Math Problem Statement

Let f (x) = cos x and g(x) = sin x. Find the center of gravity (x, y) of the region between the graphs of f and g on [0, πœ‹/ 2]

Solution

To find the center of gravity (or centroid) (xc,yc)(x_c, y_c) of the region between the graphs of f(x)=cos⁑xf(x) = \cos x and g(x)=sin⁑xg(x) = \sin x on the interval [0,Ο€2][0, \frac{\pi}{2}], we use the following formulas for the coordinates of the centroid:

xc=∫abxβ‹…(f(x)βˆ’g(x)) dx∫ab(f(x)βˆ’g(x)) dxx_c = \frac{\int_a^b x \cdot (f(x) - g(x)) \, dx}{\int_a^b (f(x) - g(x)) \, dx} yc=∫abf(x)+g(x)2β‹…(f(x)βˆ’g(x)) dx∫ab(f(x)βˆ’g(x)) dxy_c = \frac{\int_a^b \frac{f(x) + g(x)}{2} \cdot (f(x) - g(x)) \, dx}{\int_a^b (f(x) - g(x)) \, dx}

Step 1: Find the area of the region

The area of the region is given by: A=∫0Ο€2(cos⁑xβˆ’sin⁑x) dxA = \int_0^{\frac{\pi}{2}} (\cos x - \sin x) \, dx Let's calculate this.

∫0Ο€2(cos⁑xβˆ’sin⁑x) dx=[sin⁑x+cos⁑x]0Ο€2\int_0^{\frac{\pi}{2}} (\cos x - \sin x) \, dx = \left[\sin x + \cos x\right]_0^{\frac{\pi}{2}} Evaluating this:

= (1 + 0) - (0 + 1) = 0$$ Thus, the total area is $$A = 0$$. Now we can proceed to calculating the coordinates of the centroid \( (x_c, y_c) \.

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Math Problem Analysis

Mathematical Concepts

Calculus
Centroid
Area between Curves
Definite Integrals

Formulas

x_c = (∫_a^b x · (f(x) - g(x)) dx) / (∫_a^b (f(x) - g(x)) dx)
y_c = (∫_a^b (f(x) + g(x))/2 · (f(x) - g(x)) dx) / (∫_a^b (f(x) - g(x)) dx)
Area = ∫_0^Ο€/2 (cos x - sin x) dx

Theorems

Centroid of a region
Definite integration

Suitable Grade Level

Undergraduate Level