If intregation from - infinity to + infinity of f(x)dx=1Show intregation of same interval of 1/(1+f(x))=infit?
457427
-inf to +inf int. 1/(1+f(x))dx\n= -inf to +inf int. inv. (1+f(x))dx\n= -inf to +inf int. (1-f(x))dx [expanding binomially]\n= -inf to +inf int. dx - -inf to +inf int. f(x)dx\n= -inf to +inf [x] - 1\n= [(+inf)-(-inf)] - 1\n= (+inf+inf)-1\n= 2xinf.-1\n= inf.-1\n= infinity.............QED
50362
It usually does:\n\nlim n->inf 1^n = 1\n\nOn the other hand,\n\nlim n->inf (1 + 1/n)^n = e
787409
that would be if it was an indefinite integral\nbut in the case of definite one, it would be the difference of the values of the function at the limits of the integral.
869377
let 1+sec x = t\n\ndifferentiate\n\nsec x tan x dx = dt\n\nnow integral = integral of dt/t = ln | t| + c = ln | sec x +1 | + c
108610
If look to the integrator more deeply, one finds that 1/x is the derivative of ln(x).\nThen this question comes up: \nWhat is the integration of uu'dx?\n\nObviously the answer is u^2/2.\n\nSo our answer is (ln(x))^2/4+c.
801052
It seems you are asking for f'(2)?  in which case, we have\nf'(x) = 1/(1+x^4)so that f'(2) = 1/9\n\nSteve
168841
Suppose that the limits on your integration are from 1 to X\n\nint{1/x dx from 1 to X} = (1-1)+int{1/x dx  from 1 to X}\n\nand one finds out that 0=0\n\nIf you want to perform an indefinite integral,\n\nint{udv}=uv-int(vdu} +C, where C is a constant.\n\nif u=1/x, and v=x, then choosing C=-1 makes this equality true.
776321
the question is unclear.\nDo you want an explicit formula for Y[n or do you just want to write out the convolution]
1130821
so for the first part, f(x)=x \nthe derivative no matter what is 1 \n\nfor the second part f(x)=1\nthe derivative no matter what is 0 \n\nso you will have combinational solution
1023164
Limit ( n to infinity) (n + i/n)^ n
1072243
Domain: (-infinity, +infinity)\nRange: (-infinity, +infinity)
109332
that's right. cause infinit is indefinit.
737185
How about something simpler, like\n\nf(x) = 0 when x<1/2, 1 when x>=1/2\n\nThen the limit from the left is 0, the limit from the right is 1, so the regular limit doesn't exist.
907364
Almost.  You would actually include the endpoints 0 and -3 (so you need to use a bracket instead of parentheses).\nD = [0, infinity)\nR = [-3, infinity)
830837
If ∫f(x)dx = y(x), then f(x) = d y(x) / dx;  So to get the answer, take the derivative of the expression x^2 - 2x + 1.  The value of the integral at limit 1 is  a constant and will not appear in the derivative.
337080
when you integrate 1/x you get In |x|\n\nso the answer is -1/2 * In |x| + constant term\n\n\nYou can use substitution, but there's not much point, because you'll get 1/u then (which is the same thing). 1/x is a well known integrand. You'd be better just memorising that the integral of 1/x = In(x)
546126
intrgrate o e^(3x)=(1/3)e^(3x)