Chapter 7, Problem 11RE

To determine

To calculate: The area of the region

Answer to Problem 11RE

The area of the region is $0.0155$

Explanation of Solution

Given information:

  $y=\sin x,$

  $y=x,$

  $x=\frac{\pi }{4}$

Calculation:

To find the area of the region which is enclosed by the lines

  $y=x,x=\frac{\pi }{4}$ and by the sinusoidal curve $y=\sin x$ .

The sketch of these graphs will help to more easily spot a given problem.

Let’s sketch the graphs into Desmos. Further by entering the code, which see in the red rectangle below, mark the required region with the color.

  

By observing the given picture notice that our area is bounded above with the line $y=x$ and below with the curve $y=\sin x$ from $x=0$ to $x=\frac{\pi }{4}$ .

The upper limit of integration it is obviously determined by the line

  $x=\frac{\pi }{4}$

And lower limit of integration will be at

  $x=0$

Because it is only value that satisfies equality $\sin x=x$ .

Now set the limits of integration, area of the region is represented as

  $\text{Area}={\displaystyle {\int }_0^{\frac{\pi }{4}}[x-\sin x]dx}$

To solve previously integral use the Power formula and Trigonometric formula and simply continue the calculation:

  $\begin{array}{c}\text{Area}={\displaystyle {\int }_0^{\frac{\pi }{4}}[x-\sin x]dx}\\ =\frac{x^2}{2}-{\left.(-\cos x)\right|}_0^{\pi /4}\\ =\frac{x^2}{2}+{\left.\cos x\right|}_0^{\pi /4}\\ =\frac{{\pi }^2}{32}+\frac{\sqrt{2}}{2}-(0-1)\\ =\frac{{\pi }^2}{32}+\frac{\sqrt{2}}{2}-1\\ \approx 0.0155\end{array}$

Therefore, the area of the region is equal $0.0155$ .