Let a and b be real constants such that the f | vedclass

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(D) For $f(x)$ to be differentiable on $\mathbb{R}$,it must be continuous at $x = 1$. Thus,$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) \implies 1^2

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(D) For $f(x)$ to be differentiable on $\mathbb{R}$,it must be continuous at $x = 1$. Thus,$\lim_{x 	o 1^-} f(x) = \lim_{x 	o 1^+} f(x) \implies 1^2 + 3(1) + a = b(1) + 2 \implies 4 + a = b + 2 \implies a = b - 2$. Also,the derivative must exist at $x = 1$. $f'(x) = 2x + 3$ for $x  1$. For differentiability at $x = 1$,$\lim_{x 	o 1^-} f'(x) = \lim_{x 	o 1^+} f'(x) \implies 2(1) + 3 = b \implies b = 5$. Substituting $b = 5$ into $a = b - 2$,we get $a = 3$. Now,we calculate the integral: $\int_{-2}^2 f(x) dx = \int_{-2}^1 (x^2 + 3x + 3) dx + \int_1^2 (5x + 2) dx$. $= \left[ \frac{x^3}{3} + \frac{3x^2}{2} + 3x ight]_{-2}^1 + \left[ \frac{5x^2}{2} + 2x ight]_1^2$. $= \left( \frac{1}{3} + \frac{3}{2} + 3 ight) - \left( \frac{-8}{3} + 6 - 6 ight) + \left( (10 + 4) - (\frac{5}{2} + 2) ight)$. $= (\frac{2 + 9 + 18}{6}) - (-\frac{8}{3}) + (14 - \frac{9}{2}) = \frac{29}{6} + \frac{16}{6} + \frac{19}{2} = \frac{45}{6} + \frac{57}{6} = \frac{102}{6} = 17$.

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

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