Question

- notaion, to evaluate the integral. Also give a sketch of the region of integration. Please use proper #1. Use the limit-sum definition notation and give details. 2 – 3x) dx (15 points)

Answer 1

Step 1)

we know that,

$\small \int _a^bf(x)dx = \lim_{n\rightarrow \infty }\sum _{i=1}^nf(x_i)\triangle x$

where,

$\small \triangle x = \frac{b-a}{n}$

$\small x_i = a+i\triangle x$

we have to evaluate,

$\small \int _{-1}^1(x^2-3x)dx$

Hence we have,

$\small f(x)=x^2-3x, a = -1 \ and\ b = 1$

we can write,

$\small \triangle x = \frac{1-(-1)}{n} = \frac{1+1}{n} = \frac{2}{n}$

$\small x_i=-1+i\left ( \frac{2}{n} \right ) = -1+\frac{2i}{n}$

Hence,

$\small f(x_i)=f\left (-1+\frac{2i}{n} \right )$

$\small f(x_i)=\left (-1+\frac{2i}{n} \right )^2-3\left ( -1+\frac{2i}{n} \right )$

$\small f(x_i)=(-1)^2-\frac{4i}{n}+\left ( \frac{2i}{n} \right )^2+3-\frac{6i}{n}$

$\small f(x_i)=1-\frac{10i}{n}+\frac{4i^2}{n^2}+3$

$\small f(x_i)=4-\frac{10i}{n}+\frac{4i^2}{n^2}$

using formula 1) we can write,

$\small \int _{-1}^1(x^2-3x)dx = \lim_{n\rightarrow \infty }\sum _{i=1}^n\left ( 4-\frac{10i}{n}+\frac{4i^2}{n^2} \right )\left ( \frac{2}{n} \right )$

$\small \int _{-1}^1(x^2-3x)dx = \lim_{n\rightarrow \infty }\sum _{i=1}^n\left ( \frac{8}{n}-\frac{20i}{n^2}+\frac{8i^2}{n^3} \right )$

$\small \int _{-1}^1(x^2-3x)dx = \lim_{n\rightarrow \infty }\left ( \sum _{i=1}^n\frac{8}{n}-\sum _{i=1}^n\frac{20i}{n^2}+\sum _{i=1}^n\frac{8i^2}{n^3} \right )$

$\small \int _{-1}^1(x^2-3x)dx = \lim_{n\rightarrow \infty }\left (\frac{8}{n}\sum _{i=1}^n1-\frac{20}{n^2}\sum _{i=1}^ni+\frac{8}{n^3}\sum _{i=1}^ni^2 \right )$

we know that,

$\small \sum _{i=1}^n1 = n, \ \sum _{i=1}^ni= \frac{n(n+1)}{2}\ and \ \sum _{i=1}^ni^2 = \frac{n(n+1)(2n+1)}{6}$

Hence,

$\small \int _{-1}^1(x^2-3x)dx = \lim_{n\rightarrow \infty }\left (\frac{8}{n}\cdot n-\frac{20}{n^2}\cdot \frac{n(n+1)}{2}+\frac{8}{n^3}\cdot \frac{n(n+1)(2n+1)}{6} \right )$

$\small \int _{-1}^1(x^2-3x)dx = \lim_{n\rightarrow \infty }\left (8-\frac{10n^2\left ( 1+\frac{1}{n} \right )}{n^2}+\frac{4n^3\left ( 1+\frac{1}{n} \right )\left ( 2+\frac{1}{n} \right )}{3n^3} \right )$

$\small \int _{-1}^1(x^2-3x)dx = \lim_{n\rightarrow \infty }\left (8-10\left ( 1+\frac{1}{n} \right )+\frac{4\left ( 1+\frac{1}{n} \right )\left ( 2+\frac{1}{n} \right )}{3} \right )$

$\small \int _{-1}^1(x^2-3x)dx = 8-10\left ( 1+\frac{1}{\infty} \right )+\frac{4\left ( 1+\frac{1}{\infty} \right )\left ( 2+\frac{1}{\infty} \right )}{3}$

$\small \int _{-1}^1(x^2-3x)dx = 8-10\left ( 1+0 \right )+\frac{4\left ( 1+0\right )\left ( 2+0 \right )}{3}$

$\small \int _{-1}^1(x^2-3x)dx = 8-10\left ( 1 \right )+\frac{4\left ( 1\right )\left ( 2 \right )}{3}$

$\small \int _{-1}^1(x^2-3x)dx = 8-10+\frac{8}{3}$

$\small \int _{-1}^1(x^2-3x)dx = -2+\frac{8}{3}$

$\small \int _{-1}^1(x^2-3x)dx =\frac{-6+8}{3}$

$\small \int _{-1}^1(x^2-3x)dx =\frac{2}{3}$

Step 2)

we have,

$\small \int _{-1}^1(x^2-3x)dx$

Hence we can say that,

x ranges from x = -1 to x = 1 it means region of integration is the region between the lines x = -1 and x = 1

we can graph the region of integration as below:

X =