Chapter 6.5, Problem 6E

(a)

To determine

Approximate the value of integral using Trapezoidal rule with $n=4$ .

Answer to Problem 6E

With $n=4$ :

  ${\displaystyle {\int }_0^{\pi }\sin xdx}\approx 1.896$

Explanation of Solution

Given information:

  ${\displaystyle {\int }_0^{\pi }\sin xdx}$

Since we are required to use the Trapezoidal rule, first we need to find the height.

  $h=\frac{b-a}{n}=\frac{\pi -0}{4}=\frac{\pi }{4}$

Function values:

  

Now,

Setup the equation:

  ${\displaystyle {\int }_0^{\pi }\sin xdx}=\frac{\pi /4}{2}\left[f\left(0\right)+2\cdot f\left(\frac{\pi }{4}\right)+2\cdot f\left(\frac{\pi }{2}\right)+2\cdot f\left(\frac{3\pi }{4}\right)+f\left(\pi \right)\right]$

Plugin the values into the equation:

  $\begin{array}{l}{\displaystyle {\int }_0^{\pi }\sin xdx}=\frac{\pi }{8}\left[0+2\cdot \frac{\sqrt{2}}{2}+2\cdot 1+2\cdot \frac{\sqrt{2}}{2}+0\right]\\ =\frac{\pi }{8}\left[0+\sqrt{2}+2+\sqrt{2}+0\right]\\ \approx 1.896\end{array}$

(b)

To determine

Whether the approximation is an overestimate or an under − estimate.

Answer to Problem 6E

The approximation is an under − estimate.

Explanation of Solution

Given information:

  ${\displaystyle {\int }_0^{\pi }\sin xdx}$

First derivative of the function:

  $f\text{'}\left(x\right)=\cos x$

Second derivative of the function:

  $f\text{'}\text{'}\left(x\right)=-\sin x<0$ for all $x\in \left[0,\pi \right]$

We know that

Second derivative gives the concavity.

If the second derivative is greater than 0,

It’s an overestimate.

If the second derivative is less than 0,

It’s an underestimate.

Note that

The value of second derivative is $-\sin x$ which is less than 0.

Therefore,

The approximation is an under − estimate.

(c)

To determine

Integral’s exact value to verify the answer.

Answer to Problem 6E

The exact value of integral is 2 which is not same as using the Trapezoidal rule with $n=4$ .

Explanation of Solution

Given information:

  ${\displaystyle {\int }_0^{\pi }\sin xdx}$

Using FTC (Fundamental Theorem of Calculus):

  $\begin{array}{l}{\displaystyle {\int }_0^{\pi }\sin xdx}={-\cos x|}_0^{\pi }\\ =-\cos \pi +\cos 0\\ =-\left(-1\right)+1\\ =2\end{array}$

Therefore,

Integral’s exact value is 2 which is not same as the value of integral obtained in Part (a).