Let a and b be real constants such that the function f defined by [f(x)=[arrayll x^2+3 x+a…

Question

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

Accepted Answer

To determine the integral of the piecewise function $f$ over the interval $$[-2, 2]$$, we first ensure that $f$ is differentiable on $$\mathbb{R}$$, as given in the problem statement. Differentiability implies continuity, so $f$ must also be continuous at $x=1$.

The condition for continuity at $x=1$ is:

$x^2 + 3x + a = bx + 2$ at $x=1$.

This simplifies to:

$1 + 3 + a = b(1) + 2$

$$\Rightarrow a = b - 2$$

The first derivative of $f$ gives us two different expressions depending on the value of $x$:

For $$x \leq 1$$, $f'(x) = 2x + 3$;
and for $x > 1$, $f'(x) = b$.

For $f$ to be differentiable at $x=1$, these derivatives must be equal at that point. Setting $f'(1)$ from both expressions equal to each other gives $2(1) + 3 = b$, thus $b = 5$. And from $a = b - 2$, we have $a = 3$.

Now, we can calculate the integral of $f$ over the specified interval:

$$\int\limits_{-2}^1 (x^2 + 3x + 3) \, dx + \int\limits_{1}^2 (5x + 2) \, dx$$

$$\begin{aligned}
& =\left[\frac{x^3}{3}+\frac{3 x^2}{2}+3 x\right]_{-2}^1+\left[\frac{5 x^2}{2}+2 x\right]_1^2 \\
& =\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right) \\
& =6+\frac{3}{2}+12-\frac{5}{2}=17
\end{aligned}$$

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