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#?

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Answer 1 · 2015-05-01T15:20:14

Approximate #int_1^3 1/x^2 dx# using trapezoids with #n=5#.

For this problems #a=1#, #b=3#, and #f(x)=1/x^2#

#Delta x = (b-a)/n = (3-1)/5 = 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 = 7/5#, #x_3 = 9/5#, #x_4 = 11/5#, #x_5 = 13/5 = 3 = b#,

The approximation is:

#T = 1/2 Delta x (f(x_0) + 2f(x_1) +2f(x_2) +2f(x_3) +2f(x_4) +f(x_5))#

Plug in the values and do the arithmetic:

#T = 1/2(2/5)(1/(1)^2 + 2(1/(7/5)^2) +2(1/(9/5)^2)+2(1/(11/5)^2) +2(1/(13/5)^2)+(1/(3)^2))#

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

Get a common denominator and simplify.
(Yes, it really was this tedious before electronic calculators.)