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

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Answer 1 · 2017-03-21T02:04:24

#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#