AP Calculus Integration Question

Question

The question is to integrate from 0 to 1 (3x-2)^2 dx. At first I got (1/9)(3x-2)^3 and then substituted in 1 to get 1/9 as the answer but for some reason this answer is incorrect. When I checked how to do it you had to FOIL out the (3x-2)^2 to get 9x^2-12x+4 and then integrate that to get 3x^3 - 6x^2 + 4x. Then when I substituted 1 into that equation I got 1 as an answer. My question is why did I get two different answers? If I take the derivative of my first way then I get (3x-2)^2 so why doesn't it work?

Answer 1

You did the integral correctly both ways.  You made a mistake when evaluating the  first integral at 1 and 0.

 

(1/9)(3x-2)3 |¹

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= (1/9)[(3·1 - 2)³ - (3·0 - 2)³] = (1/9)[1+8} = 9/9 = 1