Chapter 6.1, Problem 35E

To determine

To Calculate: The approximate area under the sine curve in the given interval.

Answer to Problem 35E

Option C

  $\frac{\pi }{4}\left[0+\frac{\sqrt{2}}{2}+1+\frac{\sqrt{2}}{2}\right]$

Explanation of Solution

Given information:Sine curve from the interval $x=0to\pi$ . Four subdivisions used to approximate the area.

Formula used:

  $\begin{array}{l}\Delta x\left[f(x_0)+f(x_1)+f(x_2)\mathrm{......}+f(x_{n-1})\right]\\ Where\\ \Delta x=\frac{b-a}{n}\end{array}$

Calculation: From the given interval, the value of $a=0,b=\pi$

  $\begin{array}{l}\Delta x=\frac{b-a}{n}\\ a=0,b=\pi \\ \Delta x=\frac{\pi -0}{4}=\frac{\pi }{4}\end{array}$

Given interval is $x=0to\pi$ . Find the intervals using delta x

  $\begin{array}{l}\Delta x=\frac{\pi }{4}\\ x=0to\pi \\ Intervalsare\\ (0,\frac{\pi }{4}),(\frac{\pi }{4},\frac{\pi }{2}),(\frac{\pi }{2},\frac{3\pi }{4}),(\frac{3\pi }{4},\pi )\end{array}$

Use Left end of the interval and find the sine value

  $\begin{array}{l}f(x)=\sin (x)\\ f(0)=\sin (0)=0\\ f(\frac{\pi }{4})=\sin (\frac{\pi }{4})=\frac{\sqrt{2}}{2}\\ f(\frac{\pi }{2})=\sin (\frac{\pi }{2})=1\\ f(\frac{3\pi }{4})=\sin (\frac{3\pi }{4})=\frac{\sqrt{2}}{2}\end{array}$

  $\begin{array}{l}Sum=\Delta x\left[f(x_0)+f(x_1)+f(x_2)\mathrm{......}+f(x_{n-1})\right]\\ Sum=\Delta x\left[f(x_0)+f(x_1)+f(x_2)+f(x_3)\right]\\ Sum=\frac{\pi }{4}\left[0+\frac{\sqrt{2}}{2}+1+\frac{\sqrt{2}}{2}\right]\end{array}$