Question

What is the left endpoint approximation S6 for f (x) = x^3 over [2, 4] ?

A. 45.44
B. 51.00
C. 60.00
D. 64.11

Answer 1

B. 51.00

Explanation:

The Riemann sum is a method by which we can approximate the area under a curve using a series of rectangles.

`\boxed{\begin{minipage}{11cm}\underline{Left Reimann Sum}\\\\$\displaystyle \int^b_a f(x)\; \text{d}x \approx \Delta x \left(f(x_0)+f(x_1)+f(x_2)+...+f(x_(n-2))+f(x_(n-1))\right)$\\\\$\text{where}\; \Delta x=(b-a)/(n)$\\\end{minipage}}`

The left Riemann sum uses the left endpoint of a subinterval to calculate the height of the rectangle on each subinterval.

The number of subintervals, n, is the number of rectangles used.

Given:

• Function: f(x) = x³
• Interval: [2, 4]
  a = 2, b = 4
• Subintervals, n = 6

`\implies \Delta x=(4-2)/(6)=(1)/(3)`

The given partition divides the interval [2, 4] into 6 subintervals:

Therefore, the endpoints are:

`2,\; (7)/(3), \;(8)/(3),\;3,\;(10)/(3),\;(11)/(3),\;4`

Substitute everything into the formula and solve:

`\implies \displaystyle \int^b_a f(x)\; \text{d}x \approx \Delta x \left(f(x_0)+f(x_1)+f(x_2)+...+f(x_(n-2))+f(x_(n-1))\right)`

`\implies \displaystyle \int^4_2 f(x)\; \text{d}x \approx (4-2)/(6)\left(f(x_0)+f(x_1)+f(x_2)+f(x_3)+f(x_4)+f(x_5)\right)`

`\implies \displaystyle \int^4_2 f(x)\; \text{d}x \approx (1)/(3)\left(f(2)+f\left((7)/(3)\right)+f\left((8)/(3)\right)+f(3)+f\left((10)/(3)\right)+f\left((11)/(3)\right)\right)`

`\implies \displaystyle \int^4_2 f(x)\; \text{d}x \approx (1)/(3)\left((2)^3+\left((7)/(3)\right)^3+\left((8)/(3)\right)^3+(3)^3+\left((10)/(3)\right)^3+\left((11)/(3)\right)^3\right)`

`\implies \displaystyle \int^4_2 f(x)\; \text{d}x \approx (1)/(3)\left(153\right)`

`\implies \displaystyle \int^4_2 f(x)\; \text{d}x \approx 51`