Find the centroid of the quarter of the unit circle lying in the fourth quadrant.

Step-by-step explanation:

In the fourth quadrant, the equation of the unit circle is:

y = -√(1 − x²), 0 ≤ x ≤ 1

The x and y coordinates of the centroid are:

cₓ = (∫ x dA) / A = (∫ xy dx) / A

cᵧ = (∫ y dA) / A = (∫ ½ y² dx) / A

For a quarter circle in the fourth quadrant, A = -π/4.

Solving each integral:

∫₀¹ xy dx

= ∫₀¹ -x √(1 − x²) dx

= ½ ∫₀¹ -2x √(1 − x²) dx

If u = 1 − x², then du = -2x dx.

When x = 0, u = 1.  When x = 1, u = 0.

= ½ ∫₁⁰ √u du

= ½ ∫₁⁰ u^½ du

= ½ (⅔ u^³/₂) |₁⁰

= (⅓ u√u) |₁⁰

= 0 − ⅓

= -⅓

∫₀¹ ½ y² dx

= ½ ∫₀¹ (1 − x²) dx

= ½ (x − ⅓ x³) |₀¹

= ½ [(1 − ⅓) − (0 − 0)]

= ⅓

Therefore, the x and y coordinates of the centroid are:

cₓ = (-⅓) / (-π/4) = 4/(3π)

cᵧ = (⅓) / (-π/4) = -4/(3π)

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!

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