-e^(-x)
or negative e to the negative x
this is because you multiply the function (e) by:
1 / (the derivative of the power ... in this case: -1)
e^(-x) * (1/-1) = -e^(-x)
Don't forget to add your constant!
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I'm not sure if you mean e^x + 17 or e^(x+17) so we'll do both. First, the integral of e^x + 17 because these terms are being added you can integrate them separately: integral((e^x)dx) + integral(17dx) integral of e^x is just e^x + C Integral of 17 is 17x + C, so we get: e^x + 17x + C Second, the integral of e^(x+17) we know how to integrate the form e^u, so just do a u substitution u=x+17 du=dx so we get integral((e^u)du)=e^u + C resubstitute for u and get e^(x+17) + C
The antiderivative, or indefinite integral, of ex, is ex + C.
This browser is not much use when it comes to mathematics but I'll try.Suppose X is a random variable with a Normal distribution and let f(x) be the probability density function of x.Then the mean is mu = E(X) = Integral of x*f(x) dx over the domain of X [which is negative infinity to positive infinity].The variance is E{[X - E(X)]2} = Integral of (x - mu)2*f(x) dx over the domain of X.
Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
integral (a^x) dx = (a^x) / ln(a)