How do you evaluate the definite integral #int (-2x^2+3x+2)dx# from [0,2]?

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Answer 1 · 2017-01-27T12:13:06

#14/3#

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Answer 2 · 2017-01-27T12:16:41

#int_0^2 (-2x^2+3x+2) \ dx = 14/3 #

We use # \ int x^n \ dx=x^(n+1)/(n+1)# giving;

#int_0^2 (-2x^2+3x+2) \ dx = [-2/3x^3+3/2x^2+2x]_0^2 #
# " " = (-2/3(2^3)+3/2(2^2)+2(2)) - (0) #
# " " = (-16/3 +6+4) - (0) #
# " " = 14/3 #