# How do you use the trapezoidal rule and five sub-intervals find approximation for this integral x=1 and x=3 for 1/x^2 dx?

May 1, 2015

Approximate ${\int}_{1}^{3} \frac{1}{x} ^ 2 \mathrm{dx}$ using trapezoids with $n = 5$.

For this problems $a = 1$, $b = 3$, and $f \left(x\right) = \frac{1}{x} ^ 2$

$\Delta x = \frac{b - a}{n} = \frac{3 - 1}{5} = \frac{2}{5} = 0.4$ (I'll use fractions.)

${x}_{0} = a = 1$, to find, ${x}_{1} , {x}_{2} , . . . {x}_{5}$ start at ${x}_{0} = a = 1$,and add $\Delta x$ successively.

${x}_{0} = a = 1$, ${x}_{2} = \frac{7}{5}$, ${x}_{3} = \frac{9}{5}$, ${x}_{4} = \frac{11}{5}$, ${x}_{5} = \frac{13}{5} = 3 = b$,

The approximation is:

$T = \frac{1}{2} \Delta x \left(f \left({x}_{0}\right) + 2 f \left({x}_{1}\right) + 2 f \left({x}_{2}\right) + 2 f \left({x}_{3}\right) + 2 f \left({x}_{4}\right) + f \left({x}_{5}\right)\right)$

Plug in the values and do the arithmetic:

$T = \frac{1}{2} \left(\frac{2}{5}\right) \left(\frac{1}{1} ^ 2 + 2 \left(\frac{1}{\frac{7}{5}} ^ 2\right) + 2 \left(\frac{1}{\frac{9}{5}} ^ 2\right) + 2 \left(\frac{1}{\frac{11}{5}} ^ 2\right) + 2 \left(\frac{1}{\frac{13}{5}} ^ 2\right) + \left(\frac{1}{3} ^ 2\right)\right)$

= (1/5)(1+2(25/49)+2(25/81)+2(25/121) +2(25/169)

Get a common denominator and simplify.