1. Since $f$ is differentiable on $\mathbb{R}$, it must be both continuous and have equal left and right derivatives at the joining point $x=1$.

Given

we apply the two conditions at $x=1$.

2. Continuity at $x=1$

Left value: $f(1)=1^2+3(1)+a=4+a$ Right value at $x=1$ from the second part: $b(1)+2=b+2$ So, $4+a=b+2 \implies a-b=-2 \implies a=b-2$

3. Differentiability at $x=1$

Left derivative: $\frac{d}{dx}(x^2+3x+a)=2x+3$ So at $x=1$, $2(1)+3=5$

Right derivative: $\frac{d}{dx}(bx+2)=b$ Hence, $b=5$ Then $a=b-2=5-2=3$

4. Therefore,

5. Now compute $\int_{-2}^{2} f(x)\,dx=\int_{-2}^{1}(x^2+3x+3)\,dx+\int_{1}^{2}(5x+2)\,dx$

First integral: $\int (x^2+3x+3)dx=\frac{x^3}{3}+\frac{3x^2}{2}+3x$ So,

At $x=1$: $\frac{1}{3}+\frac{3}{2}+3=\frac{2+9+18}{6}=\frac{29}{6}$ At $x=-2$: $\frac{-8}{3}+\frac{3\cdot 4}{2}-6=-\frac{8}{3}+6-6=-\frac{8}{3}$ Thus,

Second integral: $\int_1^2 (5x+2)dx=\left[\frac{5x^2}{2}+2x\right]_1^2$ At $x=2$: $\frac{5\cdot 4}{2}+4=10+4=14$ At $x=1$: $\frac{5}{2}+2=\frac{9}{2}$ So,

6. Total integral:

Hence the correct option is C.