# Evaluate the following integral $\underset{x\to \infty }{\mathop{\lim }}\,\int\limits_{0}^{2x}{\dfrac{x{{e}^{{{x}^{2}}}}dx}{{{e}^{4{{x}^{2}}}}}}$.

(a) 0

(b) $\infty $

(c) 2

(d) \[\dfrac{1}{2}\]

Answer

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Hint: First we have to evaluate the given integral and then evaluate the limit. The exponential term can be simplified using the property [write the formula $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$]. Looking at the terms inside the integral, we can apply the substitution method of integration to simplify and solve the integral.

The given expression is $\underset{x\to \infty }{\mathop{\lim }}\,\int\limits_{0}^{2x}{\dfrac{x{{e}^{{{x}^{2}}}}dx}{{{e}^{4{{x}^{2}}}}}}$.

We can see that the integral is not in the standard form, so we cannot integrate it directly; we have to make some changes to solve it.

So, the first step we will do is to use the formula of exponential i.e. $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$, in the expression $\dfrac{{{e}^{{{x}^{2}}}}}{{{e}^{4{{x}^{2}}}}}$.

Now after using the exponential formula, we get $\dfrac{{{e}^{{{x}^{2}}}}}{{{e}^{4{{x}^{2}}}}}={{e}^{-3{{x}^{2}}}}$. So, we will substitute it in the expression and we will get,

$\underset{x\to \infty }{\mathop{\lim }}\,\int\limits_{0}^{2x}{x{{e}^{-3{{x}^{2}}}}dx}$

Now we can see that the integral is not in the standard form, so we have to use the method of substitution to reduce it into standard form and then solve it further. The substitution method of integration can be used when we have two functions in the integral such that one of the functions is the derivative of the other. As we know that the integral consists of two functions and here one of the functions is derivative of the other so, we can use the substitution method of integration to reduce it into the standard form so that we can easily solve it.

Let us consider ${{e}^{-3{{x}^{2}}}}=t$.

Now we will differentiate with respect to $x$ on both sides and we will get,

$\dfrac{d{{e}^{-3{{x}^{2}}}}}{dx}=\dfrac{dt}{dx}$

We know that the derivative is given by $\dfrac{d{{e}^{x}}}{dx}={{e}^{x}}$ and $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$. Applying the same, we will get,

$\begin{align}

& {{e}^{-3{{x}^{2}}}}(-6x)=\dfrac{dt}{dx} \\

& x{{e}^{-3{{x}^{2}}}}dx=\dfrac{dt}{-6} \\

\end{align}$

The next step is to change the limit from x to t, so that we get the expression in the terms of variable t.

We have considered that ${{e}^{-3{{x}^{2}}}}=t$. So, for limits of x, the value of t can be taken as below,

x = 0 then t = 1

$x=\infty $then t = 0

Now we can substitute all the results in the integral. So, we can write the integral $\underset{x\to \infty }{\mathop{\lim }}\,\int\limits_{0}^{2x}{x{{e}^{-3{{x}^{2}}}}dx}$ in terms of t after substituting as below,

$\underset{t\to 0}{\mathop{\lim }}\,\int\limits_{1}^{2t}{\dfrac{dt}{-6}}$

Since we know that $\int{dx=x}$, we can write as,

$\begin{align}

& =-\dfrac{1}{6}\underset{t\to 0}{\mathop{\lim }}\,\left[ t \right]_{0}^{2t} \\

& =-\dfrac{1}{6}\underset{t\to 0}{\mathop{\lim }}\,\left[ 2t-0 \right] \\

& =-\dfrac{1}{6}\underset{t\to 0}{\mathop{\lim }}\,2t \\

\end{align}$

Now we can evaluate the limit by putting t = 0,

$=-\dfrac{1}{6}\underset{t\to 0}{\mathop{\lim }}\,2(0)$

$=0$

Therefore, we get the value of the integral as,

$\underset{x\to \infty }{\mathop{\lim }}\,\int\limits_{0}^{2x}{\dfrac{x{{e}^{{{x}^{2}}}}dx}{{{e}^{4{{x}^{2}}}}}}=0$

The final answer is zero and hence the correct option is (a).

Note: We have to integrate first instead of evaluating the limit first. The possibility of mistake in the question is putting the value of $x=\infty $ in the expression without solving its integral. Another mistake is not changing the limits when changing the variable from x to t while evaluating the integral.

The given expression is $\underset{x\to \infty }{\mathop{\lim }}\,\int\limits_{0}^{2x}{\dfrac{x{{e}^{{{x}^{2}}}}dx}{{{e}^{4{{x}^{2}}}}}}$.

We can see that the integral is not in the standard form, so we cannot integrate it directly; we have to make some changes to solve it.

So, the first step we will do is to use the formula of exponential i.e. $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$, in the expression $\dfrac{{{e}^{{{x}^{2}}}}}{{{e}^{4{{x}^{2}}}}}$.

Now after using the exponential formula, we get $\dfrac{{{e}^{{{x}^{2}}}}}{{{e}^{4{{x}^{2}}}}}={{e}^{-3{{x}^{2}}}}$. So, we will substitute it in the expression and we will get,

$\underset{x\to \infty }{\mathop{\lim }}\,\int\limits_{0}^{2x}{x{{e}^{-3{{x}^{2}}}}dx}$

Now we can see that the integral is not in the standard form, so we have to use the method of substitution to reduce it into standard form and then solve it further. The substitution method of integration can be used when we have two functions in the integral such that one of the functions is the derivative of the other. As we know that the integral consists of two functions and here one of the functions is derivative of the other so, we can use the substitution method of integration to reduce it into the standard form so that we can easily solve it.

Let us consider ${{e}^{-3{{x}^{2}}}}=t$.

Now we will differentiate with respect to $x$ on both sides and we will get,

$\dfrac{d{{e}^{-3{{x}^{2}}}}}{dx}=\dfrac{dt}{dx}$

We know that the derivative is given by $\dfrac{d{{e}^{x}}}{dx}={{e}^{x}}$ and $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$. Applying the same, we will get,

$\begin{align}

& {{e}^{-3{{x}^{2}}}}(-6x)=\dfrac{dt}{dx} \\

& x{{e}^{-3{{x}^{2}}}}dx=\dfrac{dt}{-6} \\

\end{align}$

The next step is to change the limit from x to t, so that we get the expression in the terms of variable t.

We have considered that ${{e}^{-3{{x}^{2}}}}=t$. So, for limits of x, the value of t can be taken as below,

x = 0 then t = 1

$x=\infty $then t = 0

Now we can substitute all the results in the integral. So, we can write the integral $\underset{x\to \infty }{\mathop{\lim }}\,\int\limits_{0}^{2x}{x{{e}^{-3{{x}^{2}}}}dx}$ in terms of t after substituting as below,

$\underset{t\to 0}{\mathop{\lim }}\,\int\limits_{1}^{2t}{\dfrac{dt}{-6}}$

Since we know that $\int{dx=x}$, we can write as,

$\begin{align}

& =-\dfrac{1}{6}\underset{t\to 0}{\mathop{\lim }}\,\left[ t \right]_{0}^{2t} \\

& =-\dfrac{1}{6}\underset{t\to 0}{\mathop{\lim }}\,\left[ 2t-0 \right] \\

& =-\dfrac{1}{6}\underset{t\to 0}{\mathop{\lim }}\,2t \\

\end{align}$

Now we can evaluate the limit by putting t = 0,

$=-\dfrac{1}{6}\underset{t\to 0}{\mathop{\lim }}\,2(0)$

$=0$

Therefore, we get the value of the integral as,

$\underset{x\to \infty }{\mathop{\lim }}\,\int\limits_{0}^{2x}{\dfrac{x{{e}^{{{x}^{2}}}}dx}{{{e}^{4{{x}^{2}}}}}}=0$

The final answer is zero and hence the correct option is (a).

Note: We have to integrate first instead of evaluating the limit first. The possibility of mistake in the question is putting the value of $x=\infty $ in the expression without solving its integral. Another mistake is not changing the limits when changing the variable from x to t while evaluating the integral.

Last updated date: 24th Sep 2023

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