Question

Approximate the area under the

function between a and b using a
left-hand sum with the given
number of intervals.
f(x) = x² + 2

a = 0
b= 6
6 intervals

Answer 1

Split up the interval [0, 6] into 6 equally spaced subintervals of length
`\Delta x = \frac{6-0}6 = 1`. So we have the partition

[0, 1] U [1, 2] U [2, 3] U [3, 4] U [4, 5] U [5, 6]

where the left endpoint of the
`i`-th interval is

`\ell_i = i - 1`

with
`i\in\{1,2,3,4,5,6\}`.

The area under
`f(x)=x^2+2` on the interval [0, 6] is then given by the definite integral and approximated by the Riemann sum,

`\displaystyle \int_0^6 f(x) \, dx \approx \sum_(i=1)^6 f(\ell_i) \Delta x \\\\ ~~~~~~~~ = \sum_(i=1)^6 \bigg((i-1)^2 + 2\bigg) \\\\ ~~~~~~~~ = \sum_(i=1)^6 \bigg(i^2 - 2i + 3\bigg) \\\\ ~~~~~~~~ = \frac{6\cdot7\cdot13}6 - 6\cdot7 + 3\cdot6 = \boxed{67}`

where we use the well-known sums,

`\displaystyle \sum_(i=1)^n 1 = \underbrace{1 + 1 + \cdots + 1}_(n\,\rm times) = n`

`\displaystyle \sum_(i=1)^n i = 1 + 2 + \cdots + n = \frac{n(n+1)}2`

`\displaystyle \sum_(i=1)^n i^2 = 1 + 4 + \cdots + n^2 = \frac{n(n+1)(2n+1)}6`

Answer 2

67 square units

Explanation:

The area using the left-hand sum is the sum of products of the function value at the left side of the interval and the width of the interval.

Area

The attachment shows a table of the x-value at the left side of each interval, and the corresponding function value there. The interval width is 1 unit in every case, so the desired area is simply the sum of the function values.

The approximate area is 67 square units.