How do you find the area using the trapezoid approximation method, given sin (x^2) dxsin(x2)dx, on the interval [0, 1/2] using n=4?

1 Answer
Steve M
Dec 15, 2016

int_(0)^(1/2) sin(x^2)dx ~~ 0.04274 " " (5dp) ∫120sin(x2)dx≈0.04274 (5dp)

Explanation:

The values of y=sin(x^2)y=sin(x2) 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_(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

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