Assume the expression is:
∫ sin(x)x²e^x dx
Then:
Take the integral: integral e^x x^2 sin(x) dx For the integrand e^x x^2 sin(x), integrate by parts, integral f dg = f g- integral g df, where f = x^2, dg = e^x sin(x) dx, df = 2 x dx, g = 1/2 e^x (sin(x)-cos(x)): = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral e^x x (sin(x)-cos(x)) dx Expanding the integrand e^x x (sin(x)-cos(x)) gives e^x x sin(x)-e^x x cos(x): = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral (e^x x sin(x)-e^x x cos(x)) dx Integrate the sum term by term and factor out constants: = 1/2 e^x x^2 sin(x)-1/2 (e^x x^2 cos(x))- integral e^x x sin(x) dx+ integral e^x x cos(x) dx For the integrand e^x x sin(x), integrate by parts, integral f dg = f g- integral g df, where f = x, dg = e^x sin(x) dx, df = dx, g = 1/2 e^x (sin(x)-cos(x)): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+ integral e^x x cos(x) dx+1/2 integral e^x (sin(x)-cos(x)) dx Expanding the integrand e^x (sin(x)-cos(x)) gives e^x sin(x)-e^x cos(x): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+ integral e^x x cos(x) dx+1/2 integral (e^x sin(x)-e^x cos(x)) dx Integrate the sum term by term and factor out constants: = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)+1/2 e^x x cos(x)+1/2 integral e^x sin(x) dx-1/2 integral e^x cos(x) dx+ integral e^x x cos(x) dx For the integrand e^x cos(x), use the formula integral exp(alpha x) cos(beta x) dx = (exp(alpha x) (alpha cos(beta x)+beta sin(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/4 e^x sin(x)-1/2 e^x x sin(x)-1/4 (e^x cos(x))+1/2 e^x x cos(x)+1/2 integral e^x sin(x) dx+ integral e^x x cos(x) dx For the integrand e^x sin(x), use the formula integral exp(alpha x) sin(beta x) dx = (exp(alpha x) (alpha sin(beta x)-beta cos(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x x sin(x)-1/2 (e^x cos(x))+1/2 e^x x cos(x)+ integral e^x x cos(x) dx For the integrand e^x x cos(x), integrate by parts, integral f dg = f g- integral g df, where f = x, dg = e^x cos(x) dx, df = dx, g = 1/2 e^x (sin(x)+cos(x)): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral e^x (sin(x)+cos(x)) dx Expanding the integrand e^x (sin(x)+cos(x)) gives e^x sin(x)+e^x cos(x): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral (e^x sin(x)+e^x cos(x)) dx Integrate the sum term by term: = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)+e^x x cos(x)-1/2 e^x cos(x)-1/2 integral e^x sin(x) dx-1/2 integral e^x cos(x) dx For the integrand e^x cos(x), use the formula integral exp(alpha x) cos(beta x) dx = (exp(alpha x) (alpha cos(beta x)+beta sin(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/4 e^x sin(x)+e^x x cos(x)+-3/4 e^x cos(x)-1/2 integral e^x sin(x) dx For the integrand e^x sin(x), use the formula integral exp(alpha x) sin(beta x) dx = (exp(alpha x) (alpha sin(beta x)-beta cos(beta x)))/(alpha^2+beta^2): = 1/2 e^x x^2 sin(x)-1/2 e^x x^2 cos(x)-1/2 e^x sin(x)+e^x x cos(x)-1/2 e^x cos(x)+constant Which is equal to: Answer: | | = 1/2 e^x ((x^2-1) sin(x)-(x-1)^2 cos(x))+constant
By using the chain rule. Since the derivative of exp(x) is exp(x), the derivative of exp(exp(exp(x))) is exp(exp(exp(x))) times the derivative of what is inside the parentheses, i.e., exp(exp(exp(x))) times derivate of exp(exp(x)). Continue using the chain rule once more, for this expression.
The definite integral of the function f=x*exp(k*x) is (1/k)*(x-(1/k))*exp(k*x). So you have the answer to your questions by setting k equal to 1 then 2. I derived my formula by using integration by parts, setting u=x and dv=exp(k*x)dx.
First of all, you have to have a scientific calculator, one that supports scientific notation. (As far as I know, all scientific calculators do.) The scientific calculator should have a special key labelled something like EXP. To input (for example) 2.3 million, you would type 2.3 EXP 6 (where EXP is short for "times 10 to the power...").
A square pyramid's base is a square. You can tell by the name (for exp. a triangular pyramid's base is a triangle.) Good luck with whatever you need it for :)
Negative square roots are just the opposite of positive square roots. Since square roots (of positive numbers) are real, the negative square roots are also real.Square roots of negative numbers are not real.Note that -1 = exp(Pi*i), so (-1)^(1/2) = exp((1/2)*Pi*i) = i.Note that exp(i*x) = cos(x) + i*sin(x), for instance by taking derivatives:(d/dx)(exp(i*x)) = i*exp(i*x), and(d/dx)^2(exp(i*x)) =(-1)*exp(i*x).This means that the second derivative of exp(i*x) equals -exp(i*x).The same property holds for cos(x) + i*sin(x):(d/dx)(cos(x) + i*sin(x)) = -sin(x) + i*cos(x)(d/dx)^2(cos(x) + i*sin(x)) = -cos(x) - i*sin(x) = -(cos(x) + i*sin(x)))Hence cos(x) + i*sin(x)) = C + Dx + exp(i*x), for some C and D.Comparing the values on both sides for x = 0, we find:1 = C+1, so C = 0 and for the first derivative:i = D + i, so D = 0.So cos(x) + i*sin(x)) = exp(i*x) for all x.by comparing x=0 for both functions and their first derivative. Since they coincide,
By using the chain rule. Since the derivative of exp(x) is exp(x), the derivative of exp(exp(exp(x))) is exp(exp(exp(x))) times the derivative of what is inside the parentheses, i.e., exp(exp(exp(x))) times derivate of exp(exp(x)). Continue using the chain rule once more, for this expression.
(3x-1) * exp(3*x) / 9
No. you just get the exp. points! :^)
Nothing you just get EXP from them that's it.
I suggest you give poliwrath the exp share and go through the elite 4 a bunch of times
No, it doesn't help you 'literally' but it is convenient because chemistry uses a lot of calculation: generally 'solving equations' most commonly + - : x log less commonly exp 'square root' integral/differential etc.
You can give it to a Pokemon you want to lvl up fasterit give two times the experience point
Do bounty hunting lot of times until u come to an enemy which gives ya 600 gold(maximum)
Yes one is 1 loop catch and exp times
to train your Pokemon quick you need to get a exp share by getting a red scale then trading it with mr.pokemons exp share.then give it to your Pokemon and then beat all the gym leaders/red and then beat the elite four a couple of times simple
Give him an EXP SHARE, and go trough the pokémon league (a couple of times) Use an other pokémon, not magikarp, but keep him in your team. He will get EXP without battling.
The hypoexponential distribution is two or more exponential distribution convolved together, so:hypo[x] = integral of A*exp(-A*x)*B*exp(-B*(t-x)) from 0 to t.If you have more stages you do more convolutions.