Calculus: Early Transcendentals, Chapter 5, 5.2, Section 5.2, Problem 37

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