Applications of integration

This chapter explores the fundamental applications of integration, focusing on the precise calculation of areas. Mastery of these techniques is crucial for solving a wide range of problems in calculus and constitutes a frequently tested component of the CMI examination.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Area under a curve | | 2 | Area between curves | | 3 | Simple geometric applications |

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We begin with Area under a curve.

Part 1: Area under a curve

Area under a Curve

Overview

The phrase “area under a curve” is simple to say but very easy to misuse. In calculus, the definite integral gives signed area, while many geometry-style questions ask for actual area, which is always non-negative. CMI-style questions often test whether you can decide which one is being asked, split intervals correctly, and use symmetry or intersection points to simplify the work. ---

Learning Objectives

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Core Meaning

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Main Formulas

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When to Split the Interval

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Standard Geometric Cases

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Symmetry Shortcuts

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Minimal Worked Examples

Example 1 Find the area between $y=x$ and the $x$-axis from $x=0$ to $x=3$. Since $x\ge 0$ on $[0,3]$, $\qquad \text{Area} = \int_0^3 x\,dx = \left[\dfrac{x^2}{2}\right]_0^3 = \dfrac{9}{2}$ So the area is $\boxed{\dfrac{9}{2}}$. --- Example 2 Find the area between $y=x^2-1$ and the $x$-axis on $[-1,1]$. On $[-1,1]$, we have $x^2-1\le 0$, so the graph lies below the axis. Hence, $\qquad \text{Area} = -\int_{-1}^{1}(x^2-1)\,dx = \int_{-1}^{1}(1-x^2)\,dx$ $\qquad = 2\int_0^1(1-x^2)\,dx = 2\left[x-\dfrac{x^3}{3}\right]_0^1 = 2\left(1-\dfrac13\right)=\dfrac43$ So the area is $\boxed{\dfrac43}$. ---

Area Between Curves: Main Workflow

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Which of the following always gives the geometric area between the graph of $y=f(x)$ and the $x$-axis on $[a,b]$?" options=["$\int_a^b f(x)\,dx$","$\left|\int_a^b f(x)\,dx\right|$","$\int_a^b |f(x)|\,dx$","$\int_a^b (f(x))^2\,dx$"] answer="C" hint="Geometric area must always count all contributions positively." solution="The geometric area between the curve and the $x$-axis must count area above and below the axis as positive. Therefore the correct expression is $\qquad \int_a^b |f(x)|\,dx$ Hence the correct option is $\boxed{C}$." ::: :::question type="NAT" question="Find the area enclosed by the graph of $y=2-x$ and the coordinate axes in the first quadrant." answer="2" hint="Identify the intercepts first." solution="The line $y=2-x$ meets the axes at:

• $x=0 \Rightarrow y=2$

• $y=0 \Rightarrow x=2$

So the enclosed region in the first quadrant is a right triangle with base $2$ and height $2$. Hence the area is $\qquad \dfrac12 \cdot 2 \cdot 2 = 2$ Therefore the answer is $\boxed{2}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $f(x)\ge 0$ on $[a,b]$, then the area under the curve is $\int_a^b f(x)\,dx$","If $f(x)\le 0$ on $[a,b]$, then the geometric area equals $-\int_a^b f(x)\,dx$","For any function $f$, the geometric area on $[a,b]$ is always $\int_a^b f(x)\,dx$","To find area between $y=f(x)$ and $y=g(x)$, we use top minus bottom"] answer="A,B,D" hint="Separate signed area from actual area." solution="1. True. If the graph lies above the axis, signed area equals geometric area.

True. If the graph lies below the axis, the definite integral is negative, so we take minus of it.

False. If the graph goes below the axis, $\int_a^b f(x)\,dx$ is signed area, not geometric area.

True. For area between curves, we use upper function minus lower function.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Find the area enclosed between the curves $y=x$ and $y=x^2$." answer="$\dfrac16$" hint="Find the points of intersection first and then integrate upper minus lower." solution="We first find the points of intersection: $\qquad x = x^2$ $\qquad x^2 - x = 0$ $\qquad x(x-1)=0$ So the curves meet at $\qquad x=0,\ 1$ Now on $[0,1]$, we have $\qquad x \ge x^2$ So the area enclosed is $\qquad \int_0^1 (x-x^2)\,dx$ $\qquad = \left[\dfrac{x^2}{2} - \dfrac{x^3}{3}\right]_0^1$ $\qquad = \dfrac12 - \dfrac13 = \dfrac16$ Hence the area is $\boxed{\dfrac16}$." ::: ---

Summary

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Part 2: Area between curves

Area Between Curves

Overview

The area between curves is one of the most important applications of definite integration. In exam problems, the main difficulty is usually not integration itself, but deciding which curve is above, where the curves intersect, and with respect to which variable the area should be computed. ---

Learning Objectives

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Core Idea

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Basic Formulae

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How to Start Every Problem

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Intersection Points Matter

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When to Integrate with Respect to $x$

Example shape: If the region lies between $\qquad y=x$ and $\qquad y=x^2$ then between the intersection points, area is $\qquad \int (\text{line} - \text{parabola})\,dx$ because the line is above the parabola on that interval. ---

When to Integrate with Respect to $y$

For $dy$ integrals, always think: $\qquad \text{right curve} - \text{left curve}$ ---

Minimal Worked Examples

Example 1 Find the area between $\qquad y=x$ and $\qquad y=x^2$. First find intersection points: $\qquad x=x^2$ $\qquad x(x-1)=0$ So the curves meet at $\qquad x=0,1$ On $[0,1]$, we have $\qquad x \ge x^2$ So the area is $\qquad A=\int_0^1 (x-x^2)\,dx$ $\qquad =\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1$ $\qquad =\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}$ Hence the area is $\qquad \boxed{\dfrac{1}{6}}$ --- Example 2 Find the area enclosed by $\qquad y=4-x^2$ and $\qquad y=0$. Intersection with the $x$-axis: $\qquad 4-x^2=0 \implies x=\pm2$ So the area is $\qquad A=\int_{-2}^{2}(4-x^2)\,dx$ $\qquad =\left[4x-\dfrac{x^3}{3}\right]_{-2}^{2}$ $\qquad =\dfrac{32}{3}$ Therefore the area is $\qquad \boxed{\dfrac{32}{3}}$ ---

Symmetry Trick

This reduces calculation and avoids mistakes. ---

Common Patterns

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What Can Go Wrong

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The area enclosed between the curves $y=x$ and $y=x^2$ is" options=["$\dfrac{1}{3}$","$\dfrac{1}{6}$","$\dfrac{1}{2}$","$1$"] answer="B" hint="Find intersection points first." solution="The curves intersect where $\qquad x=x^2$ so $\qquad x(x-1)=0$ which gives $\qquad x=0,1$. On $[0,1]$, the line $y=x$ is above the parabola $y=x^2$. Hence $\qquad A=\int_0^1 (x-x^2)\,dx$ $\qquad =\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1$ $\qquad =\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}$ Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the area enclosed by the curve $y=4-x^2$ and the $x$-axis." answer="32/3" hint="Find where the parabola meets the $x$-axis." solution="The curve meets the $x$-axis when $\qquad 4-x^2=0$ which gives $\qquad x=\pm 2$. So the enclosed area is $\qquad A=\int_{-2}^{2}(4-x^2)\,dx$ $\qquad =\left[4x-\dfrac{x^3}{3}\right]_{-2}^{2}$ $\qquad =\left(8-\dfrac{8}{3}\right)-\left(-8+\dfrac{8}{3}\right)$ $\qquad =16-\dfrac{16}{3}=\dfrac{32}{3}$ Hence the answer is $\boxed{\dfrac{32}{3}}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $f(x)\ge g(x)$ on $[a,b]$, then the area between the curves is $\int_a^b (f(x)-g(x))\\,dx$","Area can be negative if the lower curve is integrated first","If the upper curve changes inside the interval, the area may need to be split into more than one integral","When integrating with respect to $y$, the area is found using right curve minus left curve"] answer="A,C,D" hint="Area is always non-negative." solution="1. True. This is the standard formula for area between curves with respect to $x$.

False. Area itself cannot be negative; a negative result means the setup was wrong.

True. If the top curve changes, the integral must usually be split.

True. For integration with respect to $y$, we use right minus left.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Find the area enclosed between the curves $y=2x$ and $y=x^2$." answer="$\dfrac{4}{3}$" hint="Find the intersection points and identify the upper curve." solution="The curves intersect where $\qquad 2x=x^2$ So $\qquad x(x-2)=0$ Hence the intersection points are $\qquad x=0,2$. On the interval $[0,2]$, the line $y=2x$ lies above the parabola $y=x^2$. Therefore $\qquad A=\int_0^2 (2x-x^2)\,dx$ $\qquad =\left[x^2-\dfrac{x^3}{3}\right]_0^2$ $\qquad =4-\dfrac{8}{3}=\dfrac{4}{3}$ Hence the enclosed area is $\boxed{\dfrac{4}{3}}$." ::: ---

Summary

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Part 3: Simple geometric applications

Simple Geometric Applications

Overview

One of the first geometric uses of integration is to convert a changing boundary into an exact area. In simple geometric applications, the main skill is to translate a shaded region into the correct integral with correct limits. In CMI-style questions, the difficulty usually lies less in integration and more in reading the region correctly. ---

Learning Objectives

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Core Idea

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Area Under a Curve

If $f(x)$ changes sign on $[a,b]$, split the interval at the points where $f(x)=0$ and use absolute value idea. ---

Area Between Two Curves

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Area with Respect to $y$

This is useful when curves are given naturally as $x=$ function of $y$, or when vertical slicing becomes messy. ---

Symmetry

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Standard Geometric Setups

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Region Reading Strategy

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Minimal Worked Examples

Example 1 Find the area under $y=x^2$ from $x=0$ to $x=2$. $\qquad \text{Area}=\int_0^2 x^2\,dx$ $\qquad =\left[\dfrac{x^3}{3}\right]_0^2=\dfrac{8}{3}$ So the area is $\boxed{\dfrac{8}{3}}$. --- Example 2 Find the area enclosed between $y=x$ and $y=x^2$. First find intersections: $\qquad x=x^2$ $\qquad x(x-1)=0$ So $x=0,1$. On $[0,1]$, we have $x\ge x^2$, so the upper curve is $y=x$. Hence, $\qquad \text{Area}=\int_0^1 (x-x^2)\,dx$ $\qquad =\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1$ $\qquad =\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}$ So the area is $\boxed{\dfrac{1}{6}}$. ---

High-Value Observations

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Common Special Cases

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Practice Questions

:::question type="MCQ" question="The area enclosed between the curves $y=x$ and $y=x^2$ is" options=["$\dfrac{1}{3}$","$\dfrac{1}{6}$","$\dfrac{1}{2}$","$\dfrac{2}{3}$"] answer="B" hint="Find the intersection points first." solution="The curves intersect where $\qquad x=x^2$ so $\qquad x(x-1)=0$ and hence $x=0,1$. On $[0,1]$, the line $y=x$ lies above the parabola $y=x^2$. Therefore, $\qquad \text{Area}=\int_0^1 (x-x^2)\,dx$ $\qquad =\left[\dfrac{x^2}{2}-\dfrac{x^3}{3}\right]_0^1=\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}$ Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the area under the curve $y=2x+1$ from $x=0$ to $x=3$." answer="12" hint="Integrate directly." solution="The required area is $\qquad \int_0^3 (2x+1)\,dx$ $\qquad =\left[x^2+x\right]_0^3=(9+3)-0=12$ Therefore the answer is $\boxed{12}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $f(x)\ge 0$ on $[a,b]$, then $\int_a^b f(x)\\,dx$ gives geometric area under the curve","The area between $y=f(x)$ and $y=g(x)$ on $[a,b]$ is always $\int_a^b (f(x)-g(x))\\,dx$ even if $g(x)>f(x)$ somewhere","If $f$ is odd, then $\int_{-a}^a f(x)\\,dx=0$","Zero signed area always means zero geometric area"] answer="A,C" hint="Distinguish signed area from geometric area." solution="1. True, provided $f(x)\ge0$ on the interval.

False, because the correct area formula is upper minus lower, and if the upper curve changes, the interval must be split.

True. This is the standard odd-function property.

False. Positive and negative signed areas may cancel while geometric area remains positive.

Hence the correct answer is $\boxed{A,C}$." ::: :::question type="SUB" question="Find the area enclosed by the curves $y=4-x^2$ and $y=0$." answer="$\dfrac{32}{3}$" hint="Find where the parabola meets the $x$-axis." solution="The curve meets the $x$-axis where $\qquad 4-x^2=0$ so $\qquad x=\pm 2$. Hence the required area is $\qquad \int_{-2}^{2} (4-x^2)\,dx$ Using symmetry, $\qquad =2\int_0^2 (4-x^2)\,dx$ $\qquad =2\left[4x-\dfrac{x^3}{3}\right]_0^2$ $\qquad =2\left(8-\dfrac{8}{3}\right)=2\cdot \dfrac{16}{3}=\dfrac{32}{3}$ Therefore the area is $\boxed{\dfrac{32}{3}}$." ::: ---

Summary

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="What is the area of the region bounded by $y = x^2$ and $y = \sqrt{x}$?" options=["1/3", "1/2", "1/6", "2/3"] answer="1/3" hint="First find the points of intersection and determine which function is greater in the interval." solution="The curves intersect when $x^2 = \sqrt{x}$. Squaring both sides gives $x^4 = x$, so $x^4 - x = 0 \implies x(x^3-1) = 0$. Thus, $x=0$ or $x=1$. In the interval $[0, 1]$, $\sqrt{x} \ge x^2$. The area is $\int_0^1 (\sqrt{x} - x^2) \, dx = \left[\frac{2}{3}x^{3/2} - \frac{1}{3}x^3\right]_0^1 = \left(\frac{2}{3} - \frac{1}{3}\right) - (0 - 0) = \frac{1}{3}$."
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:::question type="NAT" question="A region is bounded by $y = x^2$, the x-axis, and the line $x=2$. If this region is revolved about the x-axis, what is the exact volume of the resulting solid, divided by $\pi$?" answer="32/5" hint="Use the Disk Method. The radius of each disk is $y=x^2$." solution="Using the Disk Method, the volume $V = \pi \int_0^2 (x^2)^2 \, dx = \pi \int_0^2 x^4 \, dx = \pi \left[\frac{1}{5}x^5\right]_0^2 = \pi \left(\frac{1}{5}(2)^5 - \frac{1}{5}(0)^5\right) = \pi \left(\frac{32}{5}\right)$. The volume divided by $\pi$ is $32/5$."
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:::question type="MCQ" question="What is the average value of the function $f(x) = \sin(x)$ on the interval $[0, \pi]$?" options=["0", "$2/\pi$", "$1/\pi$", "$1$"] answer="$2/\pi$" hint="Recall the formula for the average value of a function over an interval." solution="The average value $f_{\operatorname{avg}} = \frac{1}{\pi - 0} \int_0^\pi \sin(x) \, dx = \frac{1}{\pi} [-\cos(x)]_0^\pi = \frac{1}{\pi} (-\cos(\pi) - (-\cos(0))) = \frac{1}{\pi} (-(-1) - (-1)) = \frac{1}{\pi} (1+1) = \frac{2}{\pi}$."
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:::question type="NAT" question="Find the area of the region enclosed by the curves $x = y^2 - 4y$ and $x = 2y - y^2$. Express your answer as a decimal." answer="9.0" hint="Integrate with respect to $y$. Find the points of intersection and determine which function yields the greater $x$-value for integration." solution="First, find the points of intersection: $y^2 - 4y = 2y - y^2 \implies 2y^2 - 6y = 0 \implies 2y(y-3) = 0$. So, $y=0$ and $y=3$. For $y \in [0, 3]$, test a value, e.g., $y=1$. $x = 1^2 - 4(1) = -3$ and $x = 2(1) - 1^2 = 1$. So $2y - y^2 \ge y^2 - 4y$. The area is $\int_0^3 ((2y - y^2) - (y^2 - 4y)) \, dy = \int_0^3 (6y - 2y^2) \, dy = \left[3y^2 - \frac{2}{3}y^3\right]_0^3 = \left(3(3)^2 - \frac{2}{3}(3)^3\right) - (0) = (27 - 18) = 9$. The answer is $9.0$."
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