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
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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
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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 $int e^t,dt=e^t+C$.