Question

The absolute area between the graph of a function f and the x-axis on [a, b] .

Short answer

$A_2=\left|{\int }_b^c f\left(x\right)dx\right|$

Step-by-step solution

Step 1. Given Information

To define: The absolute area between the graph function $f\left(x\right)$and x-axis on $[a,b]$.

Step 2. Calculation

$$\begin{gathered} \text{v-axis on an interval}[a,b]\text{, i.e.,}A\left(\text{area}\right)={\int }_a^{b}f\left(x\right)dx\text{. The function}f\left(x\right)\text{may be above or below the}x\text{-axis,} \\ \text{although the area is always a positive quantity, it can bear a sign '}+\text{' or '-'according to} \\ \left(i\right).iff\left(x\right)\ge 0\text{on all the interval}[a,b]\text{then}{A}_1\text{(area)}={\int }_a^{b}f\left(x\right)dx\ge 0\text{, i.e.}{A}_1\text{is positive.} \\ \left(ii\right).iff\left(x\right)\le 0\text{on all the interval}[b,c]\text{then}{A}_2\text{(area)}={\int }_b^c f\left(x\right)dx\le 0\text{, i.e.}{A}_2\text{is negative.} \end{gathered}$$

Step 3. Continue

The signed area (positive and negative) are shown in figure below:

Step 4. Continue

$$\begin{gathered} \text{The magnitude of the area of the function}f\left(x\right)\text{below x-axis}{A}_2=\mid {\int }_b^c f\left(x\right)dx\text{is called the absolute area of} \\ \text{the function}f\left(x\right)\text{on the interval}[b,c]\text{. This is the area without the sign.} \\ \text{Since, the area A between curve and}x\text{-axis can have positive and negative values, however negative are in} \\ \text{actual calculations has no meaning or sometimes may be absurd, e.g. Area of square is}-49\text{sq.cm then its} \\ \text{side will be}\sqrt{-49}=\pm 7i\mathrm{cm}\text{an imaginary value. Therefore, to avoid such an absurdity the computed are is} \\ \mathrm{taken}{A}_2=\left|{\int }_b^c f\left(x\right)dx\right| \end{gathered}$$