Let p(x)  be a function defined on R such that  p'(x)=p'(1-x), for all x\in [0,1],p(0)=1\; and\; p(1)=41.Then\; \int_{0}^{1}p(x)\, dx\; equals

  • Option 1)

    \sqrt{41}

  • Option 2)

    21

  • Option 3)

    41

  • Option 4)

    42

 

Answers (1)

As we learnt in 

Constant of integration: -

\frac{\mathrm{d} }{\mathrm{d} x}\left ( F\left ( x \right )+C \right )=\frac{\mathrm{d} }{\mathrm{d} x}F \left ( x \right )+0=f\left ( x \right )

Hence    \int f\left ( x \right )dx=F\left ( x \right )+C

- wherein

Where C is the constant of integration .

 

 P{}'(x)=P{}'(1-x)\\ Integrating\:both \:sides \\ P(x)=p(1-x)+C)\\ P(x)+P(1-x)=C

Integrate again P(x) - P(1 - x) = Cx

So put x=0

P(0)+P(1)=C=42

and P(1)-P(0)=42

\int_{0}^{1}P(x)dx=P(1)-P(0)=42 

Correct option is 4.


Option 1)

\sqrt{41}

Incorrect

Option 2)

21

Incorrect

Option 3)

41

Incorrect

Option 4)

42

Correct

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