First slide

Introduction to definite integration

Question

The value of $\int_{0}^{1} lim_{n \to \infty} \sum_{k = 0}^{n} \frac{x^{k + 2} 2^{k}}{k!} d x$ is

Easy

A

$e^{2} - 1$
B

2
C

$\frac{e^{2} - 1}{2}$
D

$\frac{e^{2} - 1}{4}$

Solution

As $lim_{n \to \infty} \sum_{k = 0}^{n} \frac{x^{k + 2} 2^{k}}{k!} = x^{2} \sum_{k = 0}^{\infty} \frac{(2 x)^{k}}{k!} = x^{2} e^{2 x}$
Thus, $\int_{0}^{1} lim_{n \to \infty} (\sum_{k = 0}^{n} \frac{x^{k + 2} 2^{k}}{k!}) d x$
$\begin{array}{l} = \int_{0}^{1} x^{2} e^{2 x} d x = \frac{1}{4} (2 x^{2} - 2 x + 1) e^{2 x}] \\ = \frac{1}{4} (e^{2} - 1) \end{array}$

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