1. ## Question about area under a curve using midpoint rule

okay, the question says to use the midpoint rule with n=4 to approximate the area of the region bounded by the graph of the function and the x axis over the indicated interval

function is f(x)= x^2 + 4x interval is [0,4]

midpoint rule

sigma from i=1 to n of f((xi-xi-1)/2) times delta x

thanks very much

2. Originally Posted by doctorgk
okay, the question says to use the midpoint rule with n=4 to approximate the area of the region bounded by the graph of the function and the x axis over the indicated interval

function is f(x)= x^2 + 4x interval is [0,4]

midpoint rule

sigma from i=1 to n of f((xi-xi-1)/2) times delta x

thanks very much
why is this in the pre-calc section?

divide [0,4] into 4 equal sub-intervals, we get the intervals: $\displaystyle [0,1],~[1,2],~[2,3],~[3,4]$

obviously (i hope) we have $\displaystyle \Delta x = 1$ here. if you're not sure, remember $\displaystyle \Delta x = \frac {b - a}{n}$ where [a,b] is the interval we are considering, and n, of course, is the number of sub-intervals

the midpoints of the subintervals respectively are: $\displaystyle x_1 = \frac 12,~x_2 = \frac 32,~ x_3 = \frac 52,~x_4 = \frac 72$

so, by the midpoint rule, our estimate for the integral is given by:

$\displaystyle \int_0^4f(x)~dx = \sum_{i = 1}^{4} f(x_i) \Delta x = \Delta x \left( f \left( \frac 12 \right) + f \left( \frac 32 \right) + f \left( \frac 52 \right) + f \left( \frac 72 \right) \right)$

$\displaystyle = f \left( \frac 12 \right) + f \left( \frac 32 \right) + f \left( \frac 52 \right) + f \left( \frac 72 \right)$

i leave the rest to you