How do you use the midpoint rule to approximate the integral #-3x-8x^2dx# from the interval [-1,4] with #n=3#?

Question

Elijah Abram · Accepted Answer

Divide the interval #[-1,4]# into #3# equal width strips.
Find the mid point of each strip and evaluate #f(x)# at that point.
Multiply #f#(strip midpoint#)  xx #strip-width# to get strip area.
Add the strip areas to get approximation of integral. If #[-1,4]# is divided into #3# equal strips then each strip will be
#5/3# wide. The strip mid points will be at
#-0.1667, 1.5,# and # 3.1667# #f(x) = - 3x - 8x^2#
#f(#midpoint of first strip#) = f(-0.1667) = 0.2778#
#f(#midpoint of second strip#) = f(1.5) = - 22.5#
#f(#midpoint of third strip#) = f(3.1667) = - 89.722# Area of first strip #= 0.2778 xx 5/3 = 0.4630#
Area of second strip #= (- 22.5) xx 5/3 = -37.5#
Area of third strip #= (- 89.722) xx 5/3 = -149.537 Total area of strips
#0.4630 + (-37.5) + (-149.537)#
# = - 186.524#
which is our approximation for the integral

Ezekiel Alexander · Answer

undefined To use the midpoint rule to approximate the integral $-3x - 8x^2 \, dx$ from the interval $[-1, 4]$ with $n = 3$, follow these steps:

1. Determine the width of each subinterval: 
   $ \Delta x = \frac{b - a}{n} = \frac{4 - (-1)}{3} = \frac{5}{3} $.

2. Identify the midpoints of each subinterval:
   $ x_1 = -1 + \frac{\Delta x}{2} = -1 + \frac{5/3}{2} = -\frac{1}{3} $,
   $ x_2 = -\frac{1}{3} + \frac{5}{3} = \frac{4}{3} $,
   $ x_3 = \frac{4}{3} + \frac{5}{3} = 3 $.

3. Evaluate the function at each midpoint:
   $ f(x_1) = -3(-1/3) - 8(-1/3)^2 $,
   $ f(x_2) = -3(4/3) - 8(4/3)^2 $,
   $ f(x_3) = -3(3) - 8(3)^2 $.

4. Calculate the sum of these function values multiplied by the width of each subinterval:
   $ \text{Approximation} = \Delta x \left[ f(x_1) + f(x_2) + f(x_3) \right] $.

5. Simplify the expression to find the numerical approximation of the integral.