Let f(x)=array|c|c| (1)/(3), & x /2 b(1- x)( -2x)^2, & x /2 array.

Question

Let $f(x)=\begin{array}{|c|c|} \frac{1}{3}, & x\le \pi /2 \ \frac{b(1-\sin x)}{(\pi -2x{)}^{2}}, & x>\pi /2 \end{array}$. If $f$ is continuous at $x=\pi /2$, then the value of $\int _{0}^{3b-6}|{x}^{2}+2x-3|dx$ is :

Accepted Answer

The function $f(x)$ is continuous at $x=\frac{\pi }{2}$ where

$f(x)=\begin{gathered} \frac{1}{3} ;x\leqslant \frac{\pi }{2} \ \frac{b(1-\sin x)}{(\pi -2x{)}^{2}} ;x>\frac{\pi }{2} \end{gathered}\cdots (i)$.

We observe the LHL from equation $(i)$ :

$\underset{x	o {(\frac{\pi }{2})}^{-}}{lim}f(x)=f(\frac{\pi }{2})=\frac{1}{3}\ldots \ldots (ii)$

We must have the RHL from equation (i) to be $\frac{1}{3}$ for the function $f(x)$ to be continuous at $x=\frac{\pi }{2}$ :

$\underset{x	o {(\frac{\pi }{2})}^{+}}{lim}f(x)=\frac{1}{3}\Rightarrow \underset{x	o {(\frac{\pi }{2})}^{+}}{lim}\frac{b(1-\sin x)}{(\pi -2x{)}^{2}}=\frac{1}{3}\ldots \ldots (iii)$

$\Rightarrow$ $\underset{x	o {(\frac{\pi }{2})}^{+}}{lim}\frac{b(1-\sin x)}{(\pi -2x{)}^{2}}=\frac{0}{0}$ format $\leftarrow $ Apply L'hopital's rule

$\Rightarrow$ $\underset{x	o {(\frac{\pi }{2})}^{+}}{lim}\frac{b\cos x}{4(\pi -2x)}=\frac{0}{0}$ format  Apply L'hopital's rule once more

$\Rightarrow$ $\underset{x	o {(\frac{\pi }{2})}^{+}}{lim}\frac{b\sin x}{8}=\frac{b}{8}\ldots \ldots (iv)$

We can mathematically demonstrate conditional continuity of $f(x)$ at $x=\frac{\pi }{2}$ from equation (iii) and (iv) :

$\frac{b}{8}=\frac{1}{3}\Rightarrow b=\frac{8}{3}\ldots \ldots (v)$

Now, I = $\int _{0}^{3b-6}|{x}^{2}+2x-3|dx$

$=\int _{0}^{(3)(\frac{8}{3})-6}|{x}^{2}+2x-3|dx$

$=\int _{0}^{2}|{x}^{2}+2x-3|dx$

$=\int _{0}^{2}|(x+3)(x-1)|dx$

$\begin{gathered} (x+3)(x-1)\geqslant 0;x\in (-\infty,-3] \ (x+3)(x-1)\leqslant 0;x\in [-3,1] \ (x+3)(x-1)\geqslant 0;x\in [1,\infty) \end{gathered}$

$=\int _{0}^{1}(x+3)(1-x)dx$ $+\int _{1}^{2}(x+3)(x-1)dx$

$=\int _{0}^{1}(3-2x-{x}^{2})dx$ $+\int _{1}^{2}({x}^{2}+2x-3)dx$

$\begin{gathered} ={[3x-{x}^{2}-\frac{{x}^{3}}{3}]}_{1}^{0}+{[\frac{{x}^{3}}{3}+{x}^{2}-3x]}_{2}^{1} \end{gathered}$

$\begin{gathered} =(3-1-\frac{1}{3})+(\frac{8}{3}+4-6)-(\frac{1}{3}+1-3) \end{gathered}$

= 4

Solutions · 1