Find the area of the region between y = x^2 and y = 4-3x from x = -1 to x = 4/3
$\displaystyle \int_{-1}^1 (4-3x-x^2)\,dx+\int_1^{\frac{4}{3}}(x^2+3x-4)\,dx$
$\displaystyle \implies\left.\bigg[4x-\tfrac{3}{2}x^2-\tfrac{1}{3}x^3\bigg]\right|_{-1}^1+\left.\bigg[\tfrac{1}{3}x^3+\tfrac{3}{2}x^2-4x\bigg]\right|_1^{\frac{4}{3}}$
$\displaystyle \implies\bigg[\left(4-\tfrac{3}{2}-\tfrac{1}{3}\right)-\left(-4-\tfrac{3}{2}+\tfrac{1}{3}\right)\bigg]+\bigg[\left(\tfrac{64}{81}+\tfrac{8}{3}-\tfrac{16}{3}\right)-\left(\tfrac{1}{3}+\tfrac{3}{2}-4\right)\bigg]=\color{red}\boxed{\frac{1235}{162}}$
Does this make sense?
--Chris