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

Dec 28, 2016

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,

${\int}_{x = 1}^{4} \frac{x + 4}{x} \text{ } \mathrm{dx}$

will give us the value we seek.

This function can be rewritten as $y = 1 + \frac{4}{x} \text{ = } 1 + 4 {x}^{-} 1$ to make integration easier—in this form, the variable $x$ only appears once.

Now, we integrate:

${\int}_{x = 1}^{4} \frac{x + 4}{x} \text{ "dx = int_(x=1)^4 (1+4x^-1) " } \mathrm{dx}$

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)]
$\textcolor{w h i t e}{{\int}_{x = 1}^{4} \frac{x + 4}{x} \text{ } \mathrm{dx}} = 3 + 4 \ln 4$

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

${\int}_{a}^{c} y \text{ "dx" "-" "int_c^b y" } \mathrm{dx}$

$= {\int}_{a}^{c} y \text{ "dx" "+" "int_b^c y" } \mathrm{dx}$ (note the $+$, and $b$ & $c$ are switched)

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