Q. 9 - Calculus

Question

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

Step-by-Step Solution

Answer

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

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

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

Step 3. Continue

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

Step 4. Continue

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