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

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Answer 1 · 2017-04-19T21:19:47

#area=260.58 " unit"#

#area=int_2^5 f(x) d x#

#area=int _2^5 (x^3+4x^2-2) d x#

#area=|1/4 x^4+4*1/3 x^3-2x|_2^5#

#area=|1/4 x^4+4/3x^3-2x|_2^5#

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

#area=(625/4+500/3-10)-(4+32/3-4)#

#area=625/4+500/4-10-4-32/3+4#

#area=1125/4-10-32/3#

#area=(3375-120-128)/12#

#area=260.58 " unit"#