Question

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

Short answer

For any real number a and b, ${\int }_a^{b}xdx=\frac{1}{2}\left(b^2-a^2\right)$ is verified as,

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

Step-by-step solution

Step 1. Given information 

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

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 ${\int }_a^{b}xdx.$

role="math" localid="1648579869591" $$\begin{gathered} {\int }_a^{b}xdx=\underset{n\to \infty }{\lim }\sum _{k=1}^{n}f\left(\stackrel{*}{x_k}\right)∆x \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^n\left[a+k\left(\frac{b-a}{n}\right)\right]\left(\frac{b-a}{n}\right) \\ =\underset{n\to \infty }{\lim }\sum _{k=1}^{n}a\left(\frac{b-a}{n}\right)+\underset{n\to \infty }{\lim }\sum _{k=1}^{n}k{\left(\frac{b-a}{n}\right)}^2 \\ =\underset{n\to \infty }{\lim }an\left(\frac{b-a}{n}\right)+\underset{n\to \infty }{\lim }{\left(\frac{b-a}{n}\right)}^2\left[\frac{n\left(n+1\right)}{2}\right] \\ =\frac{1}{2}\left(b^2-a^2\right) \end{gathered}$$

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