Question

Evaluate the following integrals as limit of a sum:

\[\int\limits_0^2 (3x^2 - 1)\cdot dx\]

Answer 1

Let f(x) = 3x² – 1, for 0 ≤ x ≤ 2.

Divide the closed interval [0, 2] into n subintervals each of length h at the points.

0, 0 + h, 0 + 2h, ..., 0 + rh, .., 0 + nh = 2

i.e. 0, h, 2h, ..., rh, ..., nh = 2

∴ h = `(2)/n  "and as " n -> oo, h -> 0`

Here, a = 0

∴ f(a + rh) = f(0 + rh) = f(rh) = 3(rh)² – 1 = 3r²h² – 1

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

= \[\int\limits_0^2 (3x^2 - 1)\cdot dx = \lim\limits_{n\to \infty}\displaystyle\sum_{r=1}^{n} (3r^2h^2 - 1)\cdot h\]

= \[\lim\limits_{n\to \infty}\displaystyle\sum_{r=1}^{n} (3r^2 ×\frac{4}{n^2} - 1)\cdot \frac{2}{n}   ...[∵ h = \frac{2}{n}]\]

= \[\lim\limits_{n\to \infty}\displaystyle\sum_{r=1}^{n} (\frac{24r^2}{n^3} - \frac{2}{n})\]

= \[\lim\limits_{n\to \infty}[\frac{24}{n^3} \displaystyle\sum_{r=1}^{n} r^2 - \frac{2}{n} \displaystyle\sum_{r=1}^{n} 1]\]

= \[\lim\limits_{n\to \infty}[\frac{24}{n^3} \cdot \frac{n(n + 1)(2n + 1)}{6} - \frac{2}{n} \cdot n]\]

= \[\lim\limits_{n\to \infty}[4 \cdot ( \frac{n + 1}{n})(\frac{2n + 1}{n}) - 2]\]

= \[\lim\limits_{n\to \infty}[4(1 + \frac{1}{n})(2 + \frac{1}{n}) - 2]\]

= \[4(1 + 0)(2 + 0) - 2   ...[∵ \lim\limits_{n\to \infty} \frac{1}{n} = 0]\]

= 8 – 2

= 6