$ \therefore 4+\mathrm{a}=\mathrm{b}+2$
$ \text { b }, x>1 $
$ \mathrm{a}=\mathrm{b}-2 \quad \mathrm{f} \text { is differentiable } $
$ \therefore \mathrm{b}=5 $
$ \therefore \quad \mathrm{a}=3 $
$ \int_{-2}^1\left(x^2+3 x+3\right) d x+\int_1^2(5 x+2) d x $
$ =\left[\frac{\mathrm{x}^3}{3}+\frac{3 \mathrm{x}^2}{2}+3 \mathrm{x}\right]_{-2}^1+\left[\frac{5 \mathrm{x}^2}{2}+2 \mathrm{x}\right]_1^2$
$ =\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right)$
$ =6+\frac{3}{2}+12-\frac{5}{2}=17 $
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