Question

Evaluate the following integrals as limit of a sum : \[\int\limits_1^3 x^3 \cdot dx\]

Answer 1

Let f(x) = x³, for 1 ≤ x ≤ 3.
DIvide the closed interval [1, 3] into n equal 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) = )1 + rh)³
= 1 + 3rh + 3r²h² + r³h³

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

∴ \[\int\limits_1^3 x^3 \cdot dx = \lim\limits_{n\to \infty}\displaystyle\sum_{r=1}^{n}(1 + 3rh + 3r^2h^2 + r3h^3)\cdot h\]

= \[\lim\limits_{n\to \infty}\displaystyle\sum_{r=1}^{n} (h + 3rh^2 + 3r^2h^3 + r^3h^4)\]

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

= \[\lim\limits_{n\to \infty}\displaystyle\sum_{r=1}^{n} [\frac{2}{n} + \frac{12r}{n^2} + \frac{24r^2}{n^3} + \frac{16r^3}{n^4}]\]

= \[\lim\limits_{n\to \infty}[\frac{2}{n}\displaystyle\sum_{r=1}^{n} 1 + \frac{12}{n^2}\displaystyle\sum_{r=1}^{n} r + \frac{24}{n^3}\displaystyle\sum_{r=1}^{n} r^2 + \frac{16}{n^4}\displaystyle\sum_{r=1}^{n} r^3]\]

= \[\lim\limits_{n\to \infty}[\frac{2}{n} \cdot n + \frac{12}{n^2}\cdot \frac{n(n + 1)}{2} + \frac{24}{n^3} \frac{n(n + 1)(2n + 1)}{6} + \frac{16}{n^4} \cdot \frac{n^2(n + 1)^2}{4}]\]

= \[\lim\limits_{n\to \infty}[2 + 6 (\frac{n + 1}{n}) + 4 (\frac{n + 1}{n})(\frac{2n + 1}{n}) + 4(\frac{n + 1}{n})^2]\]

= \[\lim\limits_{n\to \infty}[2 + 6(1 + \frac{1}{n}) + 4(1 + \frac{1}{n})(2 + \frac{1}{n}) + 4(1 + \frac{1}{n})^2]\]

= \[[2 + 6(1 + 0) + 4(1 + 0)(2 + 0) + 4(1 + 0)^2]  ...[∵ \lim\limits_{n\to \infty} \frac{1}{n} = 0]\]

= 2 + 6 + 8 + 4
= 20.