91Ó°ÊÓ

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

(a) Find the area between the graphs of role="math" localid="1649093822136" $f\left(x\right)\mathrm{and}g\left(x\right)=\sin x$on $[0,\mathrm{Ï€}]$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$[0,\mathrm{Ï€}]$shown next at the right.

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

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

Let

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

$$\begin{gathered} {∫}_0^{\mathrm{Ï€}}f\left(x\right)dx={∫}_0^{\mathrm{Ï€}}udu \\ {∫}_0^{\mathrm{Ï€}}f\left(x\right)dx={\left[\frac{u^{1+1}}{1+1}\right]}_0^{\mathrm{Ï€}} \\ {∫}_0^{\mathrm{Ï€}}f\left(x\right)dx={\left[\frac{u^2}{2}\right]}_0^{\mathrm{Ï€}} \\ {∫}_0^{\mathrm{Ï€}}f\left(x\right)dx=\frac{1}{2}{\left[u^2\right]}_0^{\mathrm{Ï€}} \\ {∫}_0^{\mathrm{Ï€}}f\left(x\right)dx=\frac{1}{2}{\left[{\left(\sin x\right)}^2\right]}_0^{\mathrm{Ï€}} \\ {∫}_0^{\mathrm{Ï€}}f\left(x\right)dx=\frac{1}{2}\left[{\left(\mathrm{²õ¾\pm ²ÔÏ€}\right)}^2-{\left(\sin 0\right)}^2\right] \\ {∫}_0^{\mathrm{Ï€}}f\left(x\right)dx=0 \end{gathered}$$

$$\begin{gathered} {∫}_0^{\mathrm{Ï€}}g\left(x\right)dx={∫}_0^{\mathrm{Ï€}}\sin xdx \\ {∫}_0^{\mathrm{Ï€}}g\left(x\right)dx=-{\left[\cos x\right]}_0^{\mathrm{Ï€}} \\ {∫}_0^{\mathrm{Ï€}}g\left(x\right)dx=-\left[\mathrm{³¦´Ç²õÏ€}-\cos 0\right] \\ {∫}_0^{\mathrm{Ï€}}g\left(x\right)dx=-\left[-1-1\right] \\ {∫}_0^{\mathrm{Ï€}}g\left(x\right)dx=-\left[-2\right] \\ {∫}_0^{\mathrm{Ï€}}g\left(x\right)dx=2 \end{gathered}$$

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

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

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

Let

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

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

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