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The integral of e-2x is -1/2*e-2x + c but I am not sure what "for x0" in the question means.

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โˆ™ 2014-05-04 21:04:15
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Q: How do you integrate of e -2x for x0?
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How do you integrate of e power 2x plus e power minus 2x?

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How do you integrate e 2x?

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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

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How do you integrate e power minus 2x?

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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.

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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.

D dx e 2x?

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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

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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)

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