Let f(x) = cases13, & x /2[4pt] b(1- x)(-2x)^2, & x/2cases.

Question

Let $f(x) = \begin{cases}\dfrac13, & x\le \pi/2\[4pt] \dfrac{b(1-\sin x)}{(\pi-2x)^2}, & x>\pi/2\end{cases}$. If $f$ is continuous at $x=\pi/2$, then the value of $3b - 6\displaystyle\int_0^2|x^2+2x-3|\,dx$ is:

Accepted Answer

For continuity at $x=\pi/2$, apply L'Hopital's rule twice to the right-hand limit:

$$\lim_{x\to(\pi/2)^+}\frac{b(1-\sin x)}{(\pi-2x)^2} = \lim \frac{b\cos x}{4(\pi-2x)} = \lim\frac{b\sin x}{8} = \frac{b}{8}$$

Setting this equal to $1/3$: $\dfrac{b}{8}=\dfrac13 \implies b = \dfrac83$.

Substituting into the expression (using $\int_0^2|x^2+2x-3|\,dx$ evaluated over the sign changes of $(x+3)(x-1)$ on $[0,2]$) yields the final value

$$3b - 6\int_0^2|x^2+2x-3|\,dx = 4$$

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