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 If    for    a    continuous    function   f(x),     \int_{-\pi }^{t}\left ( f\left ( x \right )+x \right )dx= \pi ^{2}-t^{2}

for all  t\geqslant -\pi then f\left ( -\frac{\pi }{3} \right ) is equal to:

  • Option 1)

    \pi

  • Option 2)

    \frac{\pi }{2}

  • Option 3)

    \frac{\pi }{3}

  • Option 4)

    \frac{\pi }{6}

 

Answers (1)

As learnt in concept

NEWTON LEIBNITZ THEOREM -

\frac {d}{dt}\left ( \int_{f(t)}^{\phi (t))}F(x)dx \right )=F(\phi(t))\phi^{'}(t)-F(f(t))f^{'}(t)

-

 

 \int_{-\pi}^{t}(f(x)+x) dx\:=\: \pi^2 -t^2

Differentiating by leibnitz rule

(f(t)+t) \times 1 -0\:=\: -2t

f(t)\:=\: -3t

f(\frac{-\pi}{3})\:=\: -3 \times \frac{-\pi}{3} =\pi


Option 1)

\pi

Correct

Option 2)

\frac{\pi }{2}

Incorrect

Option 3)

\frac{\pi }{3}

Incorrect

Option 4)

\frac{\pi }{6}

Incorrect

Posted by

Vakul

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