what is the integral (from -inf. to +inf.) of 7xe^(-6x^2) dx?

Question

please explain the answer too, thanx

Answer 1

use the substitution u=x^2. Then du=2x dx, and you have the integral of (7/2)e^(-6u) du. antidifferentiating and then undoing the substitution gives

-(7/12) e^(-6x^2)

which needs to be evaluated at the limits -b and b, as b->infinity.

Use the formula:

integral from -inf to inf of e^(-cx^2) = sqrt(pi/c)

to finish the problem.

Answer 2

Hi Mac;
∫(7x)e^(-6x^2) dx = - (7/12) (e ^(-6x^2)) +c


I think its clear enough;
good luck

Answer 3

i) enable e^x = tan(u); differentiating, (e^x) dx = sec²u du additionally x = ?, tan(u) = ?; ==> u = ?/2 in addition for x = -?, tan(u) = -?; ==> u = -?/2 better, e^x + e^-x = e^x + a million/e^x = tan(u) + a million/tan(u) = (tan²u + a million)/tan(u) = sec²u/tan(u) ii) for this reason the given one reduces to: ?tan(u) du in barriers [-?/2 to ?/2] yet tan(u) is an ordinary functionality. So using the specific imperative residences, ?f(x) dx in [-a to a] = 0, if f(x) is an ordinary functionality, ?tan(u) du in barriers [-?/2 to ?/2] = 0 it is ?(e^x) dx/{(e^x) + (e^-x)} in barriers [-?, ?] = 0