Calculating the area
For functions without change of sign in the interval $[a; b]$, the area corresponds to the amount of the definite integral:
$A=|\int_a^b f(x)\,\mathrm{d}x|$
i
Hint
Here it has already been described that the area below the x-axis is negative for the definite integral.
Since there is no negative area, one must change the sign when calculating areas under the x-axis.
Since there is no negative area, one must change the sign when calculating areas under the x-axis.
Example
Find the area between the graph of the function $f(x)=x^2-6x+6$ and the x-axis over the interval $[2; 4]$
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Definite integral
Set up the definite integral with the given limits of integration
$\int_2^4 (x^2-6x+6)\,\mathrm{d}x$ -
Find the integral
Now calculate the integral. To do this calculate antiderivative first.
$\int_a^b f(x) \, \mathrm{d}x$ $= [F(x) + C]_a^b$ $= F(b) - F(a)$
$F(x)=\frac13x^3-3x^2+6x$
$\int_2^4 (x^2-6x+6)\,\mathrm{d}x$ $=[\frac13x^3-3x^2+6x]_2^4$ $=(\frac13\cdot4^3-3\cdot4^2+6\cdot4)-$ $(\frac13\cdot2^3-3\cdot2^2+6\cdot2)$
$=-\frac83-\frac83$ $=-\frac{16}3$ -
Determine area
The sketch of the graph shows that the function is negative in the interval $[2; 4]$. Therefore, the sign has to be changed
$A=|\int_2^4 f(x)\,\mathrm{d}x|$ $=|-\frac{16}3|$ $=\frac{16}3$ $\approx5.33$