Question

R is the region bounded by the functions f(x)=1−3sin(x) and

g(x)=2cos(x). Find the area of the region bounded by the functions
on the interval [0,π/2].

Answer 1

To find the area of the region bounded by the functions f(x)=1−3sin(x) and g(x)=2cos(x) on the interval [0,π/2], we need to find the points of intersection between the two functions and then calculate the area between those points.

Step-by-step explanation:

To find the area of the region bounded by the functions f(x)=1−3sin(x) and g(x)=2cos(x) on the interval [0,π/2], we need to find the points of intersection between the two functions and then calculate the area between those points. Setting f(x) equal to g(x), we have:

1−3sin(x) = 2cos(x)

3sin(x) + 2cos(x) = 1

Using trigonometric identities, we can convert the equation:

3sin(x) = 1 - 2cos(x)

3sin(x) = 1 - 2√(1 - sin^2(x))

Simplifying further, we have:

9sin^2(x) = 1 - 4 + 4sin^2(x)

- 3sin^2(x) = -3

sin^2(x) = 1

sin(x) = ±1

Since we are working on the interval [0,π/2], we can ignore the negative value

The point of intersection is at x = π/2

To calculate the area, we integrate the difference between the two functions:

A = ∫[0,π/2] (f(x) - g(x)) dx

= ∫[0,π/2] (1 - 3sin(x) - 2cos(x)) dx

We can calculate this integral using the fundamental theorem of calculus or by using numerical techniques such as Simpson's rule.