# Evaluate the following integrals as limit of a sum : 3∫1(3x-4)⋅dx - Mathematics and Statistics

Sum

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

#### Solution

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 → 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]

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

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

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

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

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

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

= 4.

Concept: Definite Integral as Limit of Sum
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