What is the net area between #f(x)=ln(x+1)# in #x in[1,2] # and the x-axis?

Question

What is the net area  between #f(x)=ln(x+1)# in #x  in[1,2]    # and the x-axis?

Emily Ammerman · Accepted Answer

#3ln3-2ln2-1# #int_1^2ln(x+1)dx# First change variables: Let #color(red)(w=x+1)# #dw=dx# Next, solve #intln(color(red)(w))dw# using integration by parts. #int (lnw)dw# #color(blue)(u=lnw" "v=w)# #color(blue)(du=1/wdw" "dv=1)# #int(lnw)dw=wlnw-int(w*1/w)dw# #=wlnw-int(1)dw# #=color(red)(w)lncolor(red)(w)-color(red)(w)# #=(x+1)ln(x+1)-(x+1)# #=(x +1)ln(x+1)-x-1# Now, go back to the definite integral: #int_1^2ln(x+1)dx# #=[(x+1)ln(x+1)-x-1]_1^2# #=[(2+1)ln(2+1)-2-1]-[(1+1)ln(1+1)-1-1]# #=3ln3-3-(2ln2-2)# #=3ln3-3-2ln2+2# #=3ln3-2ln2-1#

Ezekiel Alexander · Answer

undefined To find the net area between $ f(x) = \ln(x + 1) $ and the x-axis over the interval $[1, 2]$, you integrate the absolute value of $ f(x) $ over the interval and subtract the integral of the x-axis over the same interval.

The integral of $ f(x) = \ln(x + 1) $ from 1 to 2 is:

$$ \int_{1}^{2} \ln(x + 1) \, dx $$

The integral of the x-axis (y = 0) over the same interval is simply the integral of 0, which evaluates to 0.

So, the net area is:

$$ \int_{1}^{2} |\ln(x + 1)| \, dx $$