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

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.