Question

How do you integrate `e^(-1.5x)`?

Answer 1

Use substitution with `u = -1.5x`

Explanation:

Any integral of the form `int e^(kx) dx` where `k` is a constant, may be evaluated by using `u = kx`, so `du = k dx`

`int e^(kx) dx = 1/k int e^(kx) kdx`

`= 1/k int e^u du = 1/ke^u +C`

`= 1/ke^(kx) +C`

This answer makes sense, because the derivative of `e^(kx)` is `e^(kx) *k`
That is: we know the integral of `e^(-1.5x)` must involve `e^(-1.5x)`, but the derivative of that is `d/dx(e^(-1.5x))= -1.5 e^(-1.5x)`. We'll remove the `-1.5` by dividing:

`int e^(-1.5x) dx = 1/(-1.5) e^(-1.5x) +C`

Note: fraction with a decimal in the numerator or denominator looks strange to me. (I think: "Make up you mind! One or the other!")

`1.5 = 3/2` so `1/1.5 = 2/3` and we can write our answer:

`-2/3 e^(-1.5x) +C`