Q. 48 - Calculus

Question

For each pair of functions $f$ and $g$ and interval $[a,b]$ in Exercises $41-52$, use definite integrals and the Fundamental Theorem of Calculus to find the exact area of the region between the graphs of $f$ and $g$ from $x=a to x=b$.

$f\left(x\right)=\sin x, g\left(x\right)=\cos x, [a,b]=\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

Step-by-Step Solution

Answer

The exact area is $2.$

Step 1. Given Information.

The given function and interval is$f\left(x\right)=\sin x, g\left(x\right)=\cos x, [a,b]=\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$

Step 2. Graph of the functions.

The graph of the functions will be,

Step 3. Required area.

The exact area will be,

$$\begin{gathered} {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left|f\right(x)-g(x\left)\right|dx=={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{4}}\left(g\right(x)-f(x\left)\right)dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\left(f\right(x)-g(x\left)\right)dx \\ ={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{4}}(\cos x-\sin x)dx+{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}(\sin x-\cos x)dx \\ =[\sin x+\cos x{]}_{-\frac{\pi }{2}}^{\frac{\pi }{4}}+[-\cos x-\sin x{]}_{\frac{\pi }{4}}^{\frac{\pi }{2}} \\ =\left[\left(\sin \frac{\pi }{4}+\cos \frac{\pi }{4}\right)-\left(\sin \left(-\frac{\pi }{2}\right)+\cos \left(-\frac{\pi }{2}\right)\right)\right]-\left[\left(\cos \frac{\pi }{4}+\sin \frac{\pi }{4}\right)-\left(\cos \frac{\pi }{2}+\sin \frac{\pi }{2}\right)\right] \\ =\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-(-1+0)-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}+0+1 \\ =2 \end{gathered}$$