Question

Consider the function $f\left(x\right)={\sin }^2\left(\pi x\right)$.

(a) Find the area between the graphs of $f\left(x\right)$and $f\left(x\right)+\frac{1}{2}$on $\left[0,3\right]$, shown next at the left.

(b) Find the area between the graphs of $f\left(x\right)$and $-f\left(x\right)+\frac{1}{2}$on $\left[0,3\right]$, shown next at the right.

Short answer

Part (a) Area is $1.5$.

Part (b) Area is$2.15$.

Step-by-step solution

Part (a). Step 1. Given Information

We have$f\left(x\right)={\sin }^2\left(\mathrm{\pi x}\right)$.

Part (a). Step 2. Explanation.

To find the area between the graphs of $f\left(x\right)$and $f\left(x\right)+\frac{1}{2}$on $\left[0,3\right]$.

Area = role="math" localid="1649699213175" ${\int }_0^3\left(f\right(x)+\frac{1}{2})-\left(f\left(x\right)\right)dx$.

role="math" localid="1649699445177" ${\int }_0^3\left(f\right(x)+\frac{1}{2})-f\left(x\right)dx={\int }_0^3{\sin }^2\left(\mathrm{\pi x}\right)+\frac{1}{2}dx-{\int }_0^3{\sin }^2\left(\mathrm{\pi x}\right)dx$

Use trigonometric identities role="math" localid="1649699489325" ${\sin }^2\left(\mathrm{\pi x}\right)=\frac{1-\cos \left(2\mathrm{\pi x}\right)}{2}$

role="math" localid="1649774701028" ${\int }_0^3{\sin }^2\left(\mathrm{\pi x}\right)+\frac{1}{2}dx-{\int }_0^3{\sin }^2\left(\mathrm{\pi x}\right)dx={\int }_0^3\frac{\left(1-\cos \left(2\mathrm{\pi x}\right)\right)}{2}+\frac{1}{2}dx-{\int }_0^3\frac{\left(1-\cos \left(2\mathrm{\pi x}\right)\right)}{2}dx$

Simplify the integral.

role="math" localid="1649699518713" ${\int }_0^3\frac{1}{2}dx-{\int }_0^3\frac{\cos \left(2\mathrm{\pi x}\right)}{2}dx+{\int }_0^3\frac{1}{2}dx-\left({\int }_0^3\frac{1}{2}dx-{\int }_0^3\frac{\cos \left(2\mathrm{\pi x}\right)}{2}dx\right)$

role="math" localid="1649776061832" $$\begin{gathered} =\frac{1}{2}{\left[x\right]}_0^3-\frac{1}{2\mathrm{\pi }}{\left[\sin 2\mathrm{\pi x}\right]}_0^3+\frac{1}{2}{\left[x\right]}_0^3-\left(\frac{1}{2}{\left[x\right]}_0^3-\frac{1}{2\mathrm{\pi }}{\left[\sin 2\mathrm{\pi x}\right]}_0^3\right) \\ =\frac{3}{2}-0+\frac{3}{2}-\left(\frac{3}{2}-0\right) \\ =\frac{3}{2} \\ =1.5 \end{gathered}$$

Part (b). Step 1 . Explanation.

To find the area between the graphs of $f\left(x\right)$and $-f\left(x\right)+\frac{1}{2}$on [0, 3].

Area = role="math" localid="1649775738033" ${\int }_0^3(-f(x)+\frac{1}{2})-\left(f\left(x\right)\right)dx$

role="math" localid="1649774630073" ${\int }_0^{3}f\left(x\right)-(-f(x)+\frac{1}{2})dx={\int }_0^3{\sin }^2\left(\mathrm{\pi x}\right)dx-{\int }_0^3-{\sin }^2\left(\mathrm{\pi x}\right)+\frac{1}{2}dx$

role="math" localid="1649775914634" ${\int }_0^3\frac{1}{2}dx-{\int }_0^3\frac{\cos \left(2\mathrm{\pi x}\right)}{2}dx+{\int }_0^3\frac{1}{2}dx-{\int }_0^3\frac{1}{2}dx-{\int }_0^3\frac{\cos \left(2\mathrm{\pi x}\right)}{2}dx$

$=2.15$