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Solve it, - Integral Calculus - JEE Main-2

The value of    \int_{1}^{a}[x]f'(x)dx,\; a> 1,where\; [x]  denotes the greatest integer not exceeding x is

  • Option 1)

    af(a)- \left \{f(1)+f(2)+...+f([a]) \right \}\;

  • Option 2)

    \; \; [a]f(a)-\left \{f(1)+f(2)+...+f([a]) \right \}

  • Option 3)

    [a]f([a])-\left \{f(1)+f(2)+...+f(a) \right \}

  • Option 4)

    af([a])-\left \{f(1)+f(2)+...+f(a) \right \}

 
Answers (1)
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As learnt in concept

Fundamental Properties of Definite integration -

If the function is continuous in (a, b ) then integration of a function a to b will be same as the sum of integrals of the same function from a to c and c to b.

\int_{b}^{a}f\left ( x \right )dx= \int_{a}^{c}f\left ( x \right )dx+\int_{c}^{b}f\left ( x \right )dx
 

- wherein

 

 

 

 \left [ x \right ] has to be split into integral limits.

\int_{1}^{a}[x]f{}'(x)dx

=\int_{1}^{2}f{}'(x) dx+\int_{2}^{3}2f{}'(x)dx+----------------+\int_{[a]}^{a}[a]f{}'(x)dx

=f(2)-f(1)+2f(3)-2f(2)+---------------------------+[a]f(a)-[a]f([a])

Terms start cancelling out,

We get,

-f(1)-f(2)-f(3)----------------------------f[a]+[a]f(a)

=[a]f(a)-(f (1) +f(2) +---------------f([a]))


Option 1)

af(a)- \left \{f(1)+f(2)+...+f([a]) \right \}\;

This option is incorrect

Option 2)

\; \; [a]f(a)-\left \{f(1)+f(2)+...+f([a]) \right \}

This option is correct

Option 3)

[a]f([a])-\left \{f(1)+f(2)+...+f(a) \right \}

This option is incorrect

Option 4)

af([a])-\left \{f(1)+f(2)+...+f(a) \right \}

This option is incorrect

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