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}^{4} (x + e^{2 x}) d x

.

Accepted Answer

Let $I = \int_{0}^{4} (x + e^{2 x}) d x$

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, $a = 0, b = 4$ and $f (x) = x + e^{2 x}$

$\Rightarrow h = \frac{4 - 0}{n} = \frac{4}{n}$

$∴ \int_{0}^{4} (x + e^{2 x}) d x = (4 - 0) \lim_{n \to \infty} \frac{1}{n} [f (0) + f (h) + f (2 h) + \dots + f ((n - 1) h)]$

$= 4 \lim_{n \to \infty} \frac{1}{n} [(0 + e^{0}) + (h + e^{2 h}) + (2 h + e^{2^{2} h}) + \dots + \{(n - 1) h + e^{2 (n - 1) h}\}]$

$= 4 \lim_{n \to \infty} \frac{1}{n} [1 + (h + e^{2 h}) + (2 h + e^{4 h}) + \dots + \{(n - 1) h + e^{2 (n - 1) h}\}]$

$= 4 \lim_{n \to \infty} \frac{1}{n} [{h + 2 h + 3 h + \dots + (n - 1) h} + (1 + e^{2 h} + e^{4 h} + \dots + e^{2 (n - 1) h})]$

$= 4 \lim_{n \to \infty} \frac{1}{n} [h {1 + 2 + \dots (n - 1)} + (\frac{e^{2 h n} - 1}{e^{2 h} - 1})]$ (Second series is in G.P with first term $= 1$ and common ratio $= e^{2 h}$ , so $S_{n} = \frac{a (r^{n} - 1)}{r - 1}, r > 1$ )

$= 4 \lim_{n \to \infty} \frac{1}{n} [\frac{(h (n - 1) n)}{2} + (\frac{e^{2 h n} - 1}{e^{2 h} - 1})] [∵ \sum_{}^{} n = \frac{n (n + 1)}{2}]$

$= 4 \lim_{n \to \infty} \frac{1}{n} [\frac{4}{n} \cdot \frac{(n - 1) n}{2} + (\frac{e^{8} - 1}{e^{\frac{8}{n}} - 1})]$ (Since $h = \frac{4}{n}$ )

$= 4 \lim_{n \to \infty} \frac{1}{n} [2 \cdot (n - 1) + (\frac{e^{8} - 1}{e^{\frac{8}{n}} - 1})]$

$= 4 \lim_{n \to \infty} \frac{1}{n} \cdot 2 \cdot n (1 - \frac{1}{n}) + 4 \lim_{n \to \infty} \frac{1}{n} (\frac{e^{8} - 1}{e^{\frac{8}{n}} - 1})$

$= 4 (2) + 4 \lim_{n \to \infty} \frac{(e^{8} - 1)}{(\frac{e^{\frac{8}{n}} - 1}{\frac{8}{n}}) 8} [∵ \lim_{n \to \infty} \frac{1}{n} = 0]$

$= 8 + \frac{4 \cdot (e^{8} - 1)}{8} (∵ \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1)$

$= 8 + \frac{e^{8} - 1}{2}$

$= \frac{16 + e^{8} - 1}{2}$

$= \frac{15 + e^{8}}{2}$

Hence, $\int_{0}^{4} (x + e^{2 x}) d x = \frac{15 + e^{8}}{2}$ .

Evaluate the following integral as limit of a sum: $\int_{0}^{4} (x + e^{2 x}) d x$ .

Important Questions on Definite Integrals

Learn and Prepare for any exam you want !

Evaluate the following integral as limit of a sum: ∫04x+e2xdx.

Important Questions on Definite Integrals

Learn and Prepare for any exam you want !

Evaluate the following integral as limit of a sum: $\int_{0}^{4} (x + e^{2 x}) d x$ .