I used substitution to recompose it as $$int -dfrac12e^u, du$$ but this is also hard for me to evaluate. When I used wolfram alpha for $int e^-x^2 dx$ I gained a weird answer entailing a so referred to as error feature and also pi and such (I"m guessing it has something to perform with Euler"s identity, yet I"m sensibly particular this is above my textbook"s level).

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

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

So we have the right to integrate as complies with, and 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 outcome offered by $(2)$: $$fracddxleft(-frac 12 e^-x^2 + C ight) = -frac 12(-2x)e^-x^2 = xe^-x^2$$which is what you set out to integrate: $colorbluef xe^-x^2 eq e^-x^2$

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edited May 26 "13 at 3:25

answered May 25 "13 at 17:25

amWhyamWhy

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Let $u = -x^2$. It follows that $du = -2xdx$. So then

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

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answered May 25 "13 at 17:26

fahrbachfahrbach

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

The difficulty of finding a role $G(x)$ such that $G"(x)=e^-x^2$ is completely different, even though the 2 features $xe^-x^2$ and $e^-x^2$ are carefully associated. It turns out that tbelow is no *elementary* attribute whose derivative is $e^-x^2$. Roughly speaking, an *elementary function* is a role accumulated from the familiar features by utilizing enhancement, subtractivity, multiplication, department, and *composition* (substitution).

So $xe^-x^2$ and also $e^-x^2$ are elementary features. The first has actually an elementary indefinite integral. The second doesn"t. The second function is extremely important for many applications. Tbelow **is** a function whose derivative is $e^-x^2$, and also that attribute is valuable. It simply happens not to be an elementary attribute.

Because the function $e^-x^2$ is so essential, an antiderivative of a very closely associated function has been provided a name, the *error function*, often composed as $operatornameerf(x)$. That is what Alpha was talking about. If you ever examine probcapacity or statistics, you will certainly come to be deeply familiar through the error attribute.

**Back** to our problem! We are trying to integrate $xe^-x^2$. We already did that previously, by a "guess and also check" approach. But that is not completely satismanufacturing facility, so we currently use a conventional strategy, substitution.

As in your write-up, let $u=-x^2$. Then $du=-2x,dx$, so $x,dx=-frac12,du$. Substitute. We get$$int xe^-x^2,dx=int -frac12 e^u,du=-frac12e^u+C=-frac12e^-x^2+C.$$Finding $int e^u,du$ was easy. We recognize that the derivative of $e^t$ with respect to $t$ is $e^t$, so $int e^t,dt=e^t+C$.