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How do you integrate e power minus 2x?
∫e^(-2x) dx Let u = 2x du= 2 dx dx=(1/2) du ∫e^(-2x) dx = (1/2) ∫e^-u du = (1/2) (-e^-u) = -e^-u /2 + C = -e^-(2x) / 2 + C
How you integrate xxxx 1 dx?
.2x^5+x+C
What is the Integral of 23x e2x dx?
This browser is pathetic for mathematical answers but here's the best that I can do:Let u = 23x therefore du/dx = 23let dv/dx = e^(2x) therefore v = 1/2*e^(2x)then, integrating by parts,I = I(u*dv/dx) dx = u*v - I(du/dx*v) dx= 23x*(1/2)*e^(2x) - I(23*(1/2)*e^(2x) dx= 23/2*xe^(2x) - 23/2*I(e^(2x)) dx= 23/2*xe^(2x) - 23/2*(1/2)*e^(2x)= 23/2*xe^(2x) - 23/4*e^(2x)or 23/4*e^(2x)*(2x - 1)
What is the value of e to the power of (2x) when e to the power of x equals 2?
The power law of indices says: (x^a)^b = x^(ab) = x^(ba) = (x^b)^a → e^(2x) = (e^x)² but e^x = 2 → e^(2x) = (e^x)² = 2² = 4
Solve ex equals 2e1-2x?
meant to be e^x = 2e^1-2x
How do you integrate of e power 2x plus e power minus 2x?
int[e(2X) +e(- 2X)] integrate term by term 1/22 e(2X) - 1/22 e(- 2X) + C (1/4)e(2X) - (1/4)e(- 2X) + C ====================
How do you integrate e powerintegral2x-1?
2x
How do you integrate e-2x?
= inegrate (e-2x) / derivate -2x = (e-2x)/-2-> integral esomething = esomething , that's why (e-2x) don't change-> (-2x)' = -2
How do you integrate of e -2x for x0?
The integral of e-2x is -1/2*e-2x + c but I am not sure what "for x0" in the question means.
How do you integrate e power minus 2x?
∫e^(-2x) dx Let u = 2x du= 2 dx dx=(1/2) du ∫e^(-2x) dx = (1/2) ∫e^-u du = (1/2) (-e^-u) = -e^-u /2 + C = -e^-(2x) / 2 + C
How do you integrate of e -2x?
To integrate e^(-2x)dx, you need to take a u substitution. u=-2x du=-2dx Since the original integral does not have a -2 in it, you need to divide to get the dx alone. -(1/2)du=dx Since the integral of e^x is still e^x, you get: y = -(1/2)e^(-2x) Well, that was one method. I usually solve easier functions like this by thinking how the function looked like before it was differentiated. I let f(x) stand for the given function and F(x) stand for the primitive function; the function we had before differentiation (the integrated function). f(x)= e-2x <-- our given function F(x)= e-2x/-2 <-- our integrated function Evidence: F'(x)= -2e-2x/-2 = e-2x = f(x) Q.E.D It's as simple as that.
How do you integrate functions?
To integrate a function you find what the function you have is the derivative of. for example the derivative of x^2 is 2x. so the integral of 2x is x^2.
How to integrate of e to the power of -2x with x to the power of 2?
e-2x^2 cannot be integrated, only approximated unless there is an additional x attached to the front of the e, otherwise this function is not integrable Actually, it can be integrated, but it requires multivariable calculus and a conversion from cartesian to polar form to do so.
How you integrate xxxx 1 dx?
.2x^5+x+C
What is the integral of 2-2x with limits 0 to t?
Integrate(0->t) (2-2x) dx is the integral correct (Integrate(0->t) (2-2x) dy would be different, you must state integrate with respect to what otherwise it can be anything) So integrals preserves sum and product with constants. i.e. Integrate (2-2x) dx = Integrate 2 dx - 2integrate x dx = 2x - x^2 + C By Fundamental Theorem of Calculus, take any anti-derivative, say C = 0 would be fine, and Integral(0->t)(2-2x) dx = (2x-x^2)|(0->t) = (2t-t^2) - 0 = 2t-t^2 It is a special case of the Second Fundamental Theorem of Calculus -- integral(0->t) f(x) dx is an anti-derivative of f(x).
D dx e 2x?
d/dx(e^2x) = 2xe^2x
What is x if 2x plus 1e equals 5x?
Question: 2x + e = 5x - 2x 3x = e x = e/3
