calc 2 pg.6.1

Area Between Two Curves

Introduction to Area Between Curves

• Objective: Calculate the area of the region between two curves, specifically the curves described by the equations ( y = f(x) ) and ( y = g(x) ).
• Conditions: Both functions ( f ) and ( g ) are continuous and ( f(x) \geq g(x) ) for all ( x ) in the interval ([a, b]).

Setup of the Problem

• Consider the shaded region ( S ) that lies between the two curves from ( x = a ) to ( x = b ).
• Approximation of the area using rectangles:
  • Divide the interval ([a, b]) into ( n ) subintervals of width ( \Delta x ).
  • Height of rectangles is given by the difference in function values: ( f(xi^) - g(xi^) ), where ( xi^* ) is a point in the subinterval ( [xi, x_{i+1}] ).
  • Number of rectangles = ( n )

Riemann Sum and Definite Integral

• Area approximation using Riemann sums:
  • Area = Basis ( \Delta x \times ) Height ( (f(xi^) - g(xi^)) )
  • Summing the area contributions yields:
    [ ext{Area} \approx \sum{i=1}^{n} (f(xi^) - g(x_i^)) \Delta x ]
• Taking the limit as ( n o \infty ) leads to:
  [ A = \lim{n \to \infty} \sum{i=1}^{n} (f(xi^) - g(xi^)) \Delta x = \int_{a}^{b} (f(x) - g(x)) \, dx ]
• Thus, the area ( A ) between the two curves is calculated using the definite integral:
  [ A = \int_{a}^{b} (f(x) - g(x)) \, dx ]

Example 1: Area Between ( y = 2x ) and ( y = x^2 - 4x )

1. Find Points of Intersection:
   • Set the equations equal to find intersection points:
     [ 2x = x^2 - 4x ]
   • Rearranging yields the quadratic equation:
     [ x^2 - 6x = 0 ]
   • Factoring gives:
     [ x(x - 6) = 0 ]
   • Solutions: ( x = 0 ) or ( x = 6 )
   • Corresponding ( y ) values: ( (0, 0) ) and ( (6, 12) )
2. Setup Integration:
   • Determine upper and lower curves:
     • Top curve: ( y = 2x )
     • Bottom curve: ( y = x^2 - 4x )
   • Integrand setup:
     [ 2x - (x^2 - 4x) = 6x - x^2 ]
3. Integral Calculation:
   • Set up the definite integral:
     [ A = \int_{0}^{6} (6x - x^2) \, dx ]
   • Compute the antiderivative:
   • Plugging in values:
     • [ = \left[ 3x^2 - \frac{x^3}{3} \right]_{0}^{6} ]
     • Evaluate:
     • [ = 3(6^2) - \frac{6^3}{3} - (0) = 36 ]
   • Result: The area of the region is 36 square units.

Example 2: Area Between ( y = x - 1 ) and ( y^2 = 2x + 6 )

1. Find Points of Intersection:
   • Substitute ( y = x - 1 ) into ( y^2 = 2x + 6 ):
     [ (x - 1)^2 = 2x + 6 ]
   • Rearranging yields:
     [ x^2 - 4x - 5 = 0 ]
   • Factoring results in:
     [ (x - 5)(x + 1) = 0 ]
   • Solutions: ( x = 5 ) or ( x = -1 )
   • Corresponding ( y ) values: ( (-1, -2) ) and ( (5, 4) )
2. Determine Curves for Integration:
   • Solve for ( x ) from the parabola to express as a function of ( y ):
     [ x = \frac{1}{2}y^2 - 3 ]
3. Integral Setup:
   • Height of rectangles is found using:
   • Right curve minus left curve:
     [ \frac{1}{2}y^2 - 3 - (y + 1) ]
   • Set up integral:
     [ A = \int_{-2}^{4} \left(\frac{1}{2}y^2 - 3 - y - 1\right) \, dy ]
   • Simplification leads to:
     [ A = \int_{-2}^{4} \left(\frac{1}{2}y^2 - y - 4\right) \, dy ]
4. Calculation of Area:
   • Compute the definite integral:
   • Use the Fundamental Theorem of Calculus:
   • Final result: The area of the bounded region is 18 square units.

Summary of Areas

• The final areas calculated are:
  • Example 1: Area = 36 square units between ( y = 2x ) and ( y = x^2 - 4x ).
  • Example 2: Area = 18 square units between ( y = x - 1 ) and the parabola ( y^2 = 2x + 6 ).