The difference an $x$ can make

The integration is quite simple: We use $\int e^u du = e^u$ with $u = -x^2$ to get $\int x e^{-x^2} = -\frac{e^{-x^2}}{2}$

Evaluating the improper integral, we get $\frac{1}{2}$ as our answer.

To find this integral, we find the antiderivative of $e^{-x^2}$ .

First we try just $e^{-x^2}$ ; the derivative of this is $-2xe^{-x^2}$ . This is a constant ( $-2$ ) times what we want, so we divide our original antiderivative by this constant to obtain $-\frac{e^{-x^2}}{2}$ .

The integral is a continuous quantity, so the question can instead be considered to be $\lim_{M\rightarrow \infty}\int_{0}^{M}xe^{-x^2}\text{dx}$ . From the fundamental theorem of calculus, this is equal to $\lim_{M\rightarrow \infty}(-\frac{e^{-M^2}}{2}+\frac{e^{-0^2}}{2})=\lim_{M\rightarrow\infty}-\frac{e^{-M^2}}{2}+\lim_{M\rightarrow\infty}\frac{e^{-0^2}}{2}$ As $M$ becomes arbitrarily large, $e^{-M^2}$ becomes arbitrarily small, so the left limit is obviously $0$ . The right limit is just the limit of a constant, $\frac{1}{2}$ , which is that constant.

So, the answer is $\int_{0}^{\infty}xe^{-x^2}\text{dx}=0+\frac{1}{2}\Rightarrow1+2=\boxed{3}$

We substitute $y=x^2$ , so we have $xdx=\dfrac{dy}{2}$ , so our integral changes to $\displaystyle\dfrac{1}{2}\int_{0}^{\infty} e^{-y} \ dx$ . Recall that $\Gamma(n+1)=\displaystyle\int_{0}^{\infty} x^ne^{-y} \ dx$ . So the integral we have got is just $n=0$ . Hence $\displaystyle\dfrac{1}{2}\int_{0}^{\infty} e^{-y} \ dx = \dfrac{\Gamma(1)}{2} = \boxed{\dfrac{1}{2}}$

Just put x^2 = t. So, 2x*dx becomes dt. Now put this substitution in the integral and solve for 0 to infinity e^(-t)

$\begin{aligned} I & = \int_0^\infty x e^{-x^2} dx \\ & = \int_0^\infty \frac d{dx} \left(-\frac {e^{-x^2}}2\right) dx \\ & = - \frac 12 \int_0^\infty d \left(e^{-x^2}\right) \\ & = - \frac 12 \bigg[ e^{-x^2} \bigg]_0^\infty \\ & = - \frac 12 [0-1] \\ & = \frac 12 \end{aligned}$

$\implies A+B = 1+2 = \boxed{3}$ .

I beg your pardon, ladies and gentleman friends, since I cannot edit this properly. I'm using Google's "Daum Equation Editor". So just copy the formulae I wrote and paste them. The program will recognize it.

This is a simple gamma function. Gamma function of a variable 'n' is defined as:

$\Gamma \left( n \right) =\int _{ 0 }^{ \infty }{ { e }^{ -x }\quad { x }^{ n-1 } } dx\quad =\frac { n! }{ n }$

If you have:

$\Gamma \left( n \right) =\int _{ 0 }^{ \infty }{ { e }^{ { x }^{ 2 } }\quad x\quad } dx$

Then we must consider:

${ x }^{ 2 }=y\quad \Leftrightarrow \quad \frac { dx }{ dy } =\quad \frac { 1 }{ 2 } \quad { y }^{ -1/2 }\quad \therefore \quad dx\quad =\quad \frac { 1 }{ 2 } \quad { y }^{ -1/2 }dy\\ \\ So\quad we\quad have:\\ \\ \int _{ 0 }^{ \infty }{ dy } \quad { { y }^{ 1/2 }\quad { e }^{ -y }\quad \frac { 1 }{ 2 } \quad { y }^{ -1/2 }\quad =\quad \frac { 1 }{ 2 } \quad \int { dy } ^{ \infty }{ { e }^{ -y } }\quad =\quad \frac { 1 }{ 2 } \quad \Gamma (1) }=\quad \frac { 1 }{ 2 } \quad \times \quad \frac { 1! }{ 1 } \quad =\quad \frac { 1 }{ 2 } \\ \\ So\quad \\ \\ \frac { A }{ B } \quad =\quad \frac { 1 }{ 2 } \quad \therefore \quad A\quad +\quad B\quad =\quad 3$

Hope it helps y'all. See y'all around!

Just put x^2 = t. So, 2x*dx becomes dt. Now put this substitution in the integral and solve for 0 to infinity e^(-t)

Multiply and divide the integral with -2 to obtain the derivative of –X^2 within the integral. Now integrate the term to get the answer…!

Using Integration by Substitution: let u=x^2, then du=2xdx => dx=du/(2x) Hence, integrate(e^-u)/2 from 0 to inf -e^(-u) from 0 to inf= 1/2 Therefore, 2+1=3.

its a cakewalk, just substitute - x^{2} =t now, -2x(dx)=dt. hence the integral is (integration)(-1/2) e^{t}(dt). (limit from zero to negative infinity) which gives (-1/2)e^{t} as answer. after putting the limits we get 1/2 as our answer. hence 1+2=3..