I used substitution come rewrite it together $$\\int -\\dfrac12e^u\\, du$$ yet this is too tough for me come evaluate. Once I provided wolfram alpha because that $\\int e^-x^2 dx$ I acquired a monster answer entailing a so called error role and pi and such (I\"m guessing it has actually something to perform with Euler\"s identity, yet I\"m reasonably certain this is above my textbook\"s level).

You are watching: Integral of xe^-x^2

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

And having actually substituted, you obtained $$\\int -\\dfrac12e^u\\, du$$Great work. But I think you gave up too early!: $$-\\frac 12\\int e^u \\, du = -\\frac 12 e^u + C,\\tag1$$ 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 we can incorporate as follows, and also then back-substitute: $$\\int -\\dfrac12e^u\\, du = -\\frac 12 \\int e^u \\,du = -\\frac 12 e^u + C = -\\frac 12 e^-x^2 + C\\tag2$$

Now, to eliminate all doubts, simply identify the result given through $(2)$: $$\\fracddx\\left(-\\frac 12 e^-x^2 + C\\right) = -\\frac 12(-2x)e^-x^2 = xe^-x^2$$which is what you set out come integrate: $\\colorblue\\bf xe^-x^2 \\neq e^-x^2$

re-superstructure
cite
monitor
edited might 26 \"13 in ~ 3:25
answered might 25 \"13 at 17:25

amWhyamWhy
$\\endgroup$
9
3
$\\begingroup$
Let $u = -x^2$. It complies with that $du = -2xdx$. For this reason then

$$\\int xe^-x^2dx = -\\frac12\\int -2xe^-x^2dx = -\\frac12\\int e^udu = -\\frac12e^u + C = -\\frac12e^-x^2 + C.$$

share
point out
follow
answered might 25 \"13 in ~ 17:26

fahrbachfahrbach
$\\endgroup$
include a comment |
2

See more: Foot To Floor Ride On Toy S

$\\begingroup$
The difficulty is come find$$\\int xe^-x^2\\,dx.$$So we are trying to discover the features $F(x)$ such that$$F\"(x)=xe^-x^2.$$We can quickly verify, through differentiating, that $F(x)=-\\frac12e^-x^2$ \"works,\" and also therefore the basic $F(x)$ such the $F\"(x)=xe^-x^2$ is given by $F(x)=-\\frac12e^-x^2+C$.

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

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

Because the function $e^-x^2$ is therefore important, one antiderivative of a closely related duty has been provided a name, the error function, often written as $\\operatornameerf(x)$. The is what Alpha was talk about. If you ever before study probability or statistics, friend will end up being deeply familiar with the error function.

Back come our problem! We room trying to integrate $xe^-x^2$. We currently did that earlier, by a \"guess and also check\" method. Yet that is not entirely satisfactory, for this reason we now use a typical method, substitution.

As in her post, let $u=-x^2$. Then $du=-2x\\,dx$, therefore $x\\,dx=-\\frac12\\,du$. 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 know that the derivative the $e^t$ with respect come $t$ is $e^t$, therefore $\\int e^t\\,dt=e^t+C$.