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

##### 1 Answer

The area is

#### Explanation:

When we find a definite integral of a function between two *area* under the function, bounded by vertical lines at those two

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

will give us the value we seek.

This function can be rewritten as

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

## Note:

This works as long as the function is non-negative between the two given endpoints *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 *positive* area between

#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,