Q. 18 - Calculus

Question

In the proof of the Fundamental Theorem of Calculus, the Mean Value Theorem is used to choose values $x_k^{*}$ in each subinterval $\left[x_{k-1},x_k\right]$. Use the Mean Value Theorem in the same way to find the corresponding values $x_k^{*}$ for a Riemann sum approximation of ${\int }_0^4{x}^{2}dx$ with four rectangles. 

Step-by-Step Solution

Answer

The required ${\int }_0^2 x^{2}dx=3.75$

Step 1. Given information

Given value is $x_k^{*}$

Step 2.

Mean value theorem  with values $x_k^{*}$

A function f is an integrable on $\left[a,b\right]$

The approximation for the sum of small rectangles on the interval can be calculated as given below

$A\left(x\right)=\underset{k=1}{\overset{n}{\sum f\left(x_k\right)∆x}}$

Where,$x_k=a+k\frac{\left(b-a\right)}{n} and ,∆x=\frac{b-a}{n}$

The area on the interval $\left[a,b\right]$

           $={\int }_a^{b}f\left(x\right)dx$

Thus,

$\sum _{k=1}^{n}f\left(x_k\right)∆x={\int }_a^{b}f\left(x\right) dx$

By applying the theorem the problem given,

when $n=4 and f\left(x\right)=x^2$

$$\begin{gathered} ∆x=\frac{b-a}{n} \\ ∆x=\frac{2-0}{4} \\ ∆x=\frac{1}{2} \end{gathered}$$

And,

$$\begin{gathered} x_k=a+k\frac{\left(b-a\right)}{n} \\ {x}_k=0+k\frac{1}{2} \\ {x}_k=\frac{k}{2} \end{gathered}$$

Now,

$$\begin{gathered} \sum _{k=1}^4 f\left(x_k\right)∆x=\sum _{k=1}^4\frac{k^2}{4}\left(\frac{1}{2}\right) \\ =\frac{1}{8}+\frac{4}{8}+\frac{9}{8}+\frac{16}{8} \\ =\frac{30}{8} \\ =3.75 \end{gathered}$$

Thus,

$$\begin{gathered} A\left(x\right)=\sum _{k=1}^{4}f\left(x_k\right)∆x \\ ={\int }_0^2 x^{2}dx \\ =3.75 \end{gathered}$$