Let $a$ and $b$ be real constants such that the function $f$ defined by $$f(x)=\left\{\begin{array}{ll}x^2+3 x+a & , x \leq 1 \\ b x+2 & , x>1\end{array}\right.$$ be differentiable on $\mathbb{R}$. Then, the value of $\int_\limits{-2}^2 f(x) d x$ equals

<p>To determine the integral of the piecewise function <span class="math-container">$f$</span> over the interval <span class="math-container">$[-2, 2]$</span>, we first ensure that <span class="math-container">$f$</span> is differentiable on <span class="math-container">$\mathbb{R}$</span>, as given in the problem statement. Differentiability implies continuity, so <span class="math-container">$f$</span> must also be continuous at <span class="math-container">$x=1$</span>.</p> <p>The condition for continuity at <span class="math-container">$x=1$</span> is:</p> <p><span class="math-container">$x^2 + 3x + a = bx + 2$</span> at <span class="math-container">$x=1$</span>.</p> <p>This simplifies to:</p> <p><span class="math-container">$1 + 3 + a = b(1) + 2$</span></p> <p><span class="math-container">$\Rightarrow a = b - 2$</span></p> <p>The first derivative of <span class="math-container">$f$</span> gives us two different expressions depending on the value of <span class="math-container">$x$</span>:</p> <p>For <span class="math-container">$x \leq 1$</span>, <span class="math-container">$f'(x) = 2x + 3$</span>; and for <span class="math-container">$x > 1$</span>, <span class="math-container">$f'(x) = b$</span>.</p> <p>For <span class="math-container">$f$</span> to be differentiable at <span class="math-container">$x=1$</span>, these derivatives must be equal at that point. Setting <span class="math-container">$f'(1)$</span> from both expressions equal to each other gives <span class="math-container">$2(1) + 3 = b$</span>, thus <span class="math-container">$b = 5$</span>. And from <span class="math-container">$a = b - 2$</span>, we have <span class="math-container">$a = 3$</span>.</p> <p>Now, we can calculate the integral of <span class="math-container">$f$</span> over the specified interval:</p> <p><span class="math-container">$\int\limits_{-2}^1 (x^2 + 3x + 3) \, dx + \int\limits_{1}^2 (5x + 2) \, dx$</span></p> <p>$$\begin{aligned} & =\left[\frac{x^3}{3}+\frac{3 x^2}{2}+3 x\right]_{-2}^1+\left[\frac{5 x^2}{2}+2 x\right]_1^2 \\\\ & =\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right) \\\\ & =6+\frac{3}{2}+12-\frac{5}{2}=17 \end{aligned}$$</p>