Question

use limits to find the area between the curve of y = x² and the x-axis for the interval from x = 1 to x = 3.

Answer 1

Split up the interval [1, 3] into n equally spaced subintervals. Each one has length (3 - 1)/n = 2/n, so we're talking about the partition

[1, 1 + 2/n] U [1 + 2/n, 1 + 4/n] U [1 + 4/n, 1 + 6/n] U … U [1 + 2 (n - 1)/n, 3]

Let f(x) = x². Consider the left endpoints of each subinterval, given by the sequence x(i) = 1 + 2 (i - 1)/n for 1 ≤ i ≤ n. Then the area under the graph of f(x) over [1, 3] is approximated by the Riemann sum,

`\displaystyle S_n = \sum_(i=1)^n f(x(i)) \frac2n \\\\ S_n = \frac2n \sum_(i=1)^n \left(1 + \frac{2(i - 1)}n\right)^2 \\\\ S_n = \frac2{n^3} \sum_(i=1)^n \left(n^2 - 4n + 4 + (4n - 8)i + 4i^2\right)`

To compute the sum, recall the formulas,

`\displaystyle \sum_(i=1)^n 1 = n`

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

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

Then

`\displaystyle S_n = \frac2{n^3} \left( n(n^2-4n+4) + \frac{n(n+1)(4n-8)}2 + \frac{2n(n+1)(2n+1)}3\right) \\\\ S_n = (26n^2 - 24n + 4)/(3n^2)`

and the exact area under the curve over [1, 3] is

`\displaystyle \int_1^3 x^2 \, dx = \lim_(n\to\infty) S_n = \boxed{\frac{26}3}`