8.2 Finding the Area Between Curves Expressed as Functions of <latex>y</latex>

Cards (40)

  • What is the interval over which we find the area between f(y)=f(y) =3y2 3y^{2} and g(y) = y^{2} + 2</latex>?

    [0, 2]
  • The integral of 2y222y^{2} - 2 with respect to yy is \frac{2}{3}y^{3} - 2y
  • What must be identified to find the area between two curves expressed as functions of yy?

    Upper and lower curves
  • Sketching the curves is the first step in identifying the upper and lower curves
    True
  • If f(y)f(y) yields larger xx values than g(y)g(y), then f(y)f(y) is the upper
  • What are the integration limits for the integral A = \int_{0}^{2} (3y^{2} - 1) \, dy</latex>?
    [0, 2]
  • When identifying the upper and lower curves, we determine which function is consistently to the right
  • When finding the area between two curves expressed as functions of yy, we integrate with respect to y
  • Steps to find the area between f(y)=f(y) =3y2 3y^{2} and g(y)=g(y) =y2+ y^{2} +2 2 over the interval [0,2][0, 2]
    1️⃣ Determine the right and left functions
    2️⃣ Set up the integral
    3️⃣ Simplify the integrand
    4️⃣ Integrate
    5️⃣ Evaluate the integral
  • Match the step with its description for finding the area between curves
    Determine the right and left functions ↔️ Identify which function yields larger xx values
    Set up the integral ↔️ Subtract the left function from the right function
    Simplify the integrand ↔️ Combine like terms in the integral
    Evaluate the integral ↔️ Compute the definite integral
  • When comparing curves, the function that consistently yields larger xx values is considered the "lower" curve.

    False
  • Steps to set up the integral for calculating the area between two curves
    1️⃣ Identify the upper and lower curves
    2️⃣ Write the integrand by subtracting the lower curve from the upper curve
    3️⃣ Define the integration limits
    4️⃣ Set up the definite integral
  • The integrand for the curves f(y)=f(y) =4y2 4y^{2} and g(y)=g(y) =y2+ y^{2} +1 1 is 3y^{2} - 1
  • What is the formula for finding the area between two curves expressed as functions of yy?

    A=A =cd[f(y)g(y)]dy \int_{c}^{d} [f(y) - g(y)] \, dy
  • Match the curves with their correct descriptions:
    f(y) = 3y^{2}</latex> ↔️ Upper curve
    g(y)=g(y) =y2+ y^{2} +2 2 ↔️ Lower curve
  • When finding the area between two curves expressed as functions of yy, the lower curve is always subtracted from the upper curve

    True
  • The formula for the area between curves in terms of yy is A=A =cd[f(y)g(y)]dy \int_{c}^{d} [f(y) - g(y)] \, dy.

    True
  • To find the area between curves expressed as functions of yy, we must first identify the upper curve (to the right) and the lower curve (to the left).
  • Steps to set up the integral for area calculation in terms of yy.

    1️⃣ Identify upper and lower curves
    2️⃣ Write the integrand
    3️⃣ Define integration limits
    4️⃣ Write the integral
  • To solve the integral A=A =02(2y22)dy \int_{0}^{2} (2y^{2} - 2) \, dy, we need to find its antiderivative.
  • For the example curves x=x =y2 y^{2} and x=x =2y 2y, the correct bounds of integration are from 00 to 2.
  • What is the definite integral for finding the area between the curves f(y)=f(y) =3y2 3y^{2} and g(y)=g(y) =y2+ y^{2} +2 2 over the interval [0,2][0, 2]?

    A=A =02[3y2(y2+ \int_{0}^{2} [3y^{2} - (y^{2} +2)]dy 2)] \, dy
  • Which curve is identified as the upper curve in the example provided?
    f(y)=f(y) =3y2 3y^{2}
  • Steps to find the area between curves in terms of yy.

    1️⃣ Identify the upper and lower curves
    2️⃣ Set up the integral
    3️⃣ Evaluate the integral
  • What is the first step in identifying the upper and lower curves for area calculation in terms of yy?

    Sketch the curves
  • For f(y)=f(y) =3y2 3y^{2} and g(y) = y^{2} + 2</latex>, the integrand is 3y2(y2+3y^{2} - (y^{2} +2)= 2) =2y22 2y^{2} - 2.

    True
  • Steps to solve the integral for area calculation in terms of yy.

    1️⃣ Find the antiderivative
    2️⃣ Evaluate at upper limit
    3️⃣ Evaluate at lower limit
    4️⃣ Subtract lower from upper limit
  • Evaluating the antiderivative at the upper limit y=y =2 2 for (2y22)dy\int (2y^{2} - 2) \, dy gives 43\frac{4}{3}.

    True
  • In the formula for area between curves, f(y)f(y) represents the right function and g(y)g(y) represents the left function.

    True
  • The area between the curves f(y)=f(y) =3y2 3y^{2} and g(y)=g(y) =y2+ y^{2} +2 2 over the interval [0,2][0, 2] is 43\frac{4}{3}.

    True
  • To set up the integral for calculating the area between curves, we subtract the lower curve from the upper curve.
  • To find the area between two curves expressed as functions of yy, it is crucial to identify which curve is "upper" (right) and which is lower
  • The area between two curves is calculated by integrating the difference between the upper and lower curves

    True
  • Setting up the integral involves subtracting the lower curve from the upper curve

    True
  • The area between the curves f(y)=f(y) =3y2 3y^{2} and g(y)=g(y) =y2+ y^{2} +2 2 over the interval [0,2][0, 2] is \frac{4}{3}
  • When finding the area between two curves expressed as functions of yy, we integrate the difference between the upper curve and the lower curve.
  • What must be determined first to find the area between curves in terms of yy?

    Upper and lower curves
  • In the integral A=A =cd[f(y)g(y)]dy \int_{c}^{d} [f(y) - g(y)] \, dy, f(y)</latex> represents the upper curve and g(y)g(y) represents the lower curve.
  • What is the integral for calculating the area between f(y)=f(y) =3y2 3y^{2} and g(y)=g(y) =y2+ y^{2} +2 2 over [0,2][0, 2]?

    02(2y22)dy\int_{0}^{2} (2y^{2} - 2) \, dy
  • Why is it essential to use the correct bounds of integration when calculating the area between curves in terms of yy?

    To ensure accurate area