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How do you find the definite integral for: #(3x^2+2x+3) dx# for the intervals #[0, 6]#?

1 Answer

Aug 3, 2016

#int_0^6(3x^2+2x+3)d x=?#

#int_0^6(3x^2+2x+3)d x=|3*x^3/3+2*x^2/2+3x|_0^6#

#int_0^6(3x^2+2x+3)d x=|x^3+x^2+3x|_0^6#

#int_0^6(3x^2+2x+3)d x=[(6^3+6^2+3*6)-(0^3+0^2+3*0)]#

#int_0^6(3x^2+2x+3)d x=[(216+36+18)-(0)]#

#int_0^6(3x^2+2x+3)d x=270#

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