Question

How do you find the area using the trapezoidal approximation method, given `cos(x^2)`, on the interval [0, 1] with n=8?

Answer 1

and so we can conclude that `int_(0)^(1) cos(x^2)dx ~~ 0.90233` to 5dp

Explanation:

The values of `y=cos(x^2)` are tabulated as follows (using Excel) working to 5dp

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) cos(x^2)dx ~~ 0.125/2 { 1 + 0.5403`
`" " + 2(0.99988 + 0.99805 + 0.99013`
`" " + 0.96891 + 0.92467 + 0.84592`
`" " + 0.72095)}`
`" " = 0.0625 { 1.5403 + 2( 6.44851 )}`
`" " = 0.0625 { 1.5403 + 12.89702 }`
`" " = 0.0625 { 14.43733 }`
`" " = 0.90233`

and so we can conclude that `y=cos(x^2) = 0.90233` to 5dp#