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  • Need clarity, kindly explain! - Integral Calculus - JEE Main-3

Let f(x)  be a non-­negative continuous function such that the area bounded by the curve y=f(x),\; x-axis  and the ordinates   x=\frac{\pi }{4}\; and\; x=\beta > \frac{\pi }{4}\; is\; \left ( \beta sin\beta +\frac{\pi }{4}cos\beta +\sqrt{2}\beta \right ).Then\; f\left ( \frac{\pi }{2} \right )\; is

  • Option 1)

    \left ( \frac{\pi }{4}-\sqrt{2}+1 \right )\;

  • Option 2)

    \; \; \left ( \frac{\pi }{4}+\sqrt{2}-1 \right )\; \;

  • Option 3)

    \; \left (1- \frac{\pi }{4}+\sqrt{2} \right )\; \;

  • Option 4)

    \; \left (1- \frac{\pi }{4}-\sqrt{2} \right )

 

Answers (1)

best_answer

As we learnt in 

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)

-

 and

Introduction of area under the curve -

The area between the curve y= f(x),x axis and two ordinates at the point  x=a\, and \,x= b\left ( b>a \right ) is given by

A= \int_{a}^{b}f(x)dx=\int_{a}^{b}ydx

- wherein