Question

If f(x) = 3x2 − 8x, 0 ≤ x ≤ 3, evaluate the Riemann sum with n = 6, taking the sample points to be right endpoints.

Answer 1

Split up the interval [0, 3] into 6 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], [3/2, 2], [2, 5/2], [5/2, 3]

The right endpoints are given by the arithmetic sequence,

$r_i=0+\dfrac i2=\dfrac i2$

with $1\le i\le6$.

We approximate the integral of $f(x)$ on the interval [0, 3] by the Riemann sum,

$\displaystyle\int_0^3f(x)\,\mathrm dx=\sum_{i=1}^6f(r_i)\Delta x_i$

$\displaystyle=\frac{3-0}6\sum_{i=1}^6\left(3{r_i}^2-8r_i\right)$

$\displaystyle=\frac12\sum_{i=1}^6\left(\frac{3i^2}4-4i\right)$

$\displaystyle=\frac38\sum_{i=1}^6i^2-2\sum_{i=1}^6i$

Recall the formulas,

$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}6$

Then the value of the integral is approximately

$\displaystyle=\frac38\cdot\frac{6\cdot7\cdot13}6-2\cdot\frac{6\cdot7}2=\boxed{-\frac{63}8}=-7.875$

Compare to the exact value of the integral, -9.

Answer 2

The Riemann sum of $f(x) = 3\cdot x^{2}-8\cdot x$ with $n = 6$ is $-\frac{63}{8}$.

The formula for the right Riemann sum is described below:

$S = \frac{b-a}{n} \cdot \Sigma\limit_{i= 1}^{n} \,f(x+i\cdot \frac{b-a}{n} )$ (1)

Where:

• $S$ - Riemann sum.
• $a$ - Lower bound.
• $b$ - Upper bound.
• $n$ - Number of segments.
• $i$ - Segment index.

If we know that $f(x) = 3\cdot x^{2}-8\cdot x$, $a = 0$, $b = 3$ and $n = 6$, then the Riemann sum is:

$S = \frac{3-0}{6}\cdot [f(0.5) + f(1) + f(1.5) + f(2) + f(2.5) +f(3)]$

$S = \frac{1}{2}\cdot \left(-\frac{13}{4}-5-\frac{21}{4}-4-\frac{5}{4}+3\right)$

$S = -\frac{63}{8}$

The Riemann sum of $f(x) = 3\cdot x^{2}-8\cdot x$ with $n = 6$ is $-\frac{63}{8}$.

We kindly invite to check this question on Riemann sum: https://brainly.com/question/23960718