I supplied substitution come rewrite it as $$int -dfrac12e^u, du$$ however this is too difficult for me come evaluate. As soon as I supplied wolfram alpha because that $int e^-x^2 dx$ I acquired a monster answer including a so called error role and pi and such (I"m guessing it has actually something to do with Euler"s identity, yet I"m fairly certain this is above my textbook"s level).

You are watching: Integral of xe^x^2



Your substitution was spot on. You put $$u = -x^2 implies du = -2x;dx implies x,dx = -frac 12, du$$

And having actually substituted, you obtained $$int -dfrac12e^u, du$$Great work. But I think you gave up as well early!: $$-frac 12int e^u , du = -frac 12 e^u + C, ag1$$ and also recall, $u = f(x), ;du = f"(x),dx$, so $(1)$ is indistinguishable to $$-frac 12 int e^f(x),f"(x),dx = -frac 12 e^f(x) + C$$

So us can integrate as follows, and also then back-substitute: $$int -dfrac12e^u, du = -frac 12 int e^u ,du = -frac 12 e^u + C = -frac 12 e^-x^2 + C ag2$$

Now, to remove all doubts, simply differentiate the an outcome given by $(2)$: $$fracddxleft(-frac 12 e^-x^2 + C ight) = -frac 12(-2x)e^-x^2 = xe^-x^2$$which is what you collection out to integrate: $colorbluef xe^-x^2 eq e^-x^2$

point out
edited may 26 "13 in ~ 3:25
answered may 25 "13 at 17:25

203k146146 yellow badges261261 silver- badges484484 bronze badges
| show 4 much more comments
Let $u = -x^2$. It complies with that $du = -2xdx$. For this reason then

$$int xe^-x^2dx = -frac12int -2xe^-x^2dx = -frac12int e^udu = -frac12e^u + C = -frac12e^-x^2 + C.$$

point out
answered might 25 "13 at 17:26

1,72311 yellow badge1313 silver badges1515 bronze badges
include a comment |

See more: P 65 This Gui Includes Tabs, Groups, And Galleries., Computer Essentials 3 Flashcards

The trouble is to find$$int xe^-x^2,dx.$$So we are trying to find the functions $F(x)$ such that$$F"(x)=xe^-x^2.$$We can easily verify, by differentiating, that $F(x)=-frac12e^-x^2$ "works," and therefore the basic $F(x)$ such the $F"(x)=xe^-x^2$ is provided by $F(x)=-frac12e^-x^2+C$.

The problem of finding a role $G(x)$ such that $G"(x)=e^-x^2$ is completely different, also though the two attributes $xe^-x^2$ and $e^-x^2$ are very closely related. It transforms out that there is no elementary role whose derivative is $e^-x^2$. About speaking, one elementary function is a duty built up from the familiar functions by using addition, subtraction, multiplication, division, and also composition (substitution).

So $xe^-x^2$ and $e^-x^2$ space elementary functions. The first has one elementary unknown integral. The 2nd doesn"t. The second function is very important for countless applications. There is a duty whose derivative is $e^-x^2$, and also that role is useful. It just happens no to it is in an elementary function.

Because the role $e^-x^2$ is therefore important, an antiderivative of a carefully related function has been given a name, the error function, regularly written as $operatornameerf(x)$. That is what Alpha was talk about. If you ever before study probability or statistics, girlfriend will come to be deeply acquainted with the error function.

Back to our problem! We space trying to integrate $xe^-x^2$. We currently did that earlier, by a "guess and check" method. But that is not totally satisfactory, so we currently use a standard method, substitution.

As in her post, let $u=-x^2$. Then $du=-2x,dx$, for this reason $x,dx=-frac12,du$. Substitute. We get$$int xe^-x^2,dx=int -frac12 e^u,du=-frac12e^u+C=-frac12e^-x^2+C.$$Finding $int e^u,du$ to be easy. We recognize that the derivative that $e^t$ through respect come $t$ is $e^t$, for this reason $int e^t,dt=e^t+C$.