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Question: What is the integral of \(x{{e}^{x}}\) from \(-\infty \) to 0?...

What is the integral of xexx{{e}^{x}} from −∞-\infty to 0?

Explanation

Solution

We solve this problem by using the by parts rule of integration. The by parts rule is given when there is integration of product of two functions.
If u,vu,v are two functions then by parts rule is given as,
∫(u×v)dx=u∫v.dx−∫[u′∫v.dx].dx\int{\left( u\times v \right)dx}=u\int{v.dx}-\int{\left[ {u}'\int{v.dx} \right].dx}
Assume the functions u,vu,v from the given function and apply the by parts rule and substitute the limits to get the required answer.

Complete step by step answer:
We are asked to find the value of the integral of xexx{{e}^{x}} from −∞-\infty to 0.
Let us assume that the required integral as ′I′'I' then we get,
⇒I=∫−∞0xex.dx\Rightarrow I=\int\limits_{-\infty }^{0}{x{{e}^{x}}.dx}
Here, we can see that the above integral is having two functions in multiplication.
Let us define two functions u,vu,v as,
⇒u=x ⇒v=ex \begin{aligned} & \Rightarrow u=x \\\ & \Rightarrow v={{e}^{x}} \\\ \end{aligned}
Now, we get the required integral as,
⇒I=∫−∞0(u×v).dx\Rightarrow I=\int\limits_{-\infty }^{0}{\left( u\times v \right).dx}
For the time being let us remove the limits and calculate the integral that is,
⇒I=∫(u×v).dx\Rightarrow I=\int{\left( u\times v \right).dx}
We know that if u,vu,v are two functions then by parts rule is given as,
∫(u×v)dx=u∫v.dx−∫[u′∫v.dx].dx\int{\left( u\times v \right)dx}=u\int{v.dx}-\int{\left[ {u}'\int{v.dx} \right].dx}
By using the by parts rule to above integral then we get,
⇒I=u∫v.dx−∫[u′∫v.dx].dx ⇒I=I1+I2..............................equation(i) \begin{aligned} & \Rightarrow I=u\int{v.dx}-\int{\left[ {u}'\int{v.dx} \right].dx} \\\ & \Rightarrow I={{I}_{1}}+{{I}_{2}}..............................equation(i) \\\ \end{aligned}
Where, I1=u∫v.dx{{I}_{1}}=u\int{v.dx} and I2=∫[u′∫v.dx].dx{{I}_{2}}=\int{\left[ {u}'\int{v.dx} \right].dx}
Now, let us calculate the first integral by substituting the functions then we get,
⇒I1=x∫ex.dx\Rightarrow {{I}_{1}}=x\int{{{e}^{x}}.dx}
We know that the integral of ex{{e}^{x}} is ex{{e}^{x}}
By using the above result in the first integral then we get,
⇒I1=x(ex) ⇒I1=xex \begin{aligned} & \Rightarrow {{I}_{1}}=x\left( {{e}^{x}} \right) \\\ & \Rightarrow {{I}_{1}}=x{{e}^{x}} \\\ \end{aligned}
Now, let us calculate the second integral by substituting the functions then we get,

& \Rightarrow {{I}_{2}}=\int{\left[ \dfrac{dx}{dx}\int{{{e}^{x}}.dx} \right].dx} \\\ & \Rightarrow {{I}_{2}}=\int{{{e}^{x}}.dx}={{e}^{x}} \\\ \end{aligned}$$ Now, let us substitute the required values in equation (i) then we get, $\Rightarrow I=x{{e}^{x}}+{{e}^{x}}$ Now, let us substitute the limits from $-\infty \text{ to }0$ then we get, $$\Rightarrow I=\left[ x{{e}^{x}}+{{e}^{x}} \right]_{-\infty }^{0}$$ By substituting the limits in above equation we get, $\Rightarrow I=\left[ 0\times {{e}^{0}}+{{e}^{0}} \right]-\left[ \left( -\infty \times {{e}^{-\infty }} \right)+{{e}^{-\infty }} \right]$ We know that $e$ has value 2.71 which is greater than 1. So, we can say that as its power increases to infinity the value gradually decreases and equals to 0. By using the above results we get the required integral as, $\begin{aligned} & \Rightarrow I=\left[ 0+1 \right]-\left[ 0+0 \right] \\\ & \Rightarrow I=-1 \\\ \end{aligned}$ Therefore we can conclude that the value of integral of $x{{e}^{x}}$ from $-\infty $ to 0 is ‘-1’ that is, $\therefore \int\limits_{-\infty }^{0}{x{{e}^{x}}.dx}=-1$ **Note:** Here, we need to take care of assuming the functions $u,v$ Here, we made assumptions as, $\begin{aligned} & \Rightarrow u=x \\\ & \Rightarrow v={{e}^{x}} \\\ \end{aligned}$ But we can also take it in reverse order. But that does not solve the problem and it will go on forever as we use the function $'v'$ for integration. If $v=x$ then the second integral in equation (i) will again be the product of two functions. So, we need to choose the functions $u,v$ to make the integration simple and less number of steps.