Math Calculating areas with integrals Area between curves

# Area between curves

The definite integral can also be used to calculate the area between two function graphs.

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

1. Difference of the functions
2. Find and calculate definite integral
3. 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 $[0.5; 2]$

1. #### Difference of the functions

First, $g(x)$ is subtracted from $f(x)$ and combined.
$f(x)=\frac12x^2+1$
$g(x)=-\frac12x^2+x$

$h(x)=f(x)-g(x)$ $=(\frac12x^2+1)-$ $(-\frac12x^2+x)$ $=x^2-x+1$
2. #### Find definite integral

Insert limits of integration and $h(x)$ into the given integral and calculate integral.
$\int_a^b (f(x)-g(x))\,\mathrm{d}x$ $=\int_a^b h(x)\,\mathrm{d}x$

$\int_{0.5}^2 (x^2-x+1)\,\mathrm{d}x$
3. #### Calculate integral

$\int_a^b h(x) \, \mathrm{d}x$ $= [H(x) + C]_a^b$ $= H(b) - H(a)$

$H(x)=\frac13x^3-\frac12x^2+x$

$\int_{0.5}^2 (x^2-x+1)\,\mathrm{d}x$ $=[\frac13x^3-\frac12x^2+x]_{0.5}^2$ $=(\frac13\cdot2^3-\frac12\cdot2^2+2)-$ $(\frac13\cdot0.5^3-\frac12\cdot0.5^2+0.5)$
$=\frac83-\frac{5}{12}$ $=\frac94$
4. #### Determine area

$A=\int_{0.5}^2 (x^2-x+1)\,\mathrm{d}x$ $=\frac94$ $=2.25$