Defined areas under a function

In this section we will calculate areas determined by the graph of a function and the $x$ axis. We consider a real function $f(x)$ and its graph $( x, f(x) )$ in the plane:

We want to calculate the colored area $A$. The area $A$, determined by the graph of the function, and a certain interval $[a, b]$ of the axis of $x$, is:

$$A=\int_{a}^b f(x) \ dx$$

We are going to calculate the area under the function $f(x)=x^2$ in the interval $[-2,2]$. Let's represent the function to have a geometric notion of the problem:

Therefore, we have to integrate the function in the interval that delimits the area: $$A=\int_{-2}^2 x^2 \ dx=\Big[ \dfrac{x^3}{3} \Big]_{-2}^2= \dfrac{2^3}{3}-\dfrac{(-2)^3}{3}=\dfrac{8}{3}-\dfrac{-8}{3}= \dfrac{16}{3}\mathrm{u}^2$$

Note: When we write $u^2$ we refer to the surface units.

We are going to calculate the area under the function $f(x)=\sqrt{x}+\sin x$ in the interval $[0,4]$. As we have done in the previous example, let's draw the function first of all:

To calculate the area under $f(x)$, we have to integrate the above mentioned function in the corresponding interval: $$\begin{array}{rl} A=&\int_0^4(\sqrt{x}+\sin (x)) \ dx= \int_0^4\sqrt{x} \ dx + \int_0^4\sin(x) \ dx \\ =& \left[ \dfrac{x^{\frac{3}{2}}}{\dfrac{3}{2}} \right]_0^4+ [-\cos(x)]_0^4= \dfrac{2}{3}(8-0)-\cos(4)+1 \\ =& \dfrac{16}{3}+1.65=6.98 u^2 \end{array}$$