Question

The value of the integral $\int\limits_{0}^{1}{x{{e}^{x}}}dx$ is equal to
$\left( A \right)\text{ 0}$
$\left( B \right)\text{ 1}$
$\left( C \right)\text{ e}$
$\left( D \right)\text{ }{{\text{e}}^{-1}}$

Answer 1

Hint: In this question we have the integration of two terms in multiplication therefore, we will use the formula of integration by parts which is given by the formula $\int{uvdx}=u\int{v}dx-\int{\left( \dfrac{du}{dx}\int{vdx} \right)}dx$, where $u$ and $v$ are the two terms in multiplication. We will consider $u=x$ and $v={{e}^{x}}$, and use the formula to get the required integral and since this is a definite integral, we will substitute the values to get the required solution.

Complete step by step solution:
We have the expression given to us as:
$\Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}}dx$
Now it has two terms which are $x$ and ${{e}^{x}}$, in this question we will consider the first part to be $u=x$ and the latter part as $v={{e}^{x}}$ and apply the formula of integration by parts.
Now on using the formula for integration by parts on the expression, we get:
$\Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}}dx=x\int{{{e}^{x}}}dx-\int{\left( \dfrac{d(x)}{dx}\int{{{e}^{x}}dx} \right)}dx$
Now we know that $\int{{{e}^{x}}}dx={{e}^{x}}+c$ therefore, on integrating the first part, we get:
$\Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}}dx=x{{e}^{x}}-\int{\left( \dfrac{d(x)}{dx}\int{{{e}^{x}}dx} \right)}dx$
Now we know that $\dfrac{dx}{dx}=1$ therefore, we get:
$\Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}}dx=x{{e}^{x}}-\int{\left( \int{{{e}^{x}}dx} \right)}dx$
On integrating the term, we get:
$\Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}}dx=x{{e}^{x}}-\int{\left( {{e}^{x}} \right)}dx$
On integrating the remaining term, we get:
$\Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}}dx=x{{e}^{x}}-{{e}^{x}}$
On substituting the limits, we get:
$\Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}}dx=\left[ x{{e}^{x}}-{{e}^{x}} \right]_{0}^{1}$
On putting the values, we get:
$\Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}}dx=\left[ \left( 1 \right){{e}^{1}}-{{e}^{1}} \right]-\left[ \left( 0 \right){{e}^{0}}-{{e}^{0}} \right]$
Now we know that anything raised to $0$ is $1$ therefore, we get:
$\Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}}dx=\left[ e-e \right]-\left[ 0-1 \right]$
On simplifying, we get:
$\Rightarrow \int\limits_{0}^{1}{x{{e}^{x}}}dx=-\left[ -1 \right]=1$

So, the correct answer is “Option B”.

Note: It is to be remembered that while doing integration by parts the terms $u$ and $v$ should follow the sequence of the acronym $ILATE$, which stands for inverse, logarithm, algebraic, trigonometric and exponential respectively. It is to be remembered that integration and differentiation are inverse of each other. If the integration of $a$ is $b$ then the derivative $b$ of is $a$.