Let $$f(x) = \begin{cases} \frac{1}{3}, & x \le \frac{\pi}{2} \\ \frac{b(1 - \sin x)}{(\pi - 2x)^2}, & x > \frac{\pi}{2} \end{cases}$$. If $$f$$ is continuous at $$x = \pi/2$$, then the value of $$\int_0^{3b-6} |x^2 + 2x - 3| \, dx$$ is :

$$\text{LHL} = f\left(\frac{\pi}{2}\right) = \frac{1}{3}$$

$$\text{RHL} = \lim_{x \to \frac{\pi}{2}^+} \frac{b(1 - \sin x)}{(\pi - 2x)^2}$$

$$\text{RHL} = \lim_{h \to 0} \frac{b\left(1 - \sin\left(\frac{\pi}{2} + h\right)\right)}{\left(\pi - 2\left(\frac{\pi}{2} + h\right)\right)^2}$$

$$\text{RHL} = \lim_{h \to 0} \frac{b(1 - \cos h)}{(-2h)^2} = \lim_{h \to 0} \frac{b(1 - \cos h)}{4h^2}$$

$$\text{RHL} = \frac{b}{4} \cdot \left(\frac{1}{2}\right) = \frac{b}{8}$$

$$\frac{b}{8} = \frac{1}{3} \implies b = \frac{8}{3}$$

$$\text{Upper Limit} = 3\left(\frac{8}{3}\right) - 6 = 8 - 6 = 2$$

$$I = \int_{0}^{2} |x^2 + 2x - 3| \, dx$$

$$x^2 + 2x - 3 = (x + 3)(x - 1)$$

$$I = \int_{0}^{1} -(x^2 + 2x - 3) \, dx + \int_{1}^{2} (x^2 + 2x - 3) \, dx$$

$$\int (x^2 + 2x - 3) \, dx = \frac{x^3}{3} + x^2 - 3x$$

$$-\left[ \frac{x^3}{3} + x^2 - 3x \right]_0^1 = -\left( \frac{1}{3} + 1 - 3 - 0 \right) = -\left( \frac{1}{3} - 2 \right) = -\left( -\frac{5}{3} \right) = \frac{5}{3}$$

$$\left[ \frac{x^3}{3} + x^2 - 3x \right]_1^2 = \left( \frac{8}{3} + 4 - 6 \right) - \left( \frac{1}{3} + 1 - 3 \right)$$

$$= \left( \frac{8}{3} - 2 \right) - \left( -\frac{5}{3} \right) = \frac{2}{3} + \frac{5}{3} = \frac{7}{3}$$

$$I = \frac{5}{3} + \frac{7}{3} = \frac{12}{3} = 4$$