How do you evaluate the definite integral #int (x^3-x^2)dx# from [1,3]?

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Answer 1 · 2017-02-08T02:10:04

#int_1^3(x^3-x^2)dx=11.bar3#

#int_a^bf(x)dx=F(b)-F(a)#, where #F'(x)=f(x)#

In this case #a=1# and #b=3#

and #f(x)=x^3-x^2#

Also, since #intf(x)+g(x)dx=intf(x)dx+intg(x)dx#

#int_1^3(x^3-x^2)dx=int_1^3x^3dx-int_1^3x^2dx#

and since #intx^ndx=x^(n+1)/(n+1)# we get

#int_1^3(x^3-x^2)dx=x^4/4]_1^3-x^3/3]_1^3=(81/4-1/4)-(27/3-1/3)#

#=80/4-26/3=20-8.bar6=11.bar3#