#Study Notes: Area Between Two Curves

#Table of Contents

1. Introduction
2. Finding the Area Between Two Curves in Terms of $x$
   • Steps for Calculation
   • Example Problem
3. Finding the Area Between Two Curves in Terms of $y$
   • Steps for Calculation
   • Example Problem
4. Practice Questions
5. Glossary
6. Summary and Key Takeaways
7. Exam Strategy

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#Introduction

In calculus, finding the area between two curves is a common problem that requires understanding how to set up and evaluate definite integrals. This guide will help you master this topic by breaking down the steps and providing illustrative examples.

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#Finding the Area Between Two Curves in Terms of $x$

#Steps for Calculation

1. Identify the Curves and Bounds:

   • Consider the curves $y = f(x)$ and $y = g(x)$.
   • Identify the interval $[a, b]$ over which you want to find the area.

2. Set Up the Integral:

   • Ensure that $f(x) \ge g(x)$ over the interval $[a, b]$.
   • Calculate the area as: $$\text{Area} = \int_{a}^{b} (f(x) - g(x)) , dx$$

3. Evaluate the Integral:

   • Compute the definite integral to find the area between the curves.

#Example Problem

Find the area of the region enclosed by the curves $y = 2x^2 - 4x + 2$ and $y = -2x^2 + 8x - 6$.

1. Find the Points of Intersection:

   • Set the equations equal to each other and solve for $x$: $$2x^2 - 4x + 2 = -2x^2 + 8x - 6$$ Simplify and solve the quadratic equation: $$4x^2 - 12x + 8 = 0 \implies (x-2)(x-1) = 0$$ So, $x = 1$ and $x = 2$.

2. Set Up the Integral:

   • Between $x = 1$ and $x = 2$, the curve $y = -2x^2 + 8x - 6$ is above $y = 2x^2 - 4x + 2$, so: $$\int_{1}^{2} \left[ (-2x^2 + 8x - 6) - (2x^2 - 4x + 2) \right] , dx$$

3. Evaluate the Integral:

   • Simplify the integrand: $$\int_{1}^{2} (-4x^2 + 12x - 8) , dx$$
   • Compute the definite integral: $$\left[ -\frac{4}{3}x^3 + 6x^2 - 8x \right]_{1}^{2} = \left( -\frac{32}{3} + 24 - 16 \right) - \left( -\frac{4}{3} + 6 - 8 \right)$$ Simplify the result: $$\frac{2}{3} \text{ units squared}$$

This problem illustrates how to find the area between two intersecting curves by setting up and evaluating the appropriate definite integral.

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#Finding the Area Between Two Curves in Terms of $y$

#Steps for Calculation

1. Identify the Curves and Bounds:

   • Consider the curves $x = f(y)$ and $x = g(y)$.
   • Identify the interval $[a, b]$ over which you want to find the area.

2. Set Up the Integral:

   • Ensure that $f(y) \ge g(y)$ over the interval $[a, b]$.
   • Calculate the area as: $$\text{Area} = \int_{a}^{b} (f(y) - g(y)) , dy$$

3. Evaluate the Integral:

   • Compute the definite integral to find the area between the curves.

#Example Problem

Find the area of the region enclosed by the curves $y = \frac{1}{8}e^x$ and $y = 2e^{-x}$.

1. Find the Points of Intersection:

   • Set the equations equal to each other and solve for $y$: $$\frac{1}{8}e^x = 2e^{-x}$$ Solve for $x$: $$8y = e^x \implies \ln(8y) = x$$ $$2e^{-x} = y \implies \ln\left(\frac{2}{y}\right) = x$$ Set the equations equal to each other: $$\ln(8y) = \ln\left(\frac{2}{y}\right)$$ Solve for $y$: $$8y = \frac{2}{y} \implies 8y^2 = 2 \implies y^2 = \frac{1}{4} \implies y = \frac{1}{2}$$

2. Set Up the Integral:

   • The integral limits are from $y = \frac{1}{2}$ to $y = 2$: $$\int_{\frac{1}{2}}^{2} \left[ \ln(8y) - \ln\left(\frac{2}{y}\right) \right] , dy$$

3. Evaluate the Integral:

   • Simplify the integrand using logarithm properties: $$\ln(8y) - \ln\left(\frac{2}{y}\right) = \ln(4y^2) = 2\ln(2y)$$ Compute the definite integral: $$\int_{\frac{1}{2}}^{2} 2\ln(2y) , dy = 2.545 \text{ units squared}$$

This problem shows how to find the area between two intersecting curves by setting up and evaluating the appropriate definite integral in terms of $y$.

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#Practice Questions

#Multiple-Choice Questions

Practice Question

1. Find the area between the curves $y = x^2$ and $y = x + 2$ from $x = 0$ to $x = 2$.
   • A) $\frac{8}{3}$
   • B) $4$
   • C) $\frac{10}{3}$
   • D) $2$

Practice Question

2. Determine the area between $y = \sin(x)$ and $y = \cos(x)$ from $x = 0$ to $x = \frac{\pi}{2}$.
   • A) $0$
   • B) $\frac{1}{2}$
   • C) $1$
   • D) $\frac{\pi}{4}$

#Short-Answer Questions

Practice Question

3. Find the area enclosed by $y = x^3$ and $y = x$ from $x = -1$ to $x = 1$.

Practice Question

4. Calculate the area between $x = y^2$ and $x = y + 2$ from $y = -1$ to $y = 2$.

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#Glossary

1. Definite Integral: A type of integral that computes the accumulation of quantities, such as areas under curves, between specified limits.
2. Intersection Points: Points where two curves meet or cross each other.
3. Integrand: The function being integrated in an integral.
4. Logarithms: The inverse operation to exponentiation, indicating the power to which a number must be raised to obtain another number.

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#Summary and Key Takeaways

1. Area Calculation Between Curves:

   • Use definite integrals to find the area between two curves.
   • Ensure the correct function order in the integrand to avoid negative results.

2. Intersection Points:

   • Solving for intersection points is crucial for setting the correct limits of integration.

3. Integral Setup:

   • For curves in terms of $x$, integrate with respect to $x$.
   • For curves in terms of $y$, integrate with respect to $y$.

4. Logarithm Properties:

   • Simplifying integrands using properties of logarithms can make calculations easier.

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#Exam Strategy

• Read Questions Carefully: Ensure you understand whether you should integrate with respect to $x$ or $y$.
• Identify Intersection Points: These points are often your limits of integration.
• Double-Check Function Order: Make sure the function that is "above" or "further away" is first in the integrand.
• Simplify When Possible: Use algebraic simplifications to make the integral easier to evaluate.

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By following these guidelines and practicing with the provided problems, you'll be well-equipped to handle questions on the area between two curves in your exams.