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Answer please! - Integral Calculus - JEE Main-4

If F(x) = 3x^{2} -4 and f(x) = F'(x) then \int f(x)\;dx = ?

  • Option 1)

    3x^{2} -4

  • Option 2)

    3x^{2} -5

  • Option 3)

    3x^{2}

  • Option 4)

    Can be any of A, B, C

 
Answers (1)
118 Views
G gaurav

As we have learnt,

 

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 .

 

 

 Since,

f(x) = \frac{\mathrm{d} }{\mathrm{d} x}(3x^{2} -4) =6x

Thus,

\int f(x)dx = \int 6x dx = 3x^{2} + c, c\epsilon \mathbb R

 


Option 1)

3x^{2} -4

Option 2)

3x^{2} -5

Option 3)

3x^{2}

Option 4)

Can be any of A, B, C

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