Question

What is the area under `y=(x+4)/x` between `x=1` and `x=4`?

Answer 1

The area is `3 + 4 ln 4`.

Explanation:

When we find a definite integral of a function between two `x`-values, we're finding the area under the function, bounded by vertical lines at those two `x`-values (and the horizontal `x`-axis). Thus,

`int_(x=1)^4 (x+4)/x " "dx`

will give us the value we seek.

This function can be rewritten as `y=1+4/x" = "1+4x^-1` to make integration easier—in this form, the variable `x` only appears once.

Now, we integrate:

`int_(x=1)^4 (x+4)/x " "dx = int_(x=1)^4 (1+4x^-1) " "dx`

`color(white)(int_(x=1)^4 (x+4)/x " "dx) = int_(x=1)^4 1" "dx" + "int_(x=1)^4 4x^-1 " "dx`

`color(white)(int_(x=1)^4 (x+4)/x " "dx) = [x] _ (x=1)^4" + "4 [ln x]_(x=1)^4`

`color(white)(int_(x=1)^4 (x+4)/x " "dx) = [4-1]" + "4 [ln 4-ln 1]`
`color(white)(int_(x=1)^4 (x+4)/x " "dx) = 3" + "4 [ln 4-(0)]`
`color(white)(int_(x=1)^4 (x+4)/x " "dx) = 3+4ln 4`

So our area is `3+4 ln 4`.

Note:

This works as long as the function is non-negative between the two given endpoints `a` and `b`. If `y` is negative for `x`-values between `a` and `b` (i.e. if the graph falls below the `x`-axis within our bounds), the above process will treat the area below the `x`-axis as negative area.

Since we're usually interested in treating all areas as positive, we would split our integral up into sections with new endpoints. For example, if `y` becomes negative at `c`, where `a < c < b`, then the total positive area between `y` and the `x`-axis (and between `a` and `b`) would be

`int _a^c y" "dx" "-" "int_c^b y" "dx`

`=int _a^c y" "dx" "+" "int_b^c y" "dx` (note the `+`, and `b` & `c` are switched)

For this particular question, however, `y>0` for `x in [1,4]`, so this step was unnecessary.