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  • Solve! Area of the region bounded by the curves y = 2x, y = 2x - x2, x = 0 and x = 2 is given by

Area of the region bounded by the curves y = 2x, y = 2x - x2, x = 0 and x = 2 is given by

  • Option 1)

    \frac{3}{log2}-\frac{4}{3}

  • Option 2)

    \frac{3}{log2}+\frac{4}{3}

  • Option 3)

    3log2-\frac{4}{3}

  • Option 4)

    none

 

Answers (1)

best_answer

As we learnt

Area along x axis -

Let y_{1}= f_{1}(x)\, and \, y_{2}= f_{2}(x) be two curve then area bounded between the curves and the lines

x = a and x = b is

\left | \int_{a}^{b} \Delta y\, dx\right |= \left | \int_{a}^{b}\left ( y_{2}-y_{1} \right ) dx\right |

 

- wherein

Where \Delta y= f_{2}\left ( x \right )-f_{1}(x)

 

 Required\: Area= \int_{0}^{2}\left ( y_{2}-y_{1} \right )dx=\int_{0}^{2}\left ( 2^{x}-\left ( 2x-x^{2} \right ) \right )dx

                                  

= \left [ \frac{2^{x}}{\log 2} -x^{2}+\frac{x^{3}}{3}\right ]_{0}^{2}

= \frac{4}{\log 2}-4+\frac{8}{3}-\frac{1}{\log 2}= \frac{3}{\log 2}-\frac{4}{3}


 


Option 1)

\frac{3}{log2}-\frac{4}{3}

Option 2)

\frac{3}{log2}+\frac{4}{3}

Option 3)

3log2-\frac{4}{3}

Option 4)

none

Posted by

gaurav

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