The area ( in sq. units ) of the region bounded by the curves y=2^{x} and

y=|x+1|, in the first quadrant is :

  • Option 1)

    \log_{e}2+\frac{3}{2}

  • Option 2)

    \frac{3}{2}

  • Option 3)

    \frac{1}{2}

  • Option 4)

    \frac{3}{2}-\frac{1}{\log_{e}2}

 

Answers (1)
V Vakul

y=2^{x} and   y=|x+1|

The required area is 

=\int_{0}^{1}(x+1-2^{x})dx

=[\frac{x^{2}}{2}+x-\frac{2^{x}}{\ln 2}]^{1}_{0}

=(\frac{1}{2}+1-\frac{2}{\ln 2})-(0+0-\frac{1}{\ln 2})

=\frac{3}{2}-\frac{1}{\ln 2}

So, option (4) is correct.

 


Option 1)

\log_{e}2+\frac{3}{2}

Option 2)

\frac{3}{2}

Option 3)

\frac{1}{2}

Option 4)

\frac{3}{2}-\frac{1}{\log_{e}2}

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