Question

Question #0e712

Answer 1

`A_"Total"=8sqrt(2)~~11.31`

Explanation:

Although this question is marked as a Trigonometry question, the solution requires calculus to solve.

You are given two curves and want to find the space between them, as depicted by this graph.

Let `f(x) = 4sin(x)`.

Let `g(x)=4cos(x)`.


The `x`-coordinate of the left hand side intersection is `a=pi/4`.
The `x`-coordinate of the right hand side intersection is `b=(5pi)/4`.

The positive area above the `x`-axis is equal to the negative area below the `x`-axis and would cancel each other out. So we must stick to finding the area above the `x`-axis and then double that area.

One strategy would be to find the area, `A` under `f(x)=4sin(x)` between `x=pi/4` and `x=pi` and then subtract off the area, `B`, of `g(x)=4cos(x)` between `x=pi/4` and `x=pi/2`. Finally, we would need to double that area to find the total area, `A_"Total"`, both above and below the `x`-axis. In other words, we are finding `2(A-B).`

Area `A` is

`A=int_(pi/4)^pi4sin(x)dx=[-4cos(x)]_(pi/4)^pi`
`= 4+2sqrt(2)`

Area `B` is

`B=int_(pi/4)^(pi/2)4cos(x)dx=[4sin(x)]_(pi/4)^(pi/2)`

`B=4-2sqrt(2)`

Therefore, the total area `A_"Total"` is

`A_"Total"=2(A-B)`
`" "=2(4+2sqrt(2)-4+2sqrt(2))`
`" "=8sqrt(2)~~11.31`