Value of Definite Integral from Continuous Piecewise Function

To find the value of the given definite integral, we first determine the constant $b$ using the condition for the continuity of the function $f(x)$ at $x = \frac{\pi}{2}$.

Step 1: Determine the value of $b$

The function $f(x)$ is given as: $f(x)=\begin{cases} \frac{1}{3} & , x \le \pi / 2 \\ \frac{b(1-\sin x)}{(\pi-2 x)^2} & , x > \pi / 2 \end{cases}$

Since $f(x)$ is continuous at $x = \frac{\pi}{2}$, the left-hand limit, the value of the function at $x = \frac{\pi}{2}$, and the right-hand limit must all be equal: $f\left(\frac{\pi}{2}\right) = \lim_{x \to (\pi/2)^-} f(x) = \lim_{x \to (\pi/2)^+} f(x)$

We know that: $f\left(\frac{\pi}{2}\right) = \frac{1}{3}$

Now, let us calculate the right-hand limit at $x = \frac{\pi}{2}$: $\lim_{x \to (\pi/2)^+} f(x) = \lim_{x \to (\pi/2)^+} \frac{b(1-\sin x)}{(\pi-2 x)^2}$

Substitute $x = \frac{\pi}{2} + h$, where $h \to 0^+$: $\lim_{h \to 0^+} \frac{b\left(1-\sin\left(\frac{\pi}{2} + h\right)\right)}{\left(\pi - 2\left(\frac{\pi}{2} + h\right)\right)^2} = \lim_{h \to 0^+} \frac{b(1-\cos h)}{(-2h)^2}$

Using the standard limit $\lim_{h \to 0} \frac{1-\cos h}{h^2} = \frac{1}{2}$: $\lim_{h \to 0^+} \frac{b(1-\cos h)}{4h^2} = \frac{b}{4} \cdot \frac{1}{2} = \frac{b}{8}$

Equating the right-hand limit to $f\left(\frac{\pi}{2}\right)$: $\frac{b}{8} = \frac{1}{3} \implies b = \frac{8}{3}$

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Step 2: Evaluate the Definite Integral

The upper limit of the integral is: $3b - 6 = 3\left(\frac{8}{3}\right) - 6 = 8 - 6 = 2$

Thus, we need to calculate: $I = \int_0^2 \left|x^2+2x-3\right| dx$

Factorizing the expression inside the absolute value: $x^2 + 2x - 3 = (x + 3)(x - 1)$

Analyzing the sign of $(x^2+2x-3)$ on the interval $[0, 2]$:

• For $x \in [0, 1)$, $(x+3)(x-1) < 0$, so $\left|x^2+2x-3\right| = -(x^2+2x-3) = 3 - 2x - x^2$.
• For $x \in [1, 2]$, $(x+3)(x-1) \ge 0$, so $\left|x^2+2x-3\right| = x^2+2x-3$.

Splitting the integral at $x = 1$: $I = \int_0^1 (3 - 2x - x^2) dx + \int_1^2 (x^2 + 2x - 3) dx$

Calculating the first integral $I_1$: $I_1 = \left[ 3x - x^2 - \frac{x^3}{3} \right]_0^1 = \left(3(1) - 1^2 - \frac{1^3}{3}\right) - 0 = 3 - 1 - \frac{1}{3} = \frac{5}{3}$

Calculating the second integral $I_2$: $I_2 = \left[ \frac{x^3}{3} + x^2 - 3x \right]_1^2 = \left( \frac{2^3}{3} + 2^2 - 3(2) \right) - \left( \frac{1^3}{3} + 1^2 - 3(1) \right)$ $I_2 = \left( \frac{8}{3} + 4 - 6 \right) - \left( \frac{1}{3} + 1 - 3 \right) = \left( \frac{8}{3} - 2 \right) - \left( \frac{1}{3} - 2 \right) = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$

Adding $I_1$ and $I_2$: $I = I_1 + I_2 = \frac{5}{3} + \frac{7}{3} = \frac{12}{3} = 4$

Thus, the value of the integral is $4$, which corresponds to Option D.