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Evaluate the area enclosed between x = -3\pi and x = 3\pi; x-axis and the curve f(x) = |\sin x|

  • Option 1)

    3 sq units

  • Option 2)

    6 sq units

  • Option 3)

    12 sq units

  • Option 4)

    9 sq units

 

Answers (1)

best_answer

As we have learnt,

 

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 = 6\int_{0}^{\pi}\sin x dx = 6 \times \left [-\cos x \right ]^{\pi}_{0} = 6 \times 2 = 12 sq\; units

 


Option 1)

3 sq units

Option 2)

6 sq units

Option 3)

12 sq units

Option 4)

9 sq units

Posted by

prateek

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