How do you find the area using the trapezoidal approximation method, given #f(x)=x^2 -1#, on the interval [2,4] with n=8?

1 Answer
Mar 14, 2017

Answer:

Trapezium rule gives:

# int_2^4 \ x^2-1 \ dx ~~ 16.69 # (2dp)

Explanation:

The values of #f(x)=x^2-1# are tabulated as follows (using Excel) working to 6dp

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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_2^4 \ x^2-1 \ dx #
# " " ~~ 0.25/2 { (3 + 15) + 2(4.0625 + 5.25 + 6.5625 + #
# " " 8 + 9.5625 + 11.25 + 13.0625) } #
# " " = 0.125 { 18 + 2(57.75) } #
# " " = 0.125 { 18 + 115.5 } #
# " " = 0.125 { 133.5 } #
# " " = 16.6875 #

Let's compare this to the exact value:

# int_2^4 \ x^2-1 \ dx = [x^3/3-x]_2^4 #
# " " = (64/3-4)-(8/3-2) #
# " " = 50/3 #
# " " = 16.666666 ... #