integration help

Question

using integration, find the area of each of the enclosed regions

 

I've completed one correctly, but have no clue how to solve the other two can someone help? or set up the other two for me?

 

 

completed1) f(x)=x² and g(x)=x

 

x²=x                         1

x²-x=0                a= ∫ x-x² dx  = 1/6

                               0

 

how do I solve these two???????

 

 

1) x-axis and f(x)=2x-x²

 

 

2)f(x)=x³-3x and g(x)=x

Answer 1

You can apply the same techniques to the remaining two as well

 

1. f(x) = 2x-x^2 and g(x) = 0

Note that the points of intersection are 0 and 2. 

so the answer would be ₀ ∫² (2*x-x^2)dx = x^2-x^3/3 = 4-8/3 = 1.33333

 

2. f(x) = x^3 - 3*x and g(x) = x

Note that this would have 3 points of intersection at x = (-2,0,2)

There are two distinct regions [-2,0] and [0, 2]

So the area would be 

_-2∫⁰ (-4*x + x^2)dx + ₀∫² (4x - x^2) dx (duly noting the relative positions of the line and curve)

This evaluates to 4+4 = 8 using basic finite integration.

 

 

Answer 2

I will set up the integrals for each.

 

Problem 1: x-axis and f(x)=2x-x²

 

Take g(x) = 0, whose graph is the x-axis.

 

Now let's solve the equation f(x) = g(x)

 

2x-x² = 0

x(2-x) = 0

 

x = 0 or x = 2

 

So the interval of integration is [0,2]. In it we find 2x-x² ≥ 0 so the integral is

 

A = ∫₀² (2x - x²) dx

 

Problem 2: f(x)=x³-3x and g(x)=x

 

Let's solve the equation f(x) = g(x)

 

x³ - 3x = x

 

x³ - 4x = 0

 

x(x² - 4) = 0

 

x(x+2)(x-2) = 0

 

x = 0 or x = ±2

 

There are two regions that are congruent by symmetry.

 

One is in the interval [-2,0] where x³ - 3x ≥ x so the integrand is (x³ - 3x) - x = x³ - 4x

 

The integral is

 

A = ∫_-2⁰ (x³ - 4x) dx