How do you use the trapezoid rule for #int 2 sin x^2 dx# from x = 0 to x = 1/2 with n = 4?

Question

Charles Arocha · Accepted Answer

#int_0^(1/2) 2 sin x^2 dx~~T_4~~0.114489# First of all let's calculate 4 sub-intervals of [0,1/2]: #[0,1/8],[1/8,1/4],[1/4,3/8],[3/8,1/2]# Now, let's apply the trapezoidal rule to these interval: #T_4=[2sin0^2+2*2sin(1/8)^2+2*2sin(1/4)^2+2*2sin(3/8)^2+2sin(1/2)^2]*((1/2-0)/8)# #T_4=[2*0+4sin(1/64)+4sin(1/16)+4sin(9/64)+2sin(1/4)]*1/16# #T_4~~[0.062497+0.249837+0.560647+0.958851]*1/16# #T_4~~[1.831832]*1/16=0.114489#

Christopher Agresta · Answer

undefined To use the trapezoid rule for evaluating the integral ∫2sin(x^2) dx from $x = 0$ to $x = \frac{1}{2}$ with $n = 4$, follow these steps:

1. Determine the step size, $h$, which is given by:
   $h = \frac{b - a}{n}$, where $a$ is the lower limit of integration ($0$), $b$ is the upper limit of integration ($\frac{1}{2}$), and $n$ is the number of subintervals ($4$).
   
2. Calculate $h$:
   $h = \frac{\frac{1}{2} - 0}{4} = \frac{1}{8}$.
   
3. Compute the function values at the endpoints and interior points of the subintervals. Since $n = 4$, there are $n + 1 = 5$ points to evaluate.

4. Evaluate the function at the points $x_0, x_1, x_2, x_3, x_4$:
   $x_0 = 0$, $x_1 = \frac{1}{8}$, $x_2 = \frac{2}{8}$, $x_3 = \frac{3}{8}$, $x_4 = \frac{1}{2}$.
   Then compute $y_0 = 2\sin(0^2)$, $y_1 = 2\sin\left(\left(\frac{1}{8}\right)^2\right)$, $y_2 = 2\sin\left(\left(\frac{2}{8}\right)^2\right)$, $y_3 = 2\sin\left(\left(\frac{3}{8}\right)^2\right)$, $y_4 = 2\sin\left(\left(\frac{1}{2}\right)^2\right)$.

5. Apply the trapezoid rule formula:
   $T = \frac{h}{2} \left[ y_0 + 2(y_1 + y_2 + y_3) + y_4 \right]$.

6. Substitute the values obtained in step 4 into the trapezoid rule formula and calculate $T$.

That's the approximation of the integral using the trapezoid rule with $n = 4$.