$\int_{0}^{2} (x-1)^3 dx + 2\int_{-1}^{2} (x-1)x dx$ $\int_{0}^{2} (x-1)^3 dx + 2\int_{-1}^2 (x-1)x dx$
$\int_{0}^{2} u^3 du + 2\int_{-1}^2 x^2 - x dx$
I get 4 + 3 = 7
But the answer is 3. Where am i wrong? Any hints anyone?
$$I=(\frac{(x-1)^4}{4})|_{0}^{2}+(2\frac{x^3}{3}-x^2)|_{-1}^{2}=\frac{1}{4}-\frac{1}{4}+[(2\frac{8}{3}-4)-(2\frac{-1}{3}-1)]=\frac{4}{3}+\frac{5}{3}=3$$
$$I=(\frac{(x-1)^4}{4})|_{0}^{2}+(2\frac{x^3}{3}-x^2)|_{-1}^{2}=\frac{1}{4}-\frac{1}{4}+[(2\frac{8}{3}-4)-(2\frac{-1}{3}-1)]=\frac{4}{3}+\frac{5}{3}=3$$