What is the net area between #f(x) = x/sqrt(x+1) # and the x-axis over #x in [1, 4 ]#?

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Answer 1 · 2017-02-03T03:37:10

#2/3(2sqrt5+sqrt2)=3.924# areal units, nearly. See Socratic graph to see the enclosure.

#y=x/sqrt(x+1)=((x+1)-1)/sqrt(x+1)=(x+1)^(1/2)-(x+1)^(-1/2)#.-

The area = #inty dx#, with #y =(x+1)^(1/2)-(x+1)^(-1/2)# and x from 1 to 4

#=int((x+1)^(1/2)-(x+1)^(-1/2)) dx#, for the limits

#=[2/3(x+1)^(3/2)-2(x+1)^(1/2)]#, between 1 and 4

#=(10/3sqrt5-2sqrt5)-(4/3sqrt2-2sqrt2)#

#=4/3sqrt5+2/3sqrt2#

graph{(y-x/sqrt(x+1))y(x-1+.01y)(x-4)=0 [-10, 10, -5, 5]}