Question

Evaluate the following integrals as limit of a sum:

\[\int\limits_0^4 x^2 \cdot dx\]

Answer 1

Let f(x) = x², for 0 ≤ x ≤ 4
Divide the closed interval [0, 4] into n subintervals, each of length h at the points
0, 0 + h, 0 + 2h, ..., 0 + rh, ..., 0 + nh = 4
i.e. 0, h, 2h, ..., rh, ..., nh = 4

∴ h = `(4)/n` as n → ∞ , h → 0`
Here , a = 0
∴ f(a + rh) = f(0 + rh) = f(rh) = r²h²

`because int_a^b f(x) dx = lim_(n ->infty) sum_(r = 1)^n f(a + rh) * h`

`therefore int_0^4 x^2 dx = lim_(n ->infty) sum_(r = 1)^n r^2h^2 * h`

= `lim_(n ->infty) sum_(r = 1)^n r^2 64/n^3    ...[because h = 4/n]`

= `lim_(n ->infty) [64/n^3 sum_(r = 1)^n r^2]`

= `lim_(n ->infty) [64/n^3 * (n (n + 1)(2n + 1))/6]`

= `lim_(n ->infty) [64/6 ((n + 1)/n) ((2n + 1)/n)]`

= `lim_(n ->infty) [64/6 (1 + 1/n)(2 + 1/n)]`

= `64/6 (1 + 0)(2 + 0)    ...[because lim_(n -> infty) 1/n = 0]`

= `64/3`