What is the net area between #f(x) = sqrt(x^2+2x+1) # and the x-axis over #x in [2, 4 ]#?

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Answer 1 · 2017-07-14T01:46:07

#8# sq. units

Actually we can do this one without the need of Calculus, as follows.

#f(x) = sqrt(x^2+2x+1)#

#= sqrt((x+1)^2)#

#= abs( x+1)#

Since we are only concerned with #x in [2, 4]# we now only have to consider a straight line of slope 1 and #y-# intercept 1.

The area between #f(x)# and the #x-#axis #x in [2, 4]# can be separated into:

(i) a rectangle of sides #(4-2)# and #(2+1-0)#
and
(ii) a right triangle of sides #(4-2)# and #(4+1-(2+1-0))#

Area of rectangle = #2xx3 = 6#

Area of right triangle = #1/2xx2xx2 = 2#

Required area #= 6+2 = 8# sq. units