Let f(x) = begin{cases} frac{1}{3}, & x le pi | vedclass

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(D) For $f(x)$ to be continuous at $x = \pi/2$, the left-hand limit must equal the right-hand limit.$\lim_{x 	o \pi/2^-} f(x) = 1/3$.$\lim_{x 	o \pi

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$4$

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(D) For $f(x)$ to be continuous at $x = \pi/2$, the left-hand limit must equal the right-hand limit.$\lim_{x 	o \pi/2^-} f(x) = 1/3$.$\lim_{x 	o \pi/2^+} f(x) = \lim_{x 	o \pi/2^+} \frac{b(1-\sin x)}{4(\pi/2-x)^2}$.Let $h = \pi/2 - x$. As $x 	o \pi/2^+$, $h 	o 0^+$.$\lim_{h 	o 0^+} \frac{b(1-\cos h)}{4h^2} = \lim_{h 	o 0^+} \frac{b(2\sin^2(h/2))}{4(4(h/2)^2)} = \frac{2b}{16} = \frac{b}{8}$.Equating the limits: $b/8 = 1/3 \implies b = 8/3$.Now, $3b-6 = 3(8/3) - 6 = 8 - 6 = 2$.We need to evaluate $\int_0^2 |x^2+2x-3| dx$.Since $x^2+2x-3 = (x+3)(x-1)$, the expression is negative for $x \in [0, 1)$ and positive for $x \in (1, 2]$.Integral $= -\int_0^1 (x^2+2x-3) dx + \int_1^2 (x^2+2x-3) dx$.$= -[x^3/3 + x^2 - 3x]_0^1 + [x^3/3 + x^2 - 3x]_1^2$.$= -[1/3 + 1 - 3] + [(8/3 + 4 - 6) - (1/3 + 1 - 3)]$.$= -[-5/3] + [2/3 - (-5/3)] = 5/3 + 7/3 = 12/3 = 4$.

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

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