Answer:
[tex]\int _{-4}^0\left(3+\sqrt{16-x^2}\right)dx\:=12+4\pi[/tex]
Step-by-step explanation:
we are given
[tex]\int _{-4}^0\left(3+\sqrt{16-x^2}\right)dx\:[/tex]
Firstly, we will draw graph
We can see that from -4 to 0
we have two shapes
a rectangle and a quarter of circle
so, we can find areas and then we can combine them
[tex]\int _{-4}^0\left(3+\sqrt{16-x^2}\right)dx\:=3\times 4+\frac{1}{4} \pi \times (4)^2[/tex]
now, we can simplify it
[tex]\int _{-4}^0\left(3+\sqrt{16-x^2}\right)dx\:=12+4\pi[/tex]
