Q. 38 - Calculus

Question

The area between the graphs of $f\left(x\right)=4-x^2 and 1-2x on [-4,4].$

Step-by-Step Solution

Answer

The area enclosed by the curves of two functions $f\left(x\right)=4-x^2 and g\left(x\right)=1-2x$ on the interval $[-4,4]$ is $2.67$ units.

Step 1. Given information

The two given functions are $f\left(x\right)=4-x^2, g\left(x\right)=1-2x$ and the given interval $[-4,4].$

Step 2. Calculation

The area $A$ enclosed by these curves on the given interval $[0,\mathrm{\pi }]$ is

The graphs of functions $f\left(x\right)=4-x^2 and g\left(x\right)=1-2x$ are shown in Figure-1 here.

The curves intersect at points $A(-1,3) and F(3,-5).$

The area enclosed by the curves of functions $f\left(x\right)=4-x^2 and g\left(x\right)=1-2x$ is

$Area AGEFCA=Area BAGECB+Area CDFC-Area BACB-Area EFDE$     ....(1)

Now, $Area AGEFCA= Area BAGECB+Area CDFC-Area BACB-Area EFDE$

$$\begin{gathered} Area AGEFCA=\left|{\int }_{-1}^{2}f\left(x\right)dx\right|+\left|{\int }_{-1}^{\frac{1}{2}}g\left(x\right)dx\right|-\left|{\int }_{\frac{1}{2}}^{3}g\left(x\right)dx\right|-\left|{\int }_2^{3}f\left(x\right)dx\right| \\ Area AGEFCA=\left|{\int }_{-1}^2(4-x^2)dx\right|+\left|{\int }_{-1}^{\frac{1}{2}}(1-2x)dx\right|-\left|{\int }_{\frac{1}{2}}^3(1-2x)dx\right|-\left|{\int }_2^3(4-x^2)dx\right| \end{gathered}$$

Now, using graphing calculator

$$\begin{gathered} \left|{\int }_{-1}^2(4-x^2)dx\right|=\left|9\right|=9, \left|{\int }_2^3(4-x^2)dx\right|=\left|-2.33\right|=2.33, \\ \left|{\int }_{-1}^{\frac{1}{2}}(1-2x)dx\right|=\left|2.25\right|=2.25, \left|{\int }_{\frac{1}{2}}^3(1-2x)dx\right|=\left|-6.25\right|=6.25 \end{gathered}$$

Thus,

$$\begin{gathered} Area AGEFCA=\left|{\int }_{-1}^2(4-x^2)dx\right|+\left|{\int }_{-1}^{\frac{1}{2}}(1-2x)dx\right|-\left|{\int }_{\frac{1}{2}}^3(1-2x)dx\right|-\left|{\int }_2^3(4-x^2)dx\right| \\ =9.00+2.25-2.33-6.25 \\ =2.67 \end{gathered}$$