Given the following definite integral.
[tex]\int_{-4}^4\sqrt{4^2-x^2}dx[/tex]
We will use the substitution to solve the definite integral
Let the following:
[tex]\begin{gathered} 4sin(\theta)=x \\ 4cos(\theta)*d\theta=dx \\ And: \\ 4^2-x^2=4^2-4^2sin^2\theta=4^2(1-sin^2\theta)=4^2cos^2\theta \end{gathered}[/tex]
Substitute into the given integral:
[tex]\begin{gathered} \int_{-4}^4\sqrt{4^2-x^2}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sqrt{4^2cos^2\theta}*4cos(\theta)*d\theta \\ \\ =\int_{-\pi/2}^{\pi/2}4cos\theta *4cos\theta *d\theta=\int_{-\pi/2}^{\pi/2}16cos^2\theta *d\theta \end{gathered}[/tex]
Now, we will use the following identity:
[tex]cos^2\theta=\frac{1}{2}(1+cos2\theta)[/tex]
So, the integral will be:
[tex]\begin{gathered} =\int_{-\pi/2}^{\pi/2}\frac{16}{2}(1+cos2\theta)d\theta \\ \\ =8(\theta+\frac{1}{2}sin2\theta) \end{gathered}[/tex]
substitute θ = π/2, and θ = -π/2
So, the value of the integral =
[tex]8*(\frac{\pi}{2}-(-\frac{\pi}{2}))=8π[/tex]
So, the answer will be: Area = 8π
The graph of the given function is shown in the following picture
