Chapter 6.5, Problem 1E

(a)

To determine

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

Answer to Problem 1E

With $n=4$ :

  ${\displaystyle {\int }_0^{2}xdx}=2$

Explanation of Solution

Given information:

  ${\displaystyle {\int }_0^{2}xdx}$

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

  $h=\frac{b-a}{n}=\frac{2-0}{4}=\frac{1}{2}$

Now,

Setup the equation:

  ${\displaystyle {\int }_0^{2}xdx}\approx \frac{1/2}{2}\left[f\left(0\right)+2\cdot f\left(\frac{1}{2}\right)+2\cdot f\left(1\right)+2\cdot f\left(\frac{3}{2}\right)+f\left(2\right)\right]$

Plugin the values into the equation:

  $\begin{array}{l}{\displaystyle {\int }_0^{2}xdx}=\frac{1}{4}\left[0+2\cdot \left(\frac{1}{2}\right)+2\cdot 1+2\cdot \left(\frac{3}{2}\right)+2\right]\\ =\frac{1}{4}\left[1+2+3+2\right]\\ =\frac{1}{4}\left[8\right]\\ =2\end{array}$

(b)

To determine

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

Answer to Problem 1E

The value must be exact.

Explanation of Solution

Given information:

  ${\displaystyle {\int }_0^{2}xdx}$

First derivative of the function:

  $f\text{'}\left(x\right)=1$

Second derivative of the function:

  $f\text{'}\text{'}\left(x\right)=0$

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.

However,

We have second derivative equal to zero.

That means

The graph is neither concave up nor concave down.

Therefore,

The value must be exact.

(c)

To determine

Integral’s exact value to verify the answer.

Answer to Problem 1E

The value of integral is 2 which is same as using the Trapezoidal rule.

Explanation of Solution

Given information:

  ${\displaystyle {\int }_0^{2}xdx}$

Using FTC (Fundamental Theorem of Calculus):

  $\begin{array}{l}{\displaystyle {\int }_0^{2}xdx}={\frac{x^2}{2}|}_0^2\\ =\frac{{\left(2\right)}^2}{2}-\frac{{\left(0\right)}^2}{2}\\ =\frac{4}{2}-0\\ =2-0\\ =2\end{array}$

Therefore,

Integral’s exact value is 2 which is same as the value of integral obtained in Part (a).