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How to integrate of e to the power of -2x with x to the power of 2?
Wiki User
∙ 12y ago
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.
Wiki User
∙ 12y ago
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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-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 powerintegral2x-1?
2x
What is the derivative of e to the power ln x squared?
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 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 2x?
e 2x = (1/2) e 2x + C ============
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 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 powerintegral2x-1?
2x
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
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.
What is the second derivative of xsqaured e to the power of x?
f(x) = (x^2)(e^x)f'(x) = e^x((x^2)+2x) - i thinkf"(x) = ?--------f(x) = (x^2)(e^x)apply the power rulef'(x) = (x^2)(e^x) + (2x)(e^x)apply the power rule to the first part and apply the power rule to the second part, then add those togetherf''(x) = [(x^2)(e^x) + (2x)(e^x)] + [(2x)(e^x) + (2)(e^x)]simplifyf''(x) = (e^x)(x^2 + 4x +2)I got it right. It checked out on my calculator.
How does e raised to the power of 2x equal e raised to the power of x squared?
e2x=ex^2 basically means that x2=2x, in which case x2-2x=0, x = 0, 2. I don't think that's what the question meant. It could mean: e2x=(ex)2 . Which comes from one of the rules of exponents. Basically, look at it this way: Take the natural log of both sides: ln e2x= ln(ex)2 From rules of logs: (2x) ln e = (2) ln ex 2x ln e = (2) (x) ln e 2x = (2)(x)
How do you integrate x power x?
e^x/1-e^x
What is the derivative of x to the power of e?
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
