Question

Question #942b1

Answer 1

Mike, all three can be done by substitution. Let `u` (or `g(x)` depending on the notation you've been taught) be the exponent on `e`.

I'll go through the first one, {EDIT: I'll use notation and an approach that may be more familiar to you

`int_0^3 e^(-3x) dx`

First let's just find the indefinite integral:

`int e^(-3x) dx`

Let `u=-3x`,

That makes `du = -3 dx`.

I don't see a `-3` in front of the `dx` in the integrand, so We'll solve for `dx`.

`dx=-1/3 du`, now here is what we have:

`int e^(overbracecolor(red)(-3x)^(color(red)"u")) overbrace color(green)(dx)^color(green)(-1/3 du) = int e^color(red)(u) * color(green)((-1/3) du) = -1/3 int e^u du = -1/3 e^u +C`

Now use `u=-3x` to finish:

`int e^(-3x) dx = -1/3 e^(-3x) +C`

For the definite integral, we don't need the `+C` (it subtracts out)

`int_0^3 e^(-3x) dx = -1/3 (e^(-3x) |_0^3)`

`= -1/3(e^(-3(3)) - e^(-3(0)))`

`= -1/3(e^(-9) - e^0)`.

`= -1/3(e^(-9) - 1)`

Now do whatever algebra you like.

I prefer:

`=1/3(1-1/e^9)` or `=1/3-1/(3e^9)` or `= (e^9-1)/(3e^9)`