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8.4 Area Between Curves (with respect to 𝑥) Calculus

• Task Description: Sketch the graph of each function and set up the integral for finding the area of the region bounded by the graphs, without evaluating.

Problem Set

1. Functions:

   • $f(x) = x^2 + 2$
   • $g(x) = \frac{1}{x}$
   • Bounds: $x = 2$, and $x = 1$
   • Setup:
     $A = ext{Area} = ext{integral from 1 to 2 of } [ f(x) - g(x) ] \,dx$

2. Functions:

   • $f(x) = 6 - x^2$
   • $g(x) = x$
   • Setup:
     $A = ext{integral of } [ f(x) - g(x) ] = ext{integral from 0 to 6 } (6 - x^2) - (x) \,dx$

3. Functions:

   • $y = x$
   • $y = 2 - x$
   • $y = 0$
   • Setup:
     $A = ext{integral from intersection points of } [ (2 - x) - x] \,dx$

Practice

• Next Steps: Find the area bounded by the following graphs by showing work.

1. Functions:

   • $y = \frac{1}{2}$
   • $y = 0$
   • Vertical Lines: $x = 1, ext{ and } x = 5$
   • Setup:
     $A = ext{integral from 1 to 5 of } \frac{1}{2} \,dx$

2. Functions:

   • $y = \frac{1}{x}$
   • $y = x$
   • Setup:
     $A = ext{integral of } |g(x) - f(x)| \,dx$

3. Functions:

   • $y = \frac{1}{x}$
   • Bounds: from $x=0$ to $y=2$
   • Setup:
     $A = ext{integral from } 0 ext{ to } 4 [ \frac{1}{2} \, dx$

4. Functions:

   • $y = e^x$
   • Bounds: $y = 2$, $y = \frac{1}{2}$
   • Setup:
     $A = ext{integral from lower bound to upper bound of } [e^x - f(y)] [ dx$

5. Functions:

   • $y = 4x - 3$, $y = 3$, $y = 6 - 2x$
   • Setup:
     $A = ext{integral from point of intersection of curves } \ (4x-3) - (6 - 2x) dx$

6. Functions:

   • $y = ext{sqrt}(x)$
   • $y = 0$
   • $y = 3 - x$
   • Setup:
     $A = ext{integral from } x = 0 ext{ to where the two functions intersect. } [ ext{sqrt}(x) - (3-x)] \,dx$

7. Functions:

   • $y = 4x^2$, $y = 1$, $y = x$
   • Setup:
     $A = ext{integral from point of intersection of curves } [1 - 4x^2] [ dx$

Region Analysis

• Let $R$ be the region bounded by the curves illustrated. If the line $x = k$ divides $R$ into two regions of equal area, determine the value of $k$.

1. Functions:
   • $y = \frac{1}{2}$
   • $y = 2$
   • Vertical Lines: $x = 1$, $x = 4$

• Setup:
  • Equate integrals from $x = k$ to total area/2.

1. Functions:
   • $y = x^2 + 8x + 16$
   • $y = 2x + 4$
   • Vertical Lines: $x = 2, x = 4$

• Setup:
  • Find area via integral method and set equal to total area/2.

1. Functions:
   • $y = ext{sqrt}(x)$

• Vertical Line $y=2$, Boundary indicated at $x = 3$.

1. Calculation Problem:

   • If $0 ext{ to } k$ and the area under $y = ext{sin}(x)$ equals 0.2, find value of $k$.

2. Calculator Active Problem:

   • Area bounded by $y = ext{sqrt}(2-x^2)$ and $x = 3, x = -3, y = 2$.

Test Prep

• Review the concept of calculating area between curves with respect to $x$.

8.5 Area Between Curves (with respect to 𝑦) Calculus

• Task Description: For the given regions, set up integrals with respect to $y$ representing the area. Do not solve.

1. Functions:
   • $x = f(y)$ and $x = g(y)$

• Setup:
  • Identify intersection points and forming bounds for integral.

1. Functions:
   • $y = ext{ln}(x)$ and $y = 5 - x$, $y = 0$

• Setup:
  • Find where $y = 5 - x$ intersects with $ext{ln}(x)$.

1. Functions:
   • $y = \frac{1}{2}(x+3)$
   • $y = 2$

• Setup:
  • Set up the integrals with respect to $y$ and determine bound values.

1. Functions:
   • $y = f(x)$ and $y = g(x)$

• Show both sets of integrals in corresponding bounds for regions.

1. Functions:
   • $y = ext{sqrt}(x)$, $x = 0$, and lower bound $y = 2$.

• Sketch a graph representing curves to visually depict the areas.

Further Practice

6. Functions:
   • $y = x^3$ and $y = 5$, $x = -2, x = 1$

• Graph requirement to illustrate areas.

6. Functions:
   • $x = 4$, $y = 2 - 3x$

• Setup: Integrals require showing work for the area bounded.

6. Functions:
   • $y = x$, $y = 2 - x$, and $y = 0$

• Identify area through bounded values and integral computation.

6. Non-Calculator Problem:
   • Compute the area under curves $y = ext{sqrt}(x)$ and $y = - ext{sin}(x)$ over defined region.

• Tip: Integrate with respect to $y$ for easier calculations.