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
#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
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
#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,