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If y=f(x) makes +ve intercept of 2 and 0 unit x\; and\; y and encloses an area of 3/4 square unit with the axes then \int_{0}^{2}x\, f'(x)\, dx\; is

  • Option 1)

    3/2

  • Option 2)

    1

  • Option 3)

    5/4

  • Option 4)

    –3/4.

 

Answers (1)

best_answer

As learnt in concept

Integration By PARTS -

Let u and v are two functions then 

\int u\cdot vdx=u\int vdx-\int \left ( \frac{du}{dx}\int vdx \right )dx

- wherein

Where u is the Ist function v is he IInd function

 

 y= f(x), given to us that f(2) = 0

\int_{0}^{2}f'(x)dx=\frac{3}{4}

now \int_{0}^{2}xf'(x)dx=x\int_{0}^{2}f'(x)dx-\int_{0}^{2}f(x)dx

=[xf(x)]_{0}^{2}-\frac{3}{4}=2f(2)-\frac{3}{4}

=\frac{-3}{4}


Option 1)

3/2

This is incorrect

Option 2)

1

This is incorrect

Option 3)

5/4

This is incorrect

Option 4)

–3/4.

This is correct

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perimeter

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