# How do you find the definite integral of #int (1-2x-3x^2)dx# from #[0,2]#?

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To find the definite integral of ( \int (1 - 2x - 3x^2) , dx ) from ( x = 0 ) to ( x = 2 ), you first need to integrate the function with respect to ( x ), then evaluate the result at the upper and lower limits of integration (in this case, 2 and 0 respectively), and finally subtract the result at the lower limit from the result at the upper limit.

First, integrate the function: [ \int (1 - 2x - 3x^2) , dx = \left( x - x^2 - x^3 \right) + C ]

Next, evaluate the antiderivative at the upper and lower limits: [ F(2) = \left( 2 - 2^2 - 2^3 \right) = -6 ] [ F(0) = \left( 0 - 0^2 - 0^3 \right) = 0 ]

Finally, subtract the result at the lower limit from the result at the upper limit: [ F(2) - F(0) = -6 - 0 = -6 ]

So, the definite integral of ( \int (1 - 2x - 3x^2) , dx ) from ( x = 0 ) to ( x = 2 ) is ( -6 ).

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