`int_-5^5(x - sqrt(25 - x^2))dx` Evaluate the integral by interpreting it in terms of areas.

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Chapter 5, 5.2 - Problem 38 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`int_-5^5(x-sqrt(25-x^2))dx`

`=int_-5^5xdx-int_-5^5sqrt(25-x^2)dx`

`=I_1-I_2`

I_1 can be be interpreted as area of two triangles;one above the x-axis and the other below axis.Since they are on the opposite sides of x-axis so the net resultant area will be zero.(see attached graph)

I_2 can be interpreted as the top half of a circle with centre the origin and radius 5 and above the x-axis from x=-5 to x=5.

Area of semi circle=`pi(5^2)/2=(25pi)/2`

So,`int_-5^5(x-sqrt(25-x^2))dx=0-(25pi)/2=(-25pi)/2`

 

 

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