Question

Prove Theorem 4.13(c): For any real numbers a and b, ${\int }_a^b{x}^{2}dx=\frac{1}{3}\left(b^3-a^3\right).$Use the proof of Theorem 4.13(a) as a guide.

Short answer

For any real numbers a and b, ${\int }_a^b{x}^{2}dx=\frac{1}{3}\left(b^3-a^3\right)$as follows.

$$\begin{gathered} {\int }_a^{b}xdx=\underset{n\to \infty }{\lim }\sum _{k=1}^n{\left[a+k\left(\frac{b-a}{n}\right)\right]}^2\left(\frac{b-a}{n}\right) \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^n{a}^2\left(\frac{b-a}{n}\right)+\underset{n\to \infty }{\lim }\sum _{k=1}^n{k}^2{\left(\frac{b-a}{n}\right)}^3+\underset{n\to \infty }{\lim }\sum _{k=1}^{n}k{\left(\frac{b-a}{n}\right)}^2 \\ =\frac{1}{3}\left(b^3-a^3\right) \end{gathered}$$

Step-by-step solution

Step 1. Given information  

The given Integral is${\int }_a^b{x}^{2}dx=\frac{1}{3}\left(b^3-a^3\right).$

Step 2. Proof.

Take the interval $\left[a,b\right].$

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

Use the definition of definite integral to find ${\int }_a^b{x}^{2}dx.$

$$\begin{gathered} {\int }_a^{b}xdx=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(\stackrel{*}{x_k}\right)∆x \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^n{\left[a+k\left(\frac{b-a}{n}\right)\right]}^2\left(\frac{b-a}{n}\right) \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^n{a}^2\left(\frac{b-a}{n}\right)+\underset{n\to \infty }{\lim }\sum _{k=1}^n{k}^2{\left(\frac{b-a}{n}\right)}^3+\underset{n\to \infty }{\lim }\sum _{k=1}^{n}k{\left(\frac{b-a}{n}\right)}^2 \\ =\underset{n\to \infty }{\lim }{a}^{2}n\left(\frac{b-a}{n}\right)+\underset{n\to \infty }{\lim }{\left(\frac{b-a}{n}\right)}^3{\left[\frac{n\left(n+1\right)}{2}\right]}^2+\underset{n\to \infty }{\lim }{\left(\frac{b-a}{n}\right)}^2\left[\frac{n\left(n+1\right)}{2}\right] \\ =\frac{1}{3}\left(b^3-a^3\right) \end{gathered}$$

so ${\int }_a^b{x}^{2}dx=\frac{1}{3}\left(b^3-a^3\right)$for any real number a andb.