Question

Evaluate the following integrals as limit of a sum : `""int_1^3 (3x - 4).dx`

Answer 1

Let f(x) = 3x – 4, for 1 ≤ x ≤ 3

Divide the closed interval [1, 3] into n subintervals each of length h at the points

1, 1 + h, 1 + 2h, 1 + rh, ..., 1 + nh = 3

∴ nh = 2

∴ h = `(2)/n` and as `n → oo, h → 0`

Here, a = 1

∴ f(a + rh) = f(1 + rh) = 3(1 + rh) – 4 = 3rh – 1

∵ \[\int\limits_a^b f(x) \cdot dx = \lim\limits_{n\to \infty}\displaystyle\sum_{r=1}^{n} f(a + rh)\cdot h\]

∴ \[\int\limits_1^3 (3x-4)\cdot dx=\lim\limits_{n\to \infty}\displaystyle\sum_{r=1}^{n} f(3rh - 1)\cdot h\]

= \[\lim\limits_{n\to \infty}\displaystyle\sum_{r=1}^{n} (3r \cdot \frac{2}{n} - 1) \cdot \frac{2}{n}\]   ...[∵ h = `2/n`]

= \[\lim\limits_{n\to \infty}\displaystyle\sum_{r=1}^{n} (\frac{12}{n^2}{r} - \frac{2}{n})\]

= \[\lim\limits_{n\to \infty}\displaystyle[\frac {12}{n^2}\sum_{r=1}^{n} {r} - \frac{2}{n}\sum_{r=1}^{n} {1}]\]

= \[\lim\limits_{n\to \infty}\displaystyle[\frac {12}{n^2}\frac{n(n + 1)}{2} {-} \frac{2}{n} \cdot{n}]\]

= \[\lim\limits_{n\to \infty}[{6}(\frac{n + 1}{n}) - {2}]\]

= \[\lim\limits_{n\to \infty}[{6}({1}+ \frac{1}{n}) - {2}]\]

= 6(1 + 0) – 2                 ...[∵ \[\lim\limits_{n\to \infty} \frac{1}{n} =0] \]

= 4