What is the integral of $int e^((-1/2)*x) dx$ ?

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Answer 1 · 2016-03-25T01:33:29

One way to do this is by inspection, starting from the derivative of the exponential function which is

$d/(dx) e^(a*x) = a*e^(a*x)$

which can be multiplied by a constant of both sides. It may not be obvious just yet why this is important, but we'll see shortly:

$d/(dx) b*e^(a*x) = b*a*e^(a*x)$

We are looking for a function of $x$ which is

$f(x)=int e^((-1/2)*x) dx$

If we take the derivative of both sides we get

$d/(dx) f(x) = e^((-1/2)*x)$

where we can guess an $f(x)$ in the form of the above equation in terms of $a$ and $b$

$d/(dx) b*e^(a*x) = b*a*e^(a*x) = e^((-1/2)*x)$

Solving for $a$ and $b$ we get:

$a=-1/2$ from the exponent and

$b*a = 1$ or

$b=1/a = -2$

therefore

$int e^((-1/2)*x) dx = -2*e^((-1/2)*x)+c$

where we have added an arbitrary constant for the indefinite integral.

What is the integral of int e^((-1/2)*x) dx int e^((-1/2)*x) dx ?