Question

For $x\, \in \left( {0,\dfrac{{5\pi }}{2}} \right)$,define $f(x) = \int {\sqrt {t\,} } \sin t\,dt.$ then $f(x)$has
$(A)$Local maximum at $\pi $and $2\pi $
$(B)$Local minimum at $\pi $and local maximum at $2\pi $
$(C)$Local maximum at $\pi $and local minimum at $2\pi $
$(D)$Local minimum at $\pi $and $2\pi $

Answer 1

Hint: when the slope of a function is zero at $x$ and the second derivative at $x$ is, less than zero, then it is local maximum, greater than zero, it is a local minimum, if it is equal to zero then the test fails (there may be other ways of finding out though).

Formula Used:
Newtons-Leibnitz formula,
$\int\limits_a^b {f(x)dx = f(b) - f(a)} $ , Here $a$ and $b$ are the limits,
The formula expressing the value of a definite integral of a given integral function $f$ over an interval as the difference of the values at the endpoints of the interval of any primitive $f$ of the solution $f$.

Complete step by step answer:
Here $f(x) = \int {\sqrt {t\,} } \sin t\,dt.$ where $x\, \in \left( {0,\dfrac{{5\pi }}{2}} \right)$;
By using Newton-Leibnitz formula, the above equation becomes,
${f^{'}}(x) = \left\{ {\sqrt {x\,} \sin x - 0} \right\}$ Here we have just substituted the value of limits to the function $f(x)$,
Now this ${f^{'}}(x)$ becomes,
${f^{'}}(x)$$ = \sqrt x \sin x = 0$;
By trigonometrically$\sin x$$ = 0$;
Therefore $x = \pi ,2\pi $;
${f^{'}}(x) = \sqrt x \cos x + \dfrac{1}{{2\sqrt x }}\sin x$;

If we substitute the value of $x = \pi ,2\pi $, in the equation we can find whether it lies in local maximum or local minimum,
${f^{'}}(\pi ) = - \sqrt \pi < 0$;
Here $f(x)$is lesser than zero thus $f(x)$has local maximum at $x = \pi $.
${f^{'}}(2\pi ) = \sqrt \pi < 0$
Here $f(x)$is greater than zero thus $f(x)$has local minimum at $x = 2\pi $.

Form this we can conclude that option $(C)$ is a correct answer for this solution.

Note:
Substituting the values in the given formula must be handled safely, under certain conditions, one may interchange the integral and partial differential operators. This important result is particularly useful in the differentiation of integral transforms.