Area between curves
The definite integral can also be used to calculate the area between two function graphs.
![](https://math.lakschool.com/de/themen/integral_flaeche/images/integral_anwendung_graphen.png)
We are looking for the enclosed area of two functions in the interval $[a;b]$. To do this, subtract the smaller from the larger area ($f(x) > g(x)$ in the interval $[a;b]$):
$A=\int_a^b f(x)\,\mathrm{d}x\,-$ $\int_a^b g(x)\,\mathrm{d}x$
Now the sum rule can be applied backwards.
In order for the formula to be valid, if $f(x) < g(x)$ in the interval $[a;b]$, one uses the absolute value.
$A=|\int_a^b (f(x)-g(x))\,\mathrm{d}x|$
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Hint
The formula is independent of the position of the area regarding the x-axis. So it does not matter if the area is above or below the x-axis.
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Method
- Difference of the functions
- Find and calculate definite integral
- Determine area
Example
Calculate the area between the graphs of the functions $f(x)=\frac12x^2+1$ and $g(x)=-\frac12x^2+x$ over the interval $[0.5; 2]$
![](https://math.lakschool.com/de/themen/integral_flaeche/images/integral_anwendung_beispiel_3.png)
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Difference of the functions
First, $g(x)$ is subtracted from $f(x)$ and combined.
$f(x)=\frac12x^2+1$
$g(x)=-\frac12x^2+x$
$h(x)=f(x)-g(x)$ $=(\frac12x^2+1)-$ $(-\frac12x^2+x)$ $=x^2-x+1$ -
Find definite integral
Insert limits of integration and $h(x)$ into the given integral and calculate integral.
$\int_a^b (f(x)-g(x))\,\mathrm{d}x$ $=\int_a^b h(x)\,\mathrm{d}x$
$\int_{0.5}^2 (x^2-x+1)\,\mathrm{d}x$ -
Calculate integral
$\int_a^b h(x) \, \mathrm{d}x$ $= [H(x) + C]_a^b$ $= H(b) - H(a)$
$H(x)=\frac13x^3-\frac12x^2+x$
$\int_{0.5}^2 (x^2-x+1)\,\mathrm{d}x$ $=[\frac13x^3-\frac12x^2+x]_{0.5}^2$ $=(\frac13\cdot2^3-\frac12\cdot2^2+2)-$ $(\frac13\cdot0.5^3-\frac12\cdot0.5^2+0.5)$
$=\frac83-\frac{5}{12}$ $=\frac94$ -
Determine area
$A=\int_{0.5}^2 (x^2-x+1)\,\mathrm{d}x$ $=\frac94$ $=2.25$