$ \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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| $X$ | $c$ | $2c$ | $3c$ | $4c$ | $5c$ | $6c$ |
| $f$ | $2$ | $1$ | $1$ | $1$ | $1$ | $1$ |
$\left| {1 - {{\log }_{\frac{1}{6}}}x} \right| + \left| {{{\log }_2}x} \right| + 2 = \left| {3 - {{\log }_{\frac{1}{6}}}x + {{\log }_{\frac{1}{2}}}x} \right|$ is $\left[ {\frac{a}{b},a} \right],a,b, \in N,$ then the value of $(a + b)$ is