Question

Evaluate the integral and interpret it as the area of a region

Answer 1

Given:

`\int ^{(\pi)/(2)}_0|4\sin x-4\cos (2x)|dx`

First, eliminate absolute

This gives

`\begin{gathered} \int ^{(\pi)/(2)}_0|4\sin x-4\cos (2x)|dx \\ =\int ^{(\pi)/(6)}_0-4\sin x+4\cos (2x)dx+\int ^{(\pi)/(2)}_{(\pi)/(6)}4\sin x-4\cos (2x)dx \end{gathered}`

Integrating each part gives

`\begin{gathered} \int ^{(\pi)/(6)}_0-4\sin x+4\cos (2x)dx+\int ^{(\pi)/(2)}_{(\pi)/(6)}4\sin x-4\cos (2x)dx \\ =4\cos x+2\sin (2x)|^{(\pi)/(6)^{}_{}}_0-4\cos x-2\sin (2x)|^{(\pi)/(2)}_{(\pi)/(6)} \end{gathered}`

Simplifying further gives

`\begin{gathered} 4\cos x+2\sin (2x)|^{(\pi)/(6)^{}_{}}_0 \\ =\mleft\lbrace(4\cos ((\pi)/(6))+2\sin ((2\pi)/(6))-(4\cos 0+2\sin 0)\mright\rbrace \\ =(2\sqrt[]{3}+\sqrt[]{3}-(4+0) \\ =3\sqrt[]{3}-4 \end{gathered}`

Also

`\begin{gathered} -4\cos x-2\sin (2x)|^{(\pi)/(2)}_{(\pi)/(6)} \\ =-4\cos ((\pi)/(2))-2\sin ((2\pi)/(2))-(-4\cos ((\pi)/(6))-2\sin ((2\pi)/(6)) \\ =0+0-(-2\sqrt[]{3}-\sqrt[]{3)} \\ =2\sqrt[]{3}+\sqrt[]{3} \\ =3\sqrt[]{3} \end{gathered}`

Hence the solution is

`\begin{gathered} \int ^{(\pi)/(6)}_0-4\sin x+4\cos (2x)dx+\int ^{(\pi)/(2)}_{(\pi)/(6)}4\sin x-4\cos (2x)dx \\ =3\sqrt[]{3}-4+3\sqrt[]{3} \\ =6\sqrt[]{3}-4 \end{gathered}`

Therefore, the answer is

`6\sqrt[]{3}-4`