Question

No 13 and 14

(a) Find an approximation to the integral [** (x2 - 4x) dx using a Riemann sum with night endpoints and n = 8. Rg (b) iffis i

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Answer #1

(a)

\text{ To approximate integral }\int_0^2 (x^2-4x)dx

Here, a = 0,b=2, f(2) f(0) = 22 – 4.0

\text{ Given, }n=8

\text{ Width of each sub interval, }\bigtriangleup x =\frac{b-a}{n}=\frac{2-0}{8}=\frac{1}{4}

Sub intervals are

\left[0, \frac{1}{4}\right], \left[\frac{1}{4}, \frac{1}{2}\right], \left[\frac{1}{2}, \frac{3}{4}\right], \left[\frac{3}{4}, 1\right], \left[1, \frac{5}{4}\right], \left[\frac{5}{4}, \frac{3}{2}\right], \left[\frac{3}{2}, \frac{7}{4}\right], \left[\frac{7}{4}, 2\right]

\text{ Right end points of sub intervals are }

\frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1, \frac{5}{4}, \frac{3}{2}, \frac{7}{4}, 2

\text{ Approximating area using Reimann sum }

R_8=\bigtriangleup x \left ( f\left ( \frac{1}{4} \right )+f \left (\frac{1}{2} \right )+ f \left (\frac{3}{4} \right )+f(1)+f\left ( \frac{5}{4} \right )+ f \left ( \frac{3}{2} \right ) +f \left ( \frac{7}{4} \right )+f( 2) \right )

=> R_8=\bigtriangleup x \left (\left(\left(\frac{1}{4}\right)^2-4\left(\frac{1}{4}\right)\right)+\left(\left(\frac{1}{2}\right)^2-4\left(\frac{1}{2}\right)\right)+\left(\left(\frac{3}{4}\right)^2-4\left(\frac{3}{4}\right)\right)+\left(\left(1\right)^2-4\left(1\right)\right)+\left(\left(\frac{5}{4}\right)^2-4\left(\frac{5}{4}\right)\right)+\left(\left(\frac{3}{2}\right)^2-4\left(\frac{3}{2}\right)\right)+\left(\left(\frac{7}{4}\right)^2-4\left(\frac{7}{4}\right)\right)+\left(\left(2\right)^2-4\left(2\right)\right) \right )

=>R_8= \left ( \frac{1}{4} \right ) \left ( -23.25 \right )

=> R_8= (x^2-4x)dx \approx -5.8125

(b)

\int_a^bf(x)dx =\lim_{n\to \infty} \sum_{i=1}^n f(x_i)\bigtriangleup x

\bigtriangleup x =\frac{b-a}{n}

x_i =a+i\bigtriangleup x

\text{ To approximate integral }\int_0^2 (x^2-4x)dx

Here, a = 0,b=2, f(2) f(0) = 22 – 4.0

\bigtriangleup x =\frac{2-0}{n}=\frac{2}{n}

x_i =0+i \left ( \frac{2}{n} \right ) =\frac{2i}{n}

f(x_i) =x_i^2-4x_i =\left ( \frac{2i}{n} \right )^2-4\left ( \frac{2i}{n} \right )

\int_0^2 (x^2-4x)dx =\lim_{n\to \infty} \sum_{i=1}^n \left (\left ( \frac{2i}{n} \right )^2-4\left ( \frac{2i}{n} \right )\right )\left ( \frac{2}{n} \right )

=> \int_0^2 (x^2-4x)dx =\lim_{n\to \infty} \sum_{i=1}^n \left ( \frac{8i^2}{n^3} - \frac{16i}{n^2} \right )

=> \int_0^2 (x^2-4x)dx =\lim_{n\to \infty} \left (\frac{8}{n^3 } \sum_{i=1}^n \left (i^2 \right )-\frac{16}{n^2 }\sum_{i=1}^n \left ( i \right ) \right )

=> \int_0^2 (x^2-4x)dx =\lim_{n\to \infty} \left (\frac{8}{n^3 } \left ( \frac{n(n+1)(2n+1)}{6} \right )-\frac{16}{n^2 }\left ( \frac{n(n+1)}{2} \right ) \right )

=> \int_0^2 (x^2-4x)dx =\lim_{n\to \infty} \left ( \frac{ 4n^3\left ( 1+\frac{1}{n} \right )\left (2+\frac{1}{n} \right )}{3n^3}- \frac{ 8n^2\left ( 1+\frac{1}{n} \right )}{n^2} \right )

=> \int_0^2 (x^2-4x)dx =\lim_{n\to \infty} \left ( \frac{ 4\left ( 1+\frac{1}{n} \right )\left (2+\frac{1}{n} \right )}{3}- 8\left ( 1+\frac{1}{n} \right ) \right )

=> \int_0^2 (x^2-4x)dx =\left ( \frac{ 4\left ( 1+0\right )\left (2+0 \right )}{3}- 8\left ( 1+0 \right ) \right )

=> \int_0^2 (x^2-4x)dx =\left ( \frac{ 8}{3}- 8 \right )

=> \int_0^2 (x^2-4x)dx = \frac{ -16}{3} =-5.3333

Hence,

\int_0^2 (x^2-4x)dx=-5.3333

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