Integration: find as an exact value the enclosed area between $y=\frac{3x}{5π}$ and the curve $y=\sin x$ for $0≤x≤π$ shown shaded in the diagram.

Question

The diagram shows the line $y=\frac{3x}{5\pi}$ and the curve $y=\sin$ $x$ for $0\le x\le \pi$.

Find (as an exact value) the enclosed area shown shaded in the diagram.

I'm not sure where I've made an error, but this is my working out so far:

Area under $y=\frac{3x}{5\pi}$ from $0$ to $\frac{5\pi}{6}$ is:

$$\int_{0}^{5\pi/6} \frac{3x}{5\pi}dx=\frac{5\pi}{24}$$

The remaining unshaded "triangle" area from $\frac{5\pi}{6}$ to $\pi$:

$$\frac{1}{2} \cdot \frac{\pi}{6} \cdot \frac{1}{2} = \frac{\pi}{24}$$

Area under $y=\sin x$ for $0\le x\le\pi$:

$$\int_{0}^{\pi} \sin{x}dx=2$$

Hence, shaded area = $$2-\frac{5\pi}{24} - \frac{\pi}{24} = \frac{-\pi}{4} + 2$$

The correct answer is:

$$1+\frac{\sqrt3}{2} - \frac{5\pi}{24} units^2$$

If someone could explain how to correctly solve this, it would be greatly appreciated!

Answer 1

The area under $sin(x)$ between $x=5 \pi/6$ and $x=\pi$ is not a triangle. The top part is not a straight line, but $sin(x)$- which is curved. to find the area, you can integrate:

$$ \int_{5 \pi/6}^{\pi} sin(x) dx $$

Answer 2

The unshaded area from $5\pi/6$ to $\pi$ is not a triangle! It should be instead $$\int_{5\pi/6}^{\pi} \sin x \ dx =1-\frac{\sqrt 3}{2}$$

Answer 3

May be more easy is to calcualte $$\int\limits_{0}^{\frac{5\pi}{6}}\int\limits_{\frac{3x}{5\pi}}^{\sin x}dxdy=\int\limits_{0}^{\frac{5\pi}{6}}\left(\sin x - \frac{3x}{5\pi} \right)dx=1+\frac{\sqrt3}{2} - \frac{5\pi}{24} $$