Answer:
The area bounded by the two functions f(x) and g(x) on the interval [0, 2π] is 6π.
Step-by-step explanation:
The range of y = sin(2x) is [-1, 1].
As function f(x) = sin(2x) + 1 has been translated 1 unit up, the range of f(x) is [0, 2].
The range of y = cos(x) is [-1, 1].
As function g(x) = cos(x) - 2 has been translated 2 units down, the range of g(x) is [-3, -1].
As ranges of the functions do not overlap, the two functions do not intersect.
As the curve of f(x) is above the x-axis, and the curve of g(x) is below the x-axis, we can integrate to find the area between the curve and the x-axis for each function in the given interval, then add them together.
Note: As g(x) is below the x-axis, the evaluation of the integral will return a negative area. Therefore, we need to negate the integral so we have a positive area (since area cannot be negative).
[tex]\begin{aligned}A_1=\displaystyle \int^{2\pi}_{0} (\sin(2x)+1)\; \text{d}x&=\left[-\dfrac{1}{2}\cos(2x)+x \right]^{2\pi}_{0}\\\\&=\left(-\dfrac{1}{2}\cos(2(2\pi))+2\pi\right)-\left(-\dfrac{1}{2}\cos(2(0))+0\right)\\\\&=\left(-\dfrac{1}{2}+2\pi\right)-\left(-\dfrac{1}{2}\right)\\\\&=2\pi\end{aligned}[/tex]
As the curve is below the x-axis, remember that we need to negate the integral to find the area.
[tex]\begin{aligned}A_2=-\displaystyle \int^{2\pi}_{0} (\cos(x)-2)\; \text{d}x&=-\left[\vphantom{\dfrac12}\sin(x)-2x \right]^{2\pi}_{0}\\\\&=-\left[(\sin(2\pi)-2(2\pi))-(\sin(0)-2(0))\right]\\\\&=-\left[(0-4\pi)-(0-0)\right]\\\\&=-\left[-4\pi\right]\\\\&=4\pi\end{aligned}[/tex]
[tex]\begin{aligned}A_1+A_2&=2\pi+4\pi\\&=6\pi\end{aligned}[/tex]
Therefore, the area bounded by the two functions f(x) and g(x) on the interval [0, 2π] is 6π.
