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  • Help me solve this - The area of the region in sq.units, is : - Integral Calculus - JEE Main

The area of the region 

 A={(x,y):0\leq y\leq x|x|+1 and -1\leq x\leq 1} in sq.units, is :

  • Option 1)

     

    2/3

  • Option 2)

     

    2

  • Option 3)

     

    4/3

  • Option 4)

     

    1/3

Answers (1)

best_answer

 

Introduction of area under the curve -

The area between the curve y= f(x),x axis and two ordinates at the point  x=a\, and \,x= b\left ( b>a \right ) is given by

A= \int_{a}^{b}f(x)dx=\int_{a}^{b}ydx

- wherein

 

 

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)

 

Plot the graph for 

A = [(x,y): 0\leq y\leq x|x| + 1\; \textup{and} \; -1 \leq x \leq 1]

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


Option 1)

 

2/3

Option 2)

 

2

Option 3)

 

4/3

Option 4)

 

1/3

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