Better is [itex]2^{-x}= e^{ln(2^{-x})}= e^{x ln(2)}[/itex]Maddie1609 said:
Quite relevant, try to build on that.Maddie1609 said:
Samy_A said:
Wouldn't that be e-x ln 2?HallsofIvy said:
You lost me in this last equality, I guess here something is wrong.Maddie1609 said:
Yes, this is correct for the indefinite integral.Maddie1609 said:
Yes I made an edit on my post, I was thinking of u'/ln u. I think I've got it now.Samy_A said:
Yes. Thanks.Maddie1609 said:
The formula for the integral of 1/2^x is -(2^x ln(2))/ln(2) + C, where C is the constant of integration.
To solve the integral of 1/2^x, you can use the formula -(2^x ln(2))/ln(2) + C or you can use integration by substitution. Let u = 2^x and du = 2^x ln(2) dx. Then the integral becomes ∫1/u du, which is simply ln(u) + C. Substituting back in for u, we get -(2^x ln(2))/ln(2) + C.
The derivative of 1/2^x is -ln(2) * 1/2^x * (2^x ln(2)) = -ln(2) * 2^x * ln(2) = -2^x * ln^2(2).
The graph of the integral of 1/2^x is the area under the curve of the original function. This means that the graph of the integral will always be increasing, while the graph of 1/2^x can either be increasing or decreasing depending on the value of x. Additionally, the integral will approach a horizontal asymptote at y = 0, while the original function will approach a vertical asymptote at x = 0.
Yes, the integral of 1/2^x can also be solved using integration by parts or using the power rule for integration. However, using integration by substitution is usually the most efficient method for this particular integral.