Name: Estimating Definite Integrals
Uploaded: 2019-07-19
Duration: 8 min 20 s
Description: It is not possible to evaluate every integral. So, sometimes we estimate the range of any definite integral. Watch this video and learn more about it.

Question

Evaluate the following integral as limit of a sum:

\int_{0}^{5} (x + 1) d x

.

Accepted Answer

Let $I = \int_{0}^{5} (x + 1) d x$

We know that, $\int_{a}^{b} f (x) d x = (b - a) \lim_{n \to \infty} \frac{1}{n} [f (a) + f (a + h) + \dots + f (a + (n - 1) h)]$ , where $h = \frac{b - a}{n}$ .

Here, $a = 0, b = 5$ and $f (x) = x + 1$

$\Rightarrow h = \frac{5 - 0}{n} = \frac{5}{n}$

$∴ \int_{0}^{5} (x + 1) d x = (5 - 0) \lim_{n \to \infty} \frac{1}{n} [f (0) + f (\frac{5}{n}) + \dots + f ((n - 1) \frac{5}{n})]$

$= 5 \lim_{n \to \infty} \frac{1}{n} [1 + (\frac{5}{n} + 1) + \dots \{1 + (\frac{5 (n - 1)}{n})\}]$

$= 5 \lim_{n \to \infty} \frac{1}{n} [(1 + 1 + 1 \dots 1) + [\frac{5}{n} + 2 \cdot \frac{5}{n} + 3 \cdot \frac{5}{n} + \dots (n - 1) \frac{5}{n}]]$

$= 5 \lim_{n \to \infty} \frac{1}{n} [n + \frac{5}{n} {1 + 2 + 3 \dots (n - 1)}]$

$= 5 \lim_{n \to \infty} \frac{1}{n} [n + \frac{5}{n} \cdot \frac{(n - 1) n}{2}] [∵ \sum_{}^{} n = \frac{n (n + 1)}{2}]$

$= 5 \lim_{n \to \infty} [1 + \frac{5}{2} (1 - \frac{1}{n})]$

$= 5 [1 + \frac{5}{2}] [∵ \lim_{n \to \infty} \frac{1}{n} = 0]$

$= 5 [\frac{7}{2}]$

$= \frac{35}{2}$

Hence, $\int_{0}^{5} (x + 1) d x = \frac{35}{2}$ .

Andhra Pradesh Board Solutions for Exercise 1: Exercise 7(a)

Andhra Pradesh Board Mathematics Solutions for Exercise - Andhra Pradesh Board Solutions for Exercise 1: Exercise 7(a)

Questions from Andhra Pradesh Board Solutions for Exercise 1: Exercise 7(a) with Hints & Solutions

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