Question

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

Answer 1

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.