Using n=4 trapezoids, how do you approximate the value of #int sqrt(x+1) dx# from [1,3]?

Question

Cameron Alexander · Accepted Answer

To approximate #int_1^3sqrt(x+1) dx# by a series of 4 trapezoids
we need to evaluate #sqrt(x+1)# at 5 points #x={1.0, 1.5, 2.0, 2.5, 3.0}#
(using a constant #Delta x# of #0.5#) This gives us 4 trapezoidal areas
#A_(T1) = ((sqrt(2)+sqrt(2.5))/2)xx0.5#
#A_(T2) = ((sqrt(2.5)+sqrt(3.0))/2)xx0.5#
#A_(T3) = ((sqrt(3.0)+sqrt(3.5))/2)xx0.5#
#A_(T4) = ((sqrt(3.5)+sqrt(4.0))/2)xx0.5# Adding together the 4 trapezoidal areas (with the aid of a calculator) gives an approximation of the integral #=3.445563#

William Abdalla · Answer

undefined To approximate the value of $ \int_{1}^{3} \sqrt{x + 1} \, dx $ using $ n = 4 $ trapezoids:

1. Divide the interval $[1, 3]$ into $ n = 4 $ equal subintervals. Since we have 4 trapezoids, each subinterval width will be $ \Delta x = \frac{3 - 1}{4} = 0.5 $.

2. Calculate the function values at the endpoints of each subinterval. In this case, calculate $ \sqrt{x + 1} $ at $ x = 1, 1.5, 2, 2.5, 3 $.

3. Use the trapezoidal rule formula to find the area of each trapezoid:
   $ A_i = \frac{h}{2} [f(x_i) + f(x_{i+1})] $, where $ h $ is the width of each subinterval, $ f(x_i) $ is the function value at the left endpoint, and $ f(x_{i+1}) $ is the function value at the right endpoint.

4. Sum up the areas of all the trapezoids to get the approximate value of the integral:
   $ \int_{1}^{3} \sqrt{x + 1} \, dx \approx A_1 + A_2 + A_3 + A_4 $.

5. Calculate each $ A_i $ using the formula from step 3.

6. Add up the values of $ A_i $ obtained in step 5 to get the final approximation of the integral.