`int_-3^0(1 + sqrt(9 - x^2))dx` Evaluate the integral by interpreting it in terms of areas.

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Chapter 5, 5.2 - Problem 37 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`int_-3^0(1+sqrt(9-x^2))dx`

Consider the graph of y=f(x)=`1+sqrt(9-x^2)`

`y=1+sqrt(9-x^2)`

`y-1=sqrt(9-x^2)`

`(y-1)^2=9-x^2`

`x^2+(y-1)^2=3^2`

This is the equation of circle of radius 3 centred at (0,1)

The integral can be interpreted as the area of a quarter of the said circle and the area of the rectangle whose vertices are (0,0),(0,1),(-3,1) and (-3,0).Plot can be seen in the attached graph.

`int_-3^0(1+sqrt(9-x^2))dx`

We will use standard integral `intsqrt(a^2-x^2)dx=(xsqrt(a^2-x^2))/2+a^2/2arcsin(x/a)+C`

`=[x+(xsqrt(9-x^2))/2+9/2arcsin(x/3)]_-3^0`

`=[9/2arcsin(0)]-[-3+9/2arcsin(-1)]`

`=3-9/2arcsin(-1)`

`=3-9/2(-pi/2)`

`=3+(9pi)/4`

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