Let a and b be real constants such that the function f defined by...

Given: f\left(x\right) = \left\{\begin{cases}{x}^{2} + 3x + a, & x \le 1 \\ bx + 2, & x > 1\end{cases}\right)</p><p>Now, for function to be continuous,</p><p>⇒ {\left(1\right)}^{2} + 3 \times 1 + a = b \times 1 + 2$</p><p>$⇒ 4 + a = b + 2$</p><p>$⇒ a - b = - 2 . . . \left(i\right)$</p><p>Now, differentiating and putting the value of$x$,</p><p>$⇒ 2x + 3 = b$</p><p>$⇒ 2\left(1\right) + 3 = b$</p><p>$⇒ b = 5$</p><p>$⇒ a = 3$</p><p>$⇒ {∫}_{- 2}^{2}f\left(x\right)dx = {∫}_{- 2}^{1}\left({x}^{2} + 3x + a\right)dx + {∫}_{1}^{2}\left(bx + 2\right)dx$</p><p>$⇒ {∫}_{- 2}^{2}f\left(x\right)dx = {∫}_{- 2}^{1}\left({x}^{2} + 3x + 3\right)dx + {∫}_{1}^{2}\left(5x + 2\right)dx$</p><p>$⇒ {∫}_{- 2}^{2}f\left(x\right)dx = {\left[\dfrac{{x}^{3}}{3} + \dfrac{3{x}^{2}}{2} + 3x\right]}_{- 2}^{1} + {\left[\dfrac{5{x}^{2}}{2} + 2x\right]}_{1}^{2}$</p><p>$⇒ {∫}_{- 2}^{2}f\left(x\right)dx = \left[\left(\dfrac{1}{3} + \dfrac{3}{2} + 3\right) - \left(\dfrac{- 8}{3} + 6 - 6\right)\right] + \left[\left(\dfrac{5 \times 4}{2} + 2 \times 2\right) - \left(\dfrac{5}{2} + 2\right)\right]$</p><p>$⇒ {∫}_{- 2}^{2}f\left(x\right)dx = \left[\dfrac{1}{3} + \dfrac{3}{2} + 3 + \dfrac{8}{3}\right] + \left[14 - \dfrac{9}{2}\right]$</p><p>$⇒ {∫}_{- 2}^{2}f\left(x\right)dx = \dfrac{15}{2} + 14 - \dfrac{9}{2}$</p><p>$⇒ {∫}_{- 2}^{2}f\left(x\right)dx = 17