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What is the net area between #f(x) = (x-2)^3 # and the x-axis over #x in [1, 5 ]#?

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1 Answer

Ananda Dasgupta

Jul 8, 2018

#20.5#

Explanation:

The integrand #f(x) = (x-2)^3# crosses 0 at #x=2#. The area enclosed between #x=1# and #x=2# is given by

#A_1 = |int_1^2 (x-2)^3dx|#
#qquadquad = |(x-2)^4/4]_1^2| #
#qquadqquad = |-1/4| = 1/4#

while, the area enclosed between #x=2# and #x=5# is given by

#A_2 = |int_2^5 (x-2)^3dx|#
#qquadquad = |(x-2)^4/4]_2^5| #
#qquadqquad = |81/4| = 81/4#

Thus, the net area is #81/4+1/4=82/4=20.5#

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