Question

Integral (x^2)dx upper limit b and lower limit a by riemann’s sum

Answer 1

Answer:

Amazing fact #1: This limit really gives us the exact value of ∫ 2 6 1 5 x 2 d x \displaystyle\ int_2^6 \dfrac15 x^2\,dx ∫2651x2dx ...

Answer 2

We're asked to find integral of $x^2$ wrt $x$ limiting from $a$ to $b$ by definition of Riemann Sum.

A definite integral can be written as Riemann Sum as,

$\displaystyle\longrightarrow\int\limits_a^b f(x)\ dx=\lim_{n\to\infty}\sum_{r=0}^nf(a+r\Delta x)\cdot\Delta x$

where $\Delta x=\dfrac{b-a}{n}.$

In this question $f(x)=x^2.$ Then,

$\displaystyle\longrightarrow\int\limits_a^b x^2\ dx=\lim_{n\to\infty}\sum_{r=0}^n(a+r\Delta x)^2\cdot\Delta x$

$\displaystyle\longrightarrow\int\limits_a^b x^2\ dx=(b-a)\lim_{n\to\infty}\sum_{r=0}^n\left(a+\dfrac{r(b-a)}{n}\right)^2\cdot\dfrac{1}{n}$

$\displaystyle\longrightarrow\int\limits_a^b x^2\ dx=(b-a)\lim_{n\to\infty}\dfrac{1}{n}\sum_{r=0}^n\left(a^2+\dfrac{2ar(b-a)}{n}+\dfrac{r^2(b-a)^2}{n^2}\right)$

$\displaystyle\longrightarrow\int\limits_a^b x^2\ dx=(b-a)\lim_{n\to\infty}\dfrac{1}{n}\left[\sum_{r=0}^na^2+\dfrac{2a(b-a)}{n}\sum_{r=0}^nr+\dfrac{(b-a)^2}{n^2}\sum_{r=0}^nr^2\right]$

$\displaystyle\begin{aligned}\longrightarrow\int\limits_a^b x^2\ dx=\ \ (b-a)\lim_{n\to\infty}\dfrac{1}{n}\left[a^2&(n+1)+\dfrac{2a(b-a)}{n}\cdot\dfrac{n(n+1)}{2}\\&+\dfrac{(b-a)^2}{n^2}\cdot\dfrac{n(n+1)(2n+1)}{6}\bigg]\end{aligned}$

$\displaystyle\longrightarrow\int\limits_a^b x^2\ dx=(b-a)\lim_{n\to\infty}\dfrac{1}{n}\left[a^2(n+1)+\dfrac{2a(b-a)(n+1)}{2}+\dfrac{(b-a)^2(n+1)(2n+1)}{6n}\right]$

$\begin{aligned}\displaystyle\longrightarrow\int\limits_a^b x^2\ dx\ \ &=(b-a)\left[\lim_{n\to\infty}\dfrac{a^2(n+1)}{n}+\lim_{n\to\infty}\dfrac{2a(b-a)(n+1)}{2n}\\&+\lim_{n\to\infty}\dfrac{(b-a)^2(n+1)(2n+1)}{6n^2}\bigg]\end{aligned}$

$\begin{aligned}\displaystyle\longrightarrow\int\limits_a^b x^2\ dx\ \ &=(b-a)\left[a^2\lim_{n\to\infty}\dfrac{n+1}{n}+a(b-a)\lim_{n\to\infty}\dfrac{n+1}{n}\\&+\dfrac{(b-a)^2}{6}\lim_{n\to\infty}\dfrac{n+1}{n}\cdot\dfrac{2n+1}{n}\bigg]\end{aligned}$

$\begin{aligned}\displaystyle\longrightarrow\int\limits_a^b x^2\ dx\ \ &=(b-a)\left[a^2\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)+a(b-a)\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)\\&+\dfrac{(b-a)^2}{6}\lim_{n\to\infty}\left(1+\dfrac{1}{n}\right)\left(2+\dfrac{1}{n}\right)\bigg]\end{aligned}$

As $\displaystyle\lim_{n\to\infty}\dfrac{1}{n}=0,$

$\displaystyle\longrightarrow\int\limits_a^b x^2\ dx=(b-a)\left[a^2(1+0)+a(b-a)(1+0)+\dfrac{(b-a)^2}{6}\left(1+0\right)\left(2+0\right)\bigg]$

$\displaystyle\longrightarrow\int\limits_a^b x^2\ dx=(b-a)\left[a^2+ab-a^2+\dfrac{(b-a)^2}{3}\bigg]$

$\displaystyle\longrightarrow\int\limits_a^b x^2\ dx=(b-a)\left[ab+\dfrac{(b-a)^2}{3}\bigg]$

$\displaystyle\longrightarrow\int\limits_a^b x^2\ dx=\dfrac{(b-a)(b^2+ab+a^2)}{3}$

$\displaystyle\longrightarrow\underline{\underline{\int\limits_a^b x^2\ dx=\dfrac{b^3-a^3}{3}}}$