Question

Given that for each $a \in \left( {0,1} \right)$ ,$\mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{t^a}{{\left( {1 - t} \right)}^{a - 1}}dt}$ exists. Let this limit be $g\left( a \right)$ . In addition, it is given that the function $g\left( a \right)$ is differentiable on (0,1), then the value of $g\left( {\dfrac{1}{2}} \right)$ is?
A) $\pi$
B) $2\pi$
C) $\dfrac{\pi }{2}$
D) $\dfrac{\pi }{4}$

Answer 1

In the integral calculus, we are to find a function whose differential is given. Thus, integration is a process that is the inverse of differentiation.
Let $\dfrac{{dF\left( x \right)}}{{dx}} = f\left( x \right)$, then $\smallint f\left( x \right)dx = F\left( x \right) + C$ , where C is the integration constant.
Calculate $g\left( {\dfrac{1}{2}} \right)$ but substituting a by $\dfrac{1}{2}$ and solve the resultant definite integral.
The definite integral is calculated whenever a function $f$ is given over the range $\left[ {a,b} \right]$ , where $a$ and $b$ are called limits of integration, $a$ being the lower limit and $b$ being the upper limit. I.e.:
$\int\limits_a^b {f\left( x \right)dx} = F\left( b \right) - F\left( a \right)$
In other words, definite integral (integral whose limits are given) is given by upper limits minus lower limits.
A change in the variable of integration often reduces an integral to one of the fundamental integrals.
The method in which we change the variable to some other variable is called the method of substitution.
When the integrand involves some trigonometric functions, we use some well-known identities to find the integrals.

Complete step-by-step answer:
Step 1: Given that:
$g\left( a \right) = \mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{t^{ - a}}{{\left( {1 - t} \right)}^{a - 1}}} dt$
To find $g\left( {\dfrac{1}{2}} \right)$
Put $a = \dfrac{1}{2}$
$g\left( {\dfrac{1}{2}} \right) = \mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {{t^{ - \dfrac{1}{2}}}{{\left( {1 - t} \right)}^{ - \dfrac{1}{2}}}} dt$
Both $t$ and $\left( {1 - t} \right)$ have the same power $- \dfrac{1}{2}$, thus, they can be concluded under the one square-root.
$= \mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {\dfrac{1}{{\sqrt {t\left( {1 - t} \right)} }}} dt$
Step 2: Solving the integral by substitution:
Take $t = {\sin ^2}\theta$
On differentiating both sides
Using the differential formulas: $\dfrac{{dt}}{{dt}} = 1$ ,
$\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$ ,
$\dfrac{{d\left( {\sin x} \right)}}{{dx}} = \cos x$
$dt = 2\sin \theta \cos \theta d\theta$
Change the limits according to the $\theta$ .
As $t \to h \to 0$
$\theta \to {\sin ^{ - 1}}\sqrt 0 = 0$
As $t \to 1 - h = 1$
$\theta \to {\sin ^{ - 1}}\sqrt 1 = \dfrac{\pi }{2}$
On substituting $t$, $dt$ and limits of $\theta$. The integral becomes:
$\Rightarrow \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{2\sin \theta \cos \theta }}{{\sqrt {{{\sin }^2}\theta \left( {1 - {{\sin }^2}\theta } \right)} }}} d\theta$
Using trigonometric identity: ${\sin ^2}x + {\cos ^2}x = 1$
$\Rightarrow {\cos ^2}x = 1 - {\sin ^2}x$
Thus, $\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{2\sin \theta \cos \theta }}{{\sqrt {{{\sin }^2}\theta {{\cos }^2}\theta } }}} d\theta$
$= \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{2\sin \theta \cos \theta }}{{\sin \theta \cos \theta }}} d\theta$
$= \int\limits_0^{\dfrac{\pi }{2}} 2 d\theta$
Using the integration: $\int {dx = x}$
$= 2\left[ \theta \right]_0^{\dfrac{\pi }{2}}$
Step 3: Calculating the limits.
$= 2 \times \left[ \theta \right]_0^{\dfrac{\pi }{2}}$
The definite integral is given by upper limits minus lower limits.
Upper limits $= \dfrac{\pi }{2}$ , lower limits $= 0.$
$= 2 \times \left[ {\dfrac{\pi }{2} - 0} \right]$
$= \pi$
The value of $g\left( {\dfrac{1}{2}} \right)$ comes out to be $\pi$.

Thus option (B) is correct.

Note: The $\mathop {\lim }\limits_{h \to {0^{{\text{ }} + }}}$i.e., limit h tends to ${0^ + }$because, for $h = 0$, ${t^{ - a}}$ or $\dfrac{1}{{{t^a}}}$ will be $\dfrac{1}{{{0^a}}}$ which is undetermined form, to avoid this situation limit h tends to ${0^ + }$ instead of 0.

For solving integral $= \mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {\dfrac{1}{{\sqrt {t\left( {1 - t} \right)} }}} dt$ , students can also do the following method.
$\mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {\dfrac{1}{{\sqrt {t\left( {1 - t} \right)} }}} dt$
$\Rightarrow \mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {\dfrac{1}{{\sqrt {\left( {t - {t^2}} \right)} }}} dt$
Make the expression $t - {t^2}$ a perfect square.
$t - {t^2} = t - {t^2} + \dfrac{1}{4} - \dfrac{1}{4} \\\ \Rightarrow {\left( {\dfrac{1}{2}} \right)^2} - \left( {{t^2} - t + {{\left( {\dfrac{1}{2}} \right)}^2}} \right) \\\ \Rightarrow {\left( {\dfrac{1}{2}} \right)^2} - {\left( {t - \dfrac{1}{2}} \right)^2} \\\$
The integral becomes
$\Rightarrow \mathop {\lim }\limits_{h \to {0^ + }} \int\limits_h^{1 - h} {\dfrac{1}{{\sqrt {\left( {{{\left( {\dfrac{1}{2}} \right)}^2} - {{\left( {t - \dfrac{1}{2}} \right)}^2}} \right)} }}} dt$
Using the particular integral of type $\int {\dfrac{{dx}}{{\sqrt {{a^2} - {x^2}} }}} = {\sin ^{ - 1}}\dfrac{x}{a} + C$
On comparing $x = \left( {t - \dfrac{1}{2}} \right)$ and $a = \left( {\dfrac{1}{2}} \right)$
The integral solved as:

$$\left. {{{\sin }^{ - 1}}\dfrac{{\left( {t - \dfrac{1}{2}} \right)}}{{\dfrac{1}{2}}}} \right|_0^1 \\\ \Rightarrow \left. {{{\sin }^{ - 1}}\dfrac{{\left( {\dfrac{{2t - 1}}{2}} \right)}}{{\dfrac{1}{2}}}} \right|_0^1 = \left. {{{\sin }^{ - 1}}\left( {2t - 1} \right)} \right|_0^1 \\\$$

Solving limits:

$$\Rightarrow {\sin ^{ - 1}}1 - {\sin ^{ - 1}}( - 1) \\\ \Rightarrow {\sin ^{ - 1}}1 + {\sin ^{ - 1}}1 \\\ \Rightarrow 2{\sin ^{ - 1}}1 \\\$$

We know, ${\sin ^{ - 1}}1 = \dfrac{\pi }{2}$

$$\Rightarrow 2 \times \dfrac{\pi }{2} \\\ \Rightarrow \pi \\\$$