# Study Notes: Area Between a Curve and the Y-Axis ## Table of Contents 1. [Introduction](#introduction) 2. [Finding the Area Between a Curve and the Y-Axis](#finding-the-area-between-a-curve-and-the-y-axis) 3. [Worked Example 1](#worked-example-1) 4. [No Given Limits](#no-given-limits) 5. [Negative Area Integrals](#negative-area-integrals) 6. [Worked Example 2](#worked-example-2) 7. [Practice Questions](#practice-questions) 8. [Glossary](#glossary) 9. [Summary and Key Takeaways](#summary-and-key-takeaways) ## Introduction This section covers the method to find the area between a curve and the y-axis using definite integrals. Understanding this concept is essential for solving problems related to areas under curves in coordinate geometry. ## Finding the Area Between a Curve and the Y-Axis To find the area between a curve and the y-axis, follow these steps: The area is calculated by evaluating the **definite integral** of a function $x = g(y)$ with respect to $y$ between $y = a$ and $y = b$: $$ \int_{a}^{b} g(y) \, dy $$ ### Steps: 1. **Identify the function** $x = g(y)$. 2. **Set the integration limits** $y = a$ and $y = b$. 3. **Evaluate the definite integral**: $\int_{a}^{b} g(y) \, dy$. Always ensure the function is expressed in terms of $y$ before integrating. ### Example Given a function in terms of $x$, $y = f(x)$, rearrange it to $x = g(y)$ before integrating. ## Worked Example 1 ### Problem: Find the area of the region enclosed by the curve $y = 2 + \sqrt{x + 4}$ and the horizontal lines $y = 4$ and $y = 6$. ### Solution: 1. **Rearrange the equation**: $$ \begin{aligned} y & = 2 + \sqrt{x + 4} \\ y - 2 & = \sqrt{x + 4} \\ (y - 2)^2 & = x + 4 \\ x & = (y - 2)^2 - 4 \end{aligned} $$ 2. **Integrate** with respect to $y$ between $y = 4$ and $y = 6$: $$ \begin{aligned} \int_{4}^{6} \left[(y - 2)^2 - 4\right] \, dy & = \left[ \frac{1}{3}(y - 2)^3 - 4y \right]_{4}^{6} \\ & = \left( \frac{1}{3}(6 - 2)^3 - 4 \cdot 6 \right) - \left( \frac{1}{3}(4 - 2)^3 - 4 \cdot 4 \right) \\ & = \left( \frac{64}{3} - 24 \right) - \left( \frac{8}{3} - 16 \right) \\ & = \left( \frac{64}{3} - 24 \right) - \left( \frac{8}{3} - 16 \right) \\ & = \left( \frac{64 - 72}{3} \right) - \left( \frac{8 - 48}{3} \right) \\ & = \left( \frac{-8}{3} \right) - \left( \frac{-40}{3} \right) \\ & = \frac{32}{3} \end{aligned} $$ The area is $\frac{32}{3}$ square units. ## No Given Limits ### What if limits are not provided? If the integration limits are not provided, they are often the y-axis intercepts: 1. **Find the y-axis intercepts**: Solve for $x = 0$. 2. **Integrate** the function between these intercepts. The x-axis ($y = 0$) may also be one of the limits. ## Negative Area Integrals ### When is the area integral negative? * If the area lies **to the left of** the y-axis, the **value of the definite integral** will be **negative**. - However, an **area** cannot be negative. - The area is equal to the **absolute value** of the definite integral. * For areas that lie both to the right and left of the y-axis: - Use the method outlined in the 'Multiple Areas' study guide. Always check whether you need to find the **value of an integral** or an **area**. ## Worked Example 2 ### Problem: Find the area of the region enclosed by the curve $y = x^2 - 7x + 10$ and the y-axis. ### Solution: 1. **Find the y-intercepts** by solving for $x = 0$: $$ \begin{aligned} 0 & = y^2 - 7y + 10 \\ 0 & = (y - 2)(y - 5) \\ y & = 2 \, \text{or} \, y = 5 \end{aligned} $$ 2. **Integrate** with respect to $y$ between $y = 2$ and $y = 5$: $$ \begin{aligned} \int_{2}^{5} \left(y^2 - 7y + 10\right) \, dy & = \left[ \frac{1}{3} y^3 - \frac{7}{2} y^2 + 10y \right]_{2}^{5} \\ & = \left( \frac{1}{3} \cdot 125 - \frac{7}{2} \cdot 25 + 50 \right) - \left( \frac{1}{3} \cdot 8 - \frac{7}{2} \cdot 4 + 20 \right) \\ & = \left( \frac{125}{3} - \frac{175}{2} + 50 \right) - \left( \frac{8}{3} - 14 + 20 \right) \\ & = \left( \frac{250}{6} - \frac{525}{6} + \frac{300}{6} \right) - \left( \frac{16}{6} - \frac{84}{6} + \frac{120}{6} \right) \\ & = \left( \frac{25}{6} - \frac{26}{3} \right) \\ & = -\frac{9}{2} \end{aligned} $$ Since the area cannot be negative, the area is $\frac{9}{2}$ square units. ## Practice Questions **Question 1:** Find the area between the curve $x = y^2$ and the y-axis between $y = -1$ and $y = 2$. **Answer:** $$ \int_{-1}^{2} y^2 \, dy = \left[ \frac{1}{3} y^3 \right]_{-1}^{2} = \left( \frac{8}{3} - (-\frac{1}{3}) \right) = \frac{9}{3} = 3 \, \text{square units} $$ **Question 2:** Find the area enclosed by the curve $x = 4 - y^2$ and the y-axis between $y = -2$ and $y = 2$. **Answer:** $$ \int_{-2}^{2} (4 - y^2) \, dy = \left[ 4y - \frac{1}{3} y^3 \right]_{-2}^{2} = \left( 8 - \frac{8}{3} \right) - \left( -8 + \frac{8}{3} \right) = \frac{32}{3} \, \text{square units} $$ ## Glossary 1. **Definite Integral**: A type of integral with upper and lower limits, providing the net area under a curve. 2. **Y-Axis Intercept**: The points where a curve intersects the y-axis. 3. **Absolute Value**: The non-negative value of a number without regard to its sign. ## Summary and Key Takeaways ### Summary - The area between a curve and the y-axis can be found using definite integrals. - Rearrange the function into $x = g(y)$ if needed. - Integrate between the given limits or find y-axis intercepts if limits are not provided. - Account for negative areas by taking the absolute value of the definite integral. ### Key Takeaways - Always ensure the function is in terms of $y$ before integrating. - Check whether you are finding the value of an integral or an area. - Remember to take the absolute value for areas to the left of the y-axis. Practice solving different types of problems to become familiar with various scenarios and ensure accuracy in exams.

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