
Calculate the area
Calculate the area of the region limited by curves:
1) $\displaystyle y=x^3$ and $\displaystyle y=x^2+2x$
answer: $\displaystyle \frac{37}{12}$
2) $\displaystyle y=x^32x^25x+6$, the axisx, $\displaystyle x=1$ and $\displaystyle x=2$
answer: $\displaystyle \frac{157}{12}$

1) Check where the two curves meet:
$\displaystyle x^3 = x^2 + 2x$ which gives three points: $\displaystyle x=1, x=0, x=2$
So you need to split the integration area up into two parts:
$\displaystyle \int_{1}^0 (x^3  (x^2 + 2x))dx + \int_0^2 (x^2 + 2x  x^3) dx$
2) Check where the curve meets the xaxis:
$\displaystyle x^3  2x^2 5x +6 = 0$
Same procedure as in (1).