Question
Let $f(x) = \left\{ \begin{matrix} \frac{1}{3} & , & x \leq \pi/2 \\ \frac{b(1 - sinx)}{(\pi - 2x)^{2}} & , & x > \pi/2 \end{matrix} \right.\ $, If $f$ is continuous at $x = \pi/2$, then the value of $\int_{0}^{3\text{ }b - 6}\mspace{2mu}\left| x^{2} + 2x - 3 \right|dx$ is :
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Solution
Sol. [Key Concept: Continuity at $x = \frac{\pi}{2}$ gives $b = \frac{8}{3}$ so $3b - 6 = 2$; then $\int_{0}^{2}{|x}^{2} + 2x - 3|\ dx = 4$.]
$${\lim_{x \rightarrow \frac{\pi}{2}}\mspace{2mu}\frac{b\left( 1 - cos\left( \frac{\pi}{2} - x \right) \right)}{4\left( \frac{\pi}{2} - x \right)^{2}} = \frac{1}{3} \Rightarrow b = \frac{8}{3} }{\int_{0}^{2}\mspace{2mu}\left| x^{2} + 2x - 3 \right|dx = \int_{0}^{2}\mspace{2mu}|(x + 3)(x - 1)|dx }$$put $x - 1 = t$
$${\int_{- 1}^{1}\mspace{2mu}|(t + 4)t|dt = \int_{- 1}^{0}\mspace{2mu}\left( - t^{2} - 4t \right)dt + \int_{0}^{1}\mspace{2mu}\left( t^{2} + 4t \right)dt }{\left( \frac{- t^{3}}{3} - 2t^{2} \right)_{- 1}^{0} + \left( \frac{t^{3}}{3} + 2t^{2} \right)_{0}^{1} = 4 }$$
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