Error bound using trapezoidal and Simpson's rule

Question

Trying to help my daughter, she is not sure how to work this.

In estimating the integral -1 to 3 cos(x)dx using trapezoidal and Simpsons rule with n=4, we can estimate the error involved in the approximation using the error bound formulas.

For trapezoidal rule, the error will be less than ?

For Simpson's rule, the error will be less than ?

Give your answers accurate to at least 2 decimal places

Answer 1

sin(3) – sin(–1) is correct but its value is close to 0.98259, not 0.069788.

The trapezoidal rule was also incorrectly stated. If n = 4 is the number of intervals then the rule should be

(cos(–1) + 2cos(0) + 2cos(1) + 2cos(2) + cos(3))/2 ≈ 0.899310.

The conversion from deg to rad should happen in the argument of the trig functions, not the results.

The Simpson 1/3 rule was also incorrectly stated. It should be

(cos(–1) + 4cos(0) + 2cos(1) + 4cos(2) + cos(3))/3 ≈ 0.988776.

Answer 2

Integrating cosx gives sinx

Applying limits

sin3 - sin(-1)

= 0.069788

True value = 0.069788

Trapezoidal rule

= (1/2)h [y₀ + y₁ + y₂ + y₃ + y₄]

where h is the step length

h = x₁ - x₀ or x₂ - x₁

h=1

n=4

from formula

= (1/2)(1)[0.9998 +2(1) + 2(0.9998) + 2(0.9994) + 2(0.9986)]

= 4.4977

converting to rad

4.4977 x (π/180)

= 0.078499rad

Estimated value = 0.078499

Error= True value - Estimated value

Error = -0.00871

Simpson 1/3 rule

= (1/3)h[y₀ + 4y₁ + 2y₂ + 4y₃ + 2y₄]

=(1/3)(1)[0.9998 + 4(1) + 2(0.9998) + 4(0.9994) + 2(0.9986)]

=4.3314

=0.07559719122rad

Error= -0.0058

Simpson 3/8

=(3/8)h[(y₀ + y_n) + 2(y₃ + y₆ + y₉) + 3(y₁ + y₂ + y₄ + y₅)]

=3/8)(1)[(y₀ + y₄) + 2(y₃) + 3(y₁ + y₂)

=(3/8)(1)[(0.9998 + 0.9986) + 2(0.9994) + 3(1+0.9998)]

=3.748725

= 0.065427594rad

Error=True value - Estimated value

Error= 0.00436