Question

Evaluate the following integral as the limit of sums `int_1^4 (x^2 - x)dx`.

Answer 1

⇒ `h = (b - a)/n`

= `(4 - 1)/n`

= `3/n`

Here n → ∞ as h → 0

Then `int_1^4 (x^2 - x)dx = (4 - 1) lim_(n -> ∞) 1/n [f(1) + f(1 + h) + f(1 + 2h) + ... + f(1 + n - 1h)]`

= `3 lim_(n -> ∞) 1/n [{1^2 + (1 + h)^2 + (1 + 2h)^2 + ... + (1 + (n - 1)h)^2} - {1 + (1 + h) + (1 + 2h) + ... + (1 + (n - 1)h}]`

= `3 lim_(n -> ∞) h [h^2 sum_(r = 0)^(n - 1) r^2 + h sum_(r = 0)^(n - 1) r]`   ...[Because f(1 + rh) = (1 + rh)² – (1 + rh) = r²h² + rh]

= `3 lim_(n -> ∞) h {h^2 xx (n(n - 1)(2n - 1))/6 + h(n(n - 1))/2}`

= `3 lim_(n -> ∞) {(nh xx (nh - h)(2nh - h))/6 + (nh(nh - h))/2}`

= `(3(3 - 0)(6 - 0))/6 + (3(3 - 0))/2`

= `9 + 9/2`

= `27/2`

∴ `int_1^4 (x^2 - x)dx = 27/2`