What is the net area between #f(x)=-4x^3+2x^2-5# in #x in[1,5] # and the x-axis?

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Answer 1 · 2016-12-09T17:37:48

The answer is #=-561.4#

The area is given by

#dA=ydx#

#A=intf(x)dx=int_1^5(-4x^3+2x^2-5)dx#

#=[-x^4+2/3x^3-5x]_1^5 #

#=(-5^4+2/3*5^3-5^2)-(-1+2/3-5)#

#=-566.7+5.33=-561.4#

graph{-4x^3+2x^2-5 [-41.1, 41.1, -20.56, 20.56]}