Answer:
Please check the explanation.
Step-by-step explanation:
Let us consider
[tex]y = f(x)[/tex]
To find the area under the curve [tex]y = f(x)[/tex] between [tex]x = a[/tex] and [tex]x = b[/tex], all we need is to integrate [tex]y = f(x)[/tex] between the limits of [tex]a[/tex] and [tex]b[/tex].
For example, the area between the curve y = x² - 4 and the x-axis on an interval [2, -2] can be calculated as:
[tex]A=\int _a^b|f\left(x\right)|dx[/tex]
= [tex]\int _{-2}^2\left|x^2-4\right|dx[/tex]
[tex]\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx[/tex]
[tex]=\int _{-2}^2x^2dx-\int _{-2}^24dx[/tex]
solving
[tex]\int _{-2}^2x^2dx[/tex]
[tex]\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1[/tex]
[tex]=\left[\frac{x^{2+1}}{2+1}\right]^2_{-2}[/tex]
[tex]=\left[\frac{x^3}{3}\right]^2_{-2}[/tex]
computing the boundaries
[tex]=\frac{16}{3}[/tex]
Thus,
[tex]\int _{-2}^2x^2dx=\frac{16}{3}[/tex]
similarly solving
[tex]\int _{-2}^24dx[/tex]
[tex]\mathrm{Integral\:of\:a\:constant}:\quad \int adx=ax[/tex]
[tex]=\left[4x\right]^2_{-2}[/tex]
computing the boundaries
[tex]=16[/tex]
Thus,
[tex]\int _{-2}^24dx=16[/tex]
Therefore, the expression becomes
[tex]A=\int _a^b|f\left(x\right)|dx=\int _{-2}^2x^2dx-\int _{-2}^24dx[/tex]
[tex]=\frac{16}{3}-16[/tex]
[tex]=-\frac{32}{3}[/tex]
[tex]=-10.67[/tex] square units
Thus, the area under a curve is -10.67 square units
The area is negative because it is below the x-axis. Please check the attached figure.
