The area (in square units) of the region bounded by the curves y + 2x2 = 0 and y + 3x2 = 1 , is equal to :

  • Option 1)

    \frac{3}{5}

  • Option 2)

    \frac{3}{4}

  • Option 3)

    \frac{1}{3}

  • Option 4)

    \frac{4}{3}

 

Answers (1)

As learnt in concept

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)

 

y+2x2=0; y+3x2=1

y= -2x2; y=1-3x2

Calculate the point of intersection first

-2x2= 1-3x2

=> x2=1

=> x=\pm1

\int_{-1}^{+1}\left ( (1-3x^{2})-(-2x^{2}) \right )dx

=>\int_{-1}^{+1}(1-x^{2})dx = \left [ x-\frac{x^{3}}{3} \right ]_{-1}^{1}

=>2-\frac{2}{3}= \frac{4}{3}


Option 1)

\frac{3}{5}

this is incorrect option

Option 2)

\frac{3}{4}

this is incorrect option

Option 3)

\frac{1}{3}

this is incorrect option

Option 4)

\frac{4}{3}

this is correct option