How do you use the trapezoidal rule to approximate the Integral from 0 to 0.5 of #(1-x^2)^0.5 dx# with n=4 intervals?

Question

Cameron Alexander · Accepted Answer

See the explanation section, below.

On #[a,b]=[0, 0.5]#, with #n=4#, we get #Delta x = (b-a)/n = (0.5-0)/4 = 0.125# The endpoints of the subintervals are found by beginning at #a=0# and successively adding #Delta x# to find the points. #x_0 = 0#, #x_1 = 0.125#, #x_2 = 0.250#, #x_3 = 0.375#,, #x_4 = 0.5# Now apply the formula (do the arithmetic): #T_4=(Deltax)/2 [f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+f(x_4)] #

Emilia Aguilar · Answer

undefined To use the trapezoidal rule to approximate the integral of $ \sqrt{1-x^2} $ from 0 to 0.5 with $ n = 4 $ intervals, follow these steps:

1. Divide the interval [0, 0.5] into 4 equal subintervals. Since $ n = 4 $, the width of each subinterval is $ \Delta x = \frac{0.5 - 0}{4} = 0.125 $.

2. Compute the function values at the endpoints of each subinterval: $ f(x_0), f(x_1), \ldots, f(x_4) $, where $ x_0 = 0 $, $ x_1 = 0.125 $, $ x_2 = 0.25 $, $ x_3 = 0.375 $, and $ x_4 = 0.5 $.

3. Apply the trapezoidal rule formula for each subinterval: 
$$ T_i = \frac{\Delta x}{2} \left( f(x_i) + f(x_{i+1}) \right) $$

4. Sum up all the individual trapezoids:
$$ \text{Approximation} = T_0 + T_1 + T_2 + T_3 $$

5. Calculate the numerical value of the approximation.