
\[-\int_a^b \blue f(x) \; \dd x\]
Finally we will present a procedure on how to calculate the area enclosed by the graph of #\blue f#, the #x#-axis and the lines #x=a# and #x=b#. Here, the area can be partly above and partly below the graph.
Procedure |
Example |
|
Determine the area of a surface enclosed by the graph #\blue f#, the #x#-axis and the lines #x=a# and #x=b#. |
The area enclosed by #\blue f(x)=-(x-3)^2+4#, the #x#-axis and #x=0# and #x=6# |
|
Step 1 |
Determine the zeros of the graph of #\blue f# between #x=a# and #x=b#. We will name these zeros #x_1#, #x_2#, #\ldots#, #x_n# if there are #n# zeros. |
#x_1=1#, #x_2=5# |
Step 2 | For every interval #\ivco{a}{x_1}#, #\ivoo{x_1}{x_2}#, #\ldots#, #\ivoc{x_n}{ b}# determine if the #y#-values of #f# are positive or negative. |
\[f(x)\begin{cases}\lt0&\text{als } x \text{ in } \ivco{0}{1}\\ |
Step 3 |
The area of the surface is equal to: \[\pm \int_a^{x_1} \blue f (x)\; \dd x \pm \int_{x_1}^{x_2} \blue f (x)\; \dd x \pm \ldots \pm \int_{x_n}^{b} \blue f(x) \; \dd x \] Here, we have a plus sign in front of the integral if #f# is positive for that domain and a minus sign if #f# is negative. |
\[\begin{array}{c}-\int_{0}^{1} (x-3)^2+4 \; \dd x \\ + \int_{1}^{5} (x-3)^2+4 \; \dd x \\ - \int_{5}^6 (x-3)^2+4 \; \dd x\end{array}\] |
Step 4 |
Calculate the definite integrals and determine the area. |
#\frac{46}{3}# |
Calculate the area of the enclosed domain.
Give your answer as simplified fraction.
Step 1 | The only zero of #f(x)=x^2-6\cdot x+5# between #x=1# and #x=8# is #x_1=5#. The other zero of the polynomial is #x=1#, but this does not matter for the calculation. |
Step 2 | For #[1,5)#, #f(x)# is negative, for #[5,8)# #f(x)# is positive. |
Step 3 | The area of the surface is equal to \[-\int_{1}^{5} f(x) \, \dd x+ \int_{5}^{8}f(x) \, \dd x\] |
Step 4 | We calculate the definite integrals. \[\begin{array}{rcl}\displaystyle \int_{1}^{5} x^2-6\cdot x+5 \, \dd x &=&\displaystyle\left[{{x^3}\over{3}}-3\cdot x^2+5\cdot x\right]_{1}^{5}\\ &&\phantom{xxx}\blue{\text{definition definite integral}}\\ &=&\displaystyle -{{25}\over{3}} - {{7}\over{3}}\\ &&\phantom{xxx}\blue{\text{entering of boundaries and simplification}}\\ &=&\displaystyle -{{32}\over{3}}\\ &&\phantom{xxx}\blue{\text{simplified}} \end{array}\] \[\begin{array}{rcl}\displaystyle \int_{5}^{8} x^2-6\cdot x+5 \, \dd x &=&\displaystyle\left[{{x^3}\over{3}}-3\cdot x^2+5\cdot x\right]_{5}^{8}\\ &&\phantom{xxx}\blue{\text{definition definite integral}}\\ &=&\displaystyle {{56}\over{3}} +{{25}\over{3}}\\ &&\phantom{xxx}\blue{\text{entering of boundaries and simplification}}\\ &=&\displaystyle 27\\ &&\phantom{xxx}\blue{\text{simplified}} \end{array}\] We enter this and hence we find the area we are looking for \[\begin{array}{rcl}\displaystyle -\int_{1}^{5} f(x) \, \dd x+ \int_{5}^{8}f(x) \, \dd x&=&\displaystyle -(-{{32}\over{3}})+27\\&=&\displaystyle \frac{113}{3} \end{array}\] |
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