Q. 82 - Calculus

Question

Consider the function $f\left(x\right)=\sin x\cos x$.

(a) Find the area between the graphs of $f\left(x\right) \mathrm{and} g\left(x\right)=\sin x$ on $[0,\pi ]$ shown next at the left.

(b) Find the area between the graphs of $f\left(x\right) \mathrm{and} h\left(x\right)=\sin 2x$ on $[0,\pi ]$ shown next at the right.

Step-by-Step Solution

Answer

(a) Find the area between the graphs of $f\left(x\right) \mathrm{and} g\left(x\right)=\sin x$ on $[0,\pi ]$ is $-2$.

(b) Find the area between the graphs of $f\left(x\right) \mathrm{and} h\left(x\right)=\sin 2x$ on $[0,\pi ]$ is $0$.

Step 1. Given Information

Consider the function $f\left(x\right)=\sin x\cos x$.

(a) Find the area between the graphs of $f\left(x\right) \mathrm{and} g\left(x\right)=\sin x$ on $[0,\pi ]$ shown next at the left.

(b) Find the area between the graphs of $f\left(x\right) \mathrm{and} h\left(x\right)=\sin 2x$ on$[0,\pi ]$ shown next at the right.

Part (a) Step 1. Now finding the area between the graphs of f ( x )   and   g ( x ) = sin x on [ 0 , 2 π ] .

$$\begin{gathered} Area={\int }_0^{\mathrm{\pi }}\left[f\right(x)-g(x\left)\right]dx \\ Area={\int }_0^{\mathrm{\pi }}f\left(x\right)dx-{\int }_0^{\mathrm{\pi }}g\left(x\right)dx \end{gathered}$$

Part (a) Step 2. Firstly finding the value of ∫ 0 π f ( x ) d x

${\int }_0^{\mathrm{\pi }}f\left(x\right)dx={\int }_0^{\mathrm{\pi }}\sin x\cos xdx$

Let

$$\begin{gathered} u=\sin x \\ \frac{du}{dx}=\cos x \\ du=\cos xdx \end{gathered}$$

Part (a) Step 3. Using the information in equations, we can change variables completely:

$$\begin{gathered} {\int }_0^{\mathrm{\pi }}f\left(x\right)dx={\int }_0^{\mathrm{\pi }}udu \\ {\int }_0^{\mathrm{\pi }}f\left(x\right)dx={\left[\frac{u^{1+1}}{1+1}\right]}_0^{\mathrm{\pi }} \\ {\int }_0^{\mathrm{\pi }}f\left(x\right)dx={\left[\frac{u^2}{2}\right]}_0^{\mathrm{\pi }} \\ {\int }_0^{\mathrm{\pi }}f\left(x\right)dx=\frac{1}{2}{\left[u^2\right]}_0^{\mathrm{\pi }} \\ {\int }_0^{\mathrm{\pi }}f\left(x\right)dx=\frac{1}{2}{\left[{\left(\sin x\right)}^2\right]}_0^{\mathrm{\pi }} \\ {\int }_0^{\mathrm{\pi }}f\left(x\right)dx=\frac{1}{2}\left[{\left(\mathrm{sin\pi }\right)}^2-{\left(\sin 0\right)}^2\right] \\ {\int }_0^{\mathrm{\pi }}f\left(x\right)dx=0 \end{gathered}$$

Part (a) Step 4. Now finding the value of ∫ 0 π g ( x ) d x

$$\begin{gathered} {\int }_0^{\mathrm{\pi }}g\left(x\right)dx={\int }_0^{\mathrm{\pi }}\sin xdx \\ {\int }_0^{\mathrm{\pi }}g\left(x\right)dx=-{\left[\cos x\right]}_0^{\mathrm{\pi }} \\ {\int }_0^{\mathrm{\pi }}g\left(x\right)dx=-\left[\mathrm{cos\pi }-\cos 0\right] \\ {\int }_0^{\mathrm{\pi }}g\left(x\right)dx=-\left[-1-1\right] \\ {\int }_0^{\mathrm{\pi }}g\left(x\right)dx=-\left[-2\right] \\ {\int }_0^{\mathrm{\pi }}g\left(x\right)dx=2 \end{gathered}$$

Part (a) Step 5. Now putting the value in the A r e a = ∫ 0 π f ( x ) d x - ∫ 0 π g ( x ) d x

$$\begin{gathered} Area={\int }_0^{\mathrm{\pi }}f\left(x\right)dx-{\int }_0^{\mathrm{\pi }}g\left(x\right)dx \\ Area=0-2 \\ Area=-2 \end{gathered}$$

Part (b) Step 1. Now finding the area between the graphs of f ( x )   and   h ( x ) = sin 2 x on [ 0 , π ] .

$$\begin{gathered} Area={\int }_0^{\mathrm{\pi }}\left[f\right(x)-h(x\left)\right]dx \\ Area={\int }_0^{\mathrm{\pi }}f\left(x\right)dx-{\int }_0^{\mathrm{\pi }}h\left(x\right)dx \end{gathered}$$

Part (b) Step 2. Firstly finding the value of ∫ 0 π h ( x ) d x

${\int }_0^{\mathrm{\pi }}h\left(x\right)dx={\int }_0^{\mathrm{\pi }}\sin 2xdx$

Let

$$\begin{gathered} u=2x \\ \frac{du}{dx}=2 \\ du=2dx \\ \frac{1}{2}du=dx \end{gathered}$$

Part (b) Step 3. Now integrate the integral ∫ 0 π h ( x ) d x

$$\begin{gathered} {\int }_0^{\mathrm{\pi }}h\left(x\right)dx={\int }_0^{\mathrm{\pi }}\sin udx \\ {\int }_0^{\mathrm{\pi }}h\left(x\right)dx={\left[-\cos u\right]}_0^{\mathrm{\pi }} \\ {\int }_0^{\mathrm{\pi }}h\left(x\right)dx=-{\left[\cos 2x\right]}_0^{\mathrm{\pi }} \\ {\int }_0^{\mathrm{\pi }}h\left(x\right)dx=-\left[\cos 2\pi -\cos 0\right] \\ {\int }_0^{\mathrm{\pi }}h\left(x\right)dx=-\left[1-1\right] \\ {\int }_0^{\mathrm{\pi }}h\left(x\right)dx=0 \end{gathered}$$

Part (b) Step 4. Now putting the value in the A r e a = ∫ 0 π f ( x ) d x - ∫ 0 π h ( x ) d x

$$\begin{gathered} Area={\int }_0^{\mathrm{\pi }}f\left(x\right)dx-{\int }_0^{\mathrm{\pi }}h\left(x\right)dx \\ Area=0-0 \\ Area=0 \end{gathered}$$