Question

I have a calculus question about the definite integral, from my high school AP Calculus Class, pic included

Answer 1

Given the following definite integral.

`\int_(-4)^4√(4^2-x^2)dx`

We will use the substitution to solve the definite integral

Let the following:

`\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}`

Substitute into the given integral:

`\begin{gathered} \int_(-4)^4√(4^2-x^2)dx=\int_{-(\pi)/(2)}^{(\pi)/(2)}√(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}`

Now, we will use the following identity:

`cos^2\theta=(1)/(2)(1+cos2\theta)`

So, the integral will be:

`\begin{gathered} =\int_(-\pi/2)^(\pi/2)(16)/(2)(1+cos2\theta)d\theta \\ \\ =8(\theta+(1)/(2)sin2\theta) \end{gathered}`

substitute θ = π/2, and θ = -π/2

So, the value of the integral =

`8*((\pi)/(2)-(-(\pi)/(2)))=8π`

So, the answer will be: Area = 8π

The graph of the given function is shown in the following picture