Answer:
[tex]\int\limits^0_{-3}(1+\sqrt{9-x^{2}})\, dx=3+\frac{9\pi}{4}=10.068[/tex]
Step-by-step explanation:
We need to evaluate the following integral by interpreting it in terms of areas :
[tex]\int\limits^0_{-3}(1+\sqrt{9-x^{2}})\, dx[/tex]
The first step is to separate the integral into two easier integrals
[tex]\int\limits^0_{-3}(1+\sqrt{9-x^{2}})\, dx=\int\limits^0_{-3} 1 \, dx+\int\limits^0_{-3}\sqrt{9-x^{2}}\, dx[/tex] (Integral of the sum)
Now we can calculate each integral by studying the area below each function.
For the first integral the function is [tex]f(x)=1[/tex]
(I will attach a file with the functions)
The area below this function is the area of a rectangle with sides 1 and 3 ⇒
[tex]\int\limits^0_{-3}1 \, dx=3[/tex]
For the second integral the function is
[tex]f(x)=y=\sqrt{9-x^{2}}[/tex]
If we study this function :
[tex]y=\sqrt{9-x^{2}}[/tex]
[tex]y^{2}=9-x^{2}[/tex] (I)
[tex]x^{2}+y^{2}=9[/tex]
Which is the equation of a circle centered at (0,0) with radius equal to 3
From the equation (I)
[tex]y^{2}=9-x^{2}[/tex]
[tex]|y|=\sqrt{9-x^{2} }[/tex]
The two possible solutions are :
[tex]y=\sqrt{9-x^{2}}[/tex] (II) and [tex]y=-\sqrt{9-x^{2}}[/tex]
We will use (II) to solve the integral (which is the upper part of the circle)
The area of a circle with radius equal to 3 is
[tex]\pi.3^{2}[/tex]
In the integral we only need a quarter of circle ⇒ We divide the total area by 4 ⇒ [tex]\frac{\pi.3^{2}}{4}[/tex] ⇒ [tex]\frac{9\pi }{4}[/tex]
Finally the integral is equal to
[tex]\int\limits^0_{-3}(1+\sqrt{9-x^{2}})\, dx=3+\frac{9\pi}{4}=10.068[/tex]
