Question

1. Evaluate the Riemann sum for (x) = x3 − 6x, for 0 ≤ x ≤ 3 with six subintervals, taking the sample points, xi, to be the right endpoint of each interval. Give three decimal places in your answer.

2. Explain, using a graph of f(x), what the Riemann sum in Question #1 represents.

3. Express the given integral as the limit of a Riemann sum but do not evaluate: the integral from 0 to 3 of the quantity x cubed minus 6 times x, dx.

4. Use the Fundamental Theorem to evaluate the integral from 0 to 3 of the quantity x cubed minus 6 times x, dx.
(Your answer must include the antiderivative.)

5. Use a graph of the function to explain the geometric meaning of the value of the integral.

Answer 1

1. Dividing the interval [0, 3] into 6 intervals gives us the partition

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

Each subinterval has length 1/2. The right endpoints are then

$\left\{\dfrac12,1,\dfrac32,\ldots,3\right\}$

which are given by the sequence

$x_i=\dfrac i2\text{ for }1\le i\le6$

Then the integral is approximated by the Riemann sum,

$\displaystyle\int_0^3x^3-6x\,\mathrm dx\approx\sum_{i=1}^6\frac{{x_i}^3-6x_i}2=\dfrac{-\frac{23}8-5-\frac{45}8-4+\frac58+9}2=-\frac{63}{16}\approx-3.938$

2. The Riemann sum can be represented by as the sum of the areas of rectangles whose dimensions are determined by the chosen partition and sample points in order to approximate the area between the curve $f(x)$ and the $x$-axis.

3. With $n$ subintervals, we get the partition

$\left[0,\dfrac3n\right],\left[\dfrac3n,\dfrac6n\right],\left[\dfrac6n,\dfrac9n\right],\ldots,\left[\dfrac{3(n-1)}n,3\right]$

Each subinterval has length $\dfrac3n$, and the (right-endpoint) Riemann sum is

$\displaystyle\sum_{i=1}^n\frac3n\left(\left(\frac{3i}n\right)^3-6\left(\frac{3i}n\right)\right)$

$=\displaystyle\frac{27}{n^4}\sum_{i=1}^n3i^3-2in^2$

4. First compute the antiderivative:

$\displaystyle\int x^3-6x\,\mathrm dx=\dfrac{x^4}4-3x^2+C$

Then by the FTC, the definite integral is

$\displaystyle\int_0^3x^3-6x\,\mathrm dx=\left(\frac{3^4}4-3^3\right)-\left(\frac04-0\right)=-\dfrac{27}4$

5. The integral gives the exact area of the bounded region.