How do you find the area using the trapezoid approximation method, given #1/x^2 dx #, on the interval [1,3] with n=5?

1 Answer
May 21, 2017

Answer:

Trapezium rule gives:

# int_1^3 \ 1/x^2 \ dx ~~ 0.69 # (2dp)

Explanation:

The values of #f(x)=1/x^2-1# are tabulated as follows (using Excel) working to 5dp

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Using the trapezoidal rule:

# int_a^bydx ~~ h/2{(y_0+y_n)+2(y_1+y_2+...+y_(n-1))}#

We have:

# int_1^3 \ 1/x^2 \ dx #
# " " ~~ 0.4/2 { 1 + 0.11111 + 2(0.5102 + 0.30864 + 0.20661 + 0.14793) } #

# " " = 0.2 { 1.11111 + 2(1.17339) } #
# " " = 0.2 { 1.11111 + 2.34677 } #
# " " = 0.2 * 3.45788 #
# " " = 0.69158 #

Let's compare this to the exact value:

# int_1^3 \ 1/x \ dx = [-1/x]_1^3 #
# " " = (-1/3) - (-1) #
# " " = 2/3 #