How do you find the area using the trapezoid approximation method, given #sin (x^2) dx#, on the interval [0, 1/2] using n=4?
1 Answer
Dec 15, 2016
# int_(0)^(1/2) sin(x^2)dx ~~ 0.04274 " " (5dp) #
Explanation:
The values of
Using the trapezoidal rule:
# int_a^bydx ~~ h/2{(y_0+y_n)+2(y_1+y_2+...+y_(n-1))}#
We have:
# int_(0)^(1/2) sin(x^2)dx ~~ 0.125/2 { 0 + 0.2474 + 2(0.01562 + 0.06246 + 0.14016)}#
# " " = 0.0625 { 0.2474 + 2( 0.21825 )} #
# " " = 0.0625 { 0.2474 + 0.43649 } #
# " " = 0.0625 { 0.6839 } #
# " " = 0.04274 #
