Let #f(x) = x^3# and compute the Riemann sum of f over the interval [2, 3], n=2 intervals using midpoints?

Question

Evelyn Agard · Accepted Answer

Representing the interval #[2,3]# as #[a,b]#, let #\Delta x=\frac{b-a}{n}=(3-2)/2=0.5#, #x_{0}=a=2#, #x_{1}=x_{0}+\Delta x=2+0.5=2.5#, and #x_{2}=x_{1}+\Delta x=2.5+0.5=3=b#.

Now let #x_{1}^{\star}=2.25#, the midpoint of #[2,2.5]# and #x_{2}^{\star}=2.75#, the midpoint of #[2.5,3]#.

The midpoint Riemann sum approximation in this situation is then

#f(x_{1}^{\star})*\Delta x+f(x_{2}^{\star})*\Delta x=(f(2.25)+f(2.75))*0.5#

#=(2.25^3+2.75^3)*0.5=(11.390625+20.796875)*0.5=16.09375#.

Compare this with the exact answer:

#\int_{2}^{3}x^{3}\ dx=x^{4}/4 |_{x=2}^{x=3}=\frac{81-16}{4}=65/4=16.25#

Ezekiel Alexander · Answer

undefined To compute the Riemann sum of $ f(x) = x^3 $ over the interval $[2, 3]$ with $ n = 2 $ intervals using midpoints, follow these steps:

1. Determine the width of each subinterval:
$$ \Delta x = \frac{b - a}{n} = \frac{3 - 2}{2} = \frac{1}{2} $$

2. Calculate the midpoint of each subinterval:
$$ x_i^* = a + \frac{\Delta x}{2} + i \cdot \Delta x $$
$$ x_1^* = 2 + \frac{1}{2} \cdot \frac{1}{2} = \frac{5}{4} $$
$$ x_2^* = 2 + \frac{1}{2} \cdot \frac{3}{2} = \frac{7}{4} $$

3. Evaluate the function $ f(x) = x^3 $ at each midpoint:
$$ f(x_1^*) = \left(\frac{5}{4}\right)^3 = \frac{125}{64} $$
$$ f(x_2^*) = \left(\frac{7}{4}\right)^3 = \frac{343}{64} $$

4. Compute the Riemann sum using the formula:
$$ \text{Riemann Sum} = \Delta x \cdot \left(f(x_1^*) + f(x_2^*)\right) $$

Substituting the values:
$$ \text{Riemann Sum} = \frac{1}{2} \cdot \left(\frac{125}{64} + \frac{343}{64}\right) = \frac{468}{128} = \frac{117}{32} $$

So, the Riemann sum of $ f(x) = x^3 $ over the interval $[2, 3]$ with $ n = 2 $ intervals using midpoints is $ \frac{117}{32} $.