evaluate the definite intergral top is 3 and bottom is 1 (3x^2+2x)dx

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integral((3x^2+2x)dx)=integral(3x^2dx)+integral(2xdx) for 1<x<3: [x^3+x^2](1<x<​3)=(3^3+3^2)-(1^3+1^2)=36-2=34.

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