Question

If for a continuous function $f\left( x \right),{\text{ }}\int_{ - \pi }^t {\left( {f\left( x \right) + x} \right)dx} = {\pi ^2} - {t^2},$ for all $t \geqslant - \pi ,$ then $f\left( {\dfrac{-\pi }{3}} \right)$ is equal to:
A. $\pi$
B. $\dfrac{\pi }{2}$
C. $\dfrac{\pi }{3}$
D. $\dfrac{\pi }{6}$

Answer 1

If the equation is true for all values, i.e. the given function is an identity, then the roots of the given identity are the same as the roots of its derivative.
The given question can be easily solved by using the Leibniz integral rule.
The Leibniz integral rule gives an identity for differentiation of a definite integral whose limits are functions of the differential variable,
If the function $h\left( x \right)$ is given as: $h\left( x \right) = \int_{f\left( x \right)}^{g\left( x \right)} {m\left( {x,t} \right)} dt$
Then $\dfrac{{d\left( {h\left( x \right)} \right)}}{{dx}} = \int_{f\left( x \right)}^{g\left( x \right)} {\dfrac{{\partial m\left( {x,t} \right)}}{{\partial x}}} dt + m\left[ {x,g\left( x \right)} \right] \cdot g'\left( x \right) - m\left[ {x,f\left( x \right)} \right] \cdot f'\left( x \right)$
Mostly, we are given: $h\left( x \right) = \int_{f\left( x \right)}^{g\left( x \right)} {m\left( t \right)} dt$
Then, $\dfrac{{d\left( {h\left( x \right)} \right)}}{{dx}} = m\left[ {g\left( x \right)} \right] \cdot g'\left( x \right) - m\left[ {f\left( x \right)} \right] \cdot f'\left( x \right)$

Complete step-by-step answer:
Let the Given Integral function is $g\left( t \right)$:
${\text{ }}g\left( t \right) = \int_{ - \pi }^t {\left( {f\left( x \right) + x} \right)dx} = {\pi ^2} - {t^2}$
Analyse using Leibniz Integral rule:
Leibniz Integral rule for function:
$h\left( x \right) = \int_{f\left( x \right)}^{g\left( x \right)} {m\left( t \right)} dt$
On comparing with given integral ${\text{ }}g\left( t \right) = \int_{ - \pi }^t {\left( {f\left( x \right) + x} \right)dx}$
The identity is a function $h\left( x \right)$ , given function is $g\left( t \right)$,
In identity Function $m\left( t \right)$, here it is $\left( {f\left( x \right) + x} \right)$, is the function of ‘x’, hence integrated with respect to dx.
And limits of the given integral is the function of ‘t’, thus the given integral is also a function of ‘t’, i.e. $g\left( t \right)$
In identity upper and lower limits, $g\left( x \right)$ and $f\left( x \right)$ here are t and $- \pi$
We have $\dfrac{{d\left( {h\left( x \right)} \right)}}{{dx}} = m\left[ {g\left( x \right)} \right] \cdot g'\left( x \right) - m\left[ {f\left( x \right)} \right] \cdot f'\left( x \right)$
Applying Leibniz integral rule
Since, in given integral function, both L.H.S. and R.H.S. are a function of ‘t’, differentiating both sides with respect to ‘t’
Thus, the given integral becomes after using Leibniz Integral rule:
$g'\left( t \right) = \left( {f\left( t \right) + t} \right) \cdot 1 - \left( {f\left( \pi \right) - \pi } \right) \cdot 0 = - 2t$
$\Rightarrow f\left( t \right) + t = - 2t \\\ \Rightarrow f\left( t \right) = - 2t - t \\\ \Rightarrow f\left( t \right) = - 3t \\\$
Calculate the required $f\left( {\dfrac{-\pi }{3}} \right)$
For $t = \dfrac{-\pi }{3}$
$f\left( {\dfrac{-\pi }{3}} \right) = - 3 \times \dfrac{-\pi }{3} \\\ = \pi \\\$
$f\left( {\dfrac{-\pi }{3}} \right)$ comes out to be $\pi$.

So, the correct answer is “Option A”.

Note: For a, b, c…. are constant, and x, t, u,… are variable, then the integral $\int_a^b {g\left( x \right)dx}$ is equal to constant.
Leibniz Integral rule is also known as differentiation under the integral sign.