How do you evaluate the definite integral #int (3x^2-2x+1)dx# from [1,5]?

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Answer 1 · 2016-11-21T07:35:36

The answer is #=104#

Here, we use the integral

#intx^ndx=x^(n+1)/(n+1)+C (n!=-1)#

Therefore

#int_1^5(3x^2-2x+1)dx= [3*x^3/3-2*x^2/2+x]_1^5#

#=(5^3-5^2+5)-(1-1+1)#

#=125-25+5-1#

#=104#