Question

Integration of the expression $\int{(x-1)}{{e}^{-x}}dx$ is equal to
(A) $-x{{e}^{x}}+C$
(B) $x{{e}^{x}}+C$
(C) $-x{{e}^{-x}}+C$
(D) $x{{e}^{-x}}+C$

Answer 1

First expand the integral by multiplying through the bracket by ${{e}^{-x}}$. Now integrate the two terms that is $\int{x{{e}^{-x}}}dx$ and $\int{-{{e}^{-x}}dx}$ so obtained separately. For integrating $\int{x{{e}^{-x}}}dx$ , integrate using by parts method because $x{{e}^{-x}}$ is the product of two functions $x$ and ${{e}^{-x}}$. If $uv$ is the product of two functions $u$ and $v$ then integral of $uv$ is given by \int{(uv)}dx=[u.\int{vdx]-\int{\left\\{ \dfrac{du}{dx}.\int{vdx} \right\\}}}dx where u and v are functions of x.

Complete step by step answer:
In the given question we have to integrate the expression $\int{(x-1)}{{e}^{-x}}dx$ . So, let us take that term as I, that is,
$I=\int{(x-1)}{{e}^{-x}}dx$
Now we will expand the expression by multiplying ${{e}^{-x}}$ to the terms inside the bracket. Then the expression becomes,
$I=\int{(x}{{e}^{-x}}-{{e}^{-x}})dx........(1)$
We know that $\int{(f(x)+g(x))dx=\int{f(x)dx+\int{g(x)dx}}}$ where f(x) and g(x) are functions of x. Then the above equation (1) can be written as
$I=\int{x{{e}^{-x}}dx-\int{{{e}^{-x}}dx}}$
We will integrate the two terms separately so let ${{I}_{1}}=\int{x{{e}^{-x}}}dx$ and ${{I}_{2}}=\int{{{e}^{-x}}dx}$ . The above integral becomes,
$I={{I}_{1}}-{{I}_{2}}........(2)$
After solving ${{I}_{1}}$and ${{I}_{2}}$ we will substitute these values in equation (2). So first we have to solve ${{I}_{1}}$,
${{I}_{1}}=\int{x{{e}^{-x}}}dx$
Since $x$ and ${{e}^{-x}}$ are two functions of $x$ so we will integrate this whole expression using by parts method, which states that if integral is expressed as product of two functions then it is equal to
( ${{1}^{st}}$ function) x (integral of ${{2}^{nd}}$ function)-$\int{{}}${( derivative of ${{1}^{st}}$ function) x (integral of ${{2}^{nd}}$ function)}dx which mathematically can be expressed as \int{(uv)}dx=[u.\int{vdx]-\int{\left\\{ \dfrac{du}{dx}.\int{vdx} \right\\}}}dx where u and v are functions of x. We will use ILATE (Inverse, Logarithmic, Algebraic, Trigonometric and Exponent) rule for choosing u and v. Here$x$ is an algebraic term and ${{e}^{-x}}$ is an exponential term. According to ILATE rule preference is given to algebraic terms so here $u=x$ and $v={{e}^{-x}}$ .
So now after using by parts, expression becomes,
=x\int{{{e}^{-x}}}dx-\int{\left\\{ \dfrac{d(x)}{dx}\int{{{e}^{-x}}dx} \right\\}}dx
Integral of ${{e}^{-x}}$ is $-{{e}^{-x}}$ and derivative of x is 1 so the above expression becomes,
=-x{{e}^{-x}}-\int{\left\\{ (1)\int{{{e}^{-x}}}dx \right\\}}dx
$=-x{{e}^{-x}}-\int{(-{{e}^{-x}})dx}$
$=-x{{e}^{-x}}+\int{{{e}^{-x}}dx}$
$=-x{{e}^{-x}}+(-1){{e}^{-x}}$ as integral of ${{e}^{-x}}$ is $-{{e}^{-x}}$.
$=-x{{e}^{-x}}-{{e}^{-x}}+{{C}_{1}}........(3)$ where ${{C}_{1}}$ is the integral constant
Now we have to integrate ${{I}_{2}}$.
${{I}_{2}}=\int{{{e}^{-x}}dx}$
We know that the integral of ${{e}^{-x}}$ is $-{{e}^{-x}}$. So the above expression becomes,
$=-{{e}^{-x}}+{{C}_{2}}.........(4)$ where ${{C}_{2}}$is the integral constant
We will now put the values of equation (3) and (4) in (2), we get
$\begin{aligned} & I=(-x{{e}^{-x}}-{{e}^{-x}}+{{C}_{1}})-(-{{e}^{-x}}+{{C}_{2}}) \\\ & =-x{{e}^{-x}}-{{e}^{-x}}+{{C}_{1}}+{{e}^{-x}}-{{C}_{2}} \\\ & =-x{{e}^{-x}}+({{C}_{1}}-{{C}_{2}}) \\\ \end{aligned}$
$=-x{{e}^{-x}}+C$ , where $C={{C}_{1}}-{{C}_{2}}$
Therefore, the integral of $\int{(x-1)}{{e}^{-x}}dx$ is $-x{{e}^{-x}}+C$.

So, the correct answer is “Option C”.

Note: Care should be taken while evaluating the integral ${{I}_{1}}=\int{x{{e}^{-x}}}dx$ using by parts, this integral should be evaluated as x\int{{{e}^{-x}}}dx-\int{\left\\{ \dfrac{d(x)}{dx}\int{{{e}^{-x}}dx} \right\\}}dx , there is a minus sign between the terms $x\int{{{e}^{-x}}}dx$ and \int{\left\\{ \dfrac{d(x)}{dx}\int{{{e}^{-x}}dx} \right\\}}dx, it should not be mistakenly written as plus sign. And in integral \int{\left\\{ \dfrac{d(x)}{dx}\int{{{e}^{-x}}dx} \right\\}}dx there are two integral signs so first integrate the term which is inside the bracket and then multiply this result to the derivative of x then again integrate the result so obtained.