# Areas between curves

He may look like an idiot and talk like an idiot but don’t let that fool you. He really is an idiot, Groucho Marx.

# Recall

Antiderivatives are fundamental concepts in calculus. They are the inverse operation of derivatives.

Given a function f(x), an antiderivative, also known as indefinite integral, F, is the function that can be differentiated to obtain the original function, that is, F’ = f, e.g., 3x2 -1 is the antiderivative of x3 -x +7 because $\frac{d}{dx} (x^3-x+7) = 3x^2 -1$. Symbolically, we write F(x) = $\int f(x)dx$.

The process of finding antiderivatives is called integration.

The Fundamental Theorem of Calculus states roughly that the integral of a function f over an interval is equal to the change of any antiderivate F (F'(x) = f(x)) between the ends of the interval, i.e., $\int_{a}^{b} f(x)dx = F(b)-F(a)=F(x) \bigg|_{a}^{b}$

# Areas between curves

We want to determine the area between y = f(x) and y = g(x) on the interval [a, b]. For simplicity sake, we are also going to assume that f(x) ≥ g(x).

We could draw “small” rectangles whose areas are dx·(f(x)-g(x)) (Area = base·height). The total area on the interval [a, b] is the definite integral $\int_{a}^{b} (f(x)-g(x))dx$ -Figure 1.d.-.

In general, the area between two curves representing two arbitrary functions, say f and g, can be found by taking the definite integral of the absolute difference between the two functions over the interval of interest.

# How to calculate the area between two curves

• Find the area of the region enclosed by the curves and lines y = x2 +1, y = x, x = 0, x = 1 (Figure 4).
1. Identify the region bounded by the given curves. Determine the points of intersection, if any, as they will define the limits of integration, e.g., $x^2+1 = x ↭ x^2-x+1=0 ↭ x = \frac{1±\sqrt{1-4}}{2}=\frac{1±\sqrt{-3}}{2}$. Since the discriminant is negative, the solutions are complex, which means the curves do not intersect in the real plane. The interval of interest is from x = 0 and x = 1, therefore a = 0 and b = 1.
2. Identify which function lies above the other in each segment of the given interval, e.g., in this case, y = x2 + 1 lies above y = x.
3. Calculate the integral over the interval [a, b]. The integrand will be the absolute difference between the functions f(x) and g(x), Area = $\int_{a}^{b} |f(x)-g(x)|dx = \int_{0}^{1} x^2+1-xdx = \frac{x^3}{3}-\frac{x^2}{2}+x\bigg|_{0}^{1} = \frac{1}{3}+\frac{-1}{2}+1 = \frac{2}{6}+\frac{-3}{6}+\frac{6}{6} = \frac{5}{6}.$

# Solved exercises

• Find the area between y = x2 -4x and y = 2x (Figure 5).

First, we need to calculate the points of intersection of the two curves. $2x = x^2 -4x ⇒ x^2 -6x = 0 ⇒ x(x-6)=0 ⇒ x = 0, 6$. These are the points of intersection.

For x ∈ [0, 6], the upper function is y = 2x, and the lower function is y = x2 - 4x.

The integral to find the area is $\int_{0}^{6} 2x-(x^2-4x)dx = \int_{0}^{6} (-x^2+6x)dx = \frac{-x^3}{3}+\frac{6x^2}{2} = \frac{-x^3}{3}+3x^2\bigg|_{0}^{6} = \frac{(-6)^3}{3}+3·6^2 = 36$.

• Calculate the blue-shaded area (Figure B).

The easiest method is by subtraction. Area = $4*16-\int_{0}^{4} x^2dx = 64 -\frac{x^3}{3}\bigg|_{0}^{4} = 64 -\frac{4^3}{3} = \frac{128}{3}units^2$

We can also use horizontal slices with “small” heights dy and widths $\sqrt{y}$ from (0, 0) to (16, 4). Area = $\int_{0}^{16} \sqrt{y}dy = \frac{y^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3}y^{\frac{3}{2}}\bigg|_{0}^{16} = \frac{2}{3}16^{\frac{3}{2}} = \frac{2}{3}(16^{\frac{1}{2}})^3 = \frac{2}{3}(4^3) = \frac{64·2}{3} = \frac{128}{3}~units^2.$

• Calculate the green-shaded area (Figure C)

$y = ± \sqrt{\frac{1}{2}x} ↭ y^2 = \frac{1}{2}x ↭ x = 2y^2.~ y = ± \sqrt{x -4} ↭y^2 = x -4 ↭ x = y^2+4$

We are going to integrate with respect to y or, in other words, we will use horizontal slices with “small” heights dy and widths $y^2+4-2y^2$ from -2 to 2.

Area = $\int_{-2}^{2} (y^2+4-2y^2)dy = \int_{-2}^{2} (-y^2+4)dy = \frac{-y^3}{3}+4y\bigg|_{-2}^{2} = \frac{-8}{3}+8-(\frac{8}{3}-8) = \frac{-16}{3}+16 = \frac{-16}{3}+\frac{48}{3} = \frac{32}{3}~units^2.$

• Find the area between x = y2 and y = x -2.

Calculate the points of intersection, in our particular case where x = y2 and y = x -2 intersect. y = x -2 ⇒ x = y +2 =[x = y2] y2 ↭ y + 2 = y2 ↭ y2 -y -2 = 0, (y -2)(y +1) = 0, y = 2 (4, 2) or y = -1 (1, -1).

Area =[Figure 1.e. Using vertical slices] $\int_{0}^{1} (\sqrt{x}-(-\sqrt{x}))dx$ (Left section, green area) + $\int_{1}^{4} (\sqrt{x}-(x-2))dx$ (Right section, red area) = $\int_{0}^{1} 2\sqrt{x}dx + \int_{1}^{4} (\sqrt{x}-(x-2)) = \int_{0}^{1} 2\sqrt{x}dx + \int_{1}^{4} (\sqrt{x}-x+2) = \left[ \frac{4}{3}x^{\frac{3}{2}} \right]_{0}^{1} +$

$\left[ \frac{2}{3}x^{\frac{3}{2}} - \frac{1}{2}x^2 + 2x \right]_{1}^{4} = \frac{4}{3}(1)^{\frac{3}{2}} - \frac{2}{3}(0)^{\frac{3}{2}} + \frac{2}{3}(4)^{\frac{3}{2}} - \frac{1}{2}(4)^2 + 2(4) - \frac{2}{3}(1)^{\frac{3}{2}} + \frac{1}{2}(1)^2 - 2(1)= \frac{4}{3}+\frac{2}{3}8 + -8 + 8 -\frac{2}{3}+\frac{1}{2}-2 = \frac{2}{3} +\frac{2}{3}8 + +\frac{1}{2}-2 = \frac{4+32+3-12}{6} = \frac{27}{6} = \frac{9}{2}$

Area =[Using horizontal slices with “small” heights dy and widths (y+2) -y2 from (1, -1) to (4, 2), Figure 1.f] $\int_{-1}^{2} (y+2-y^2)dy = (\frac{y^2}{2}+2y-\frac{y^3}{3})\bigg|_{2}^{-1} = \frac{1}{2}(2)^2 + 2(2) - \frac{1}{3}(2)^3 -(\frac{1}{2}(-1)^2 + 2(-1) - \frac{1}{3}(-1)^3)= \frac{10}{3}+\frac{7}{6} = \frac{20}{6}+\frac{7}{6} = \frac{27}{6} = \frac{9}{2}.$

• Compute the area between f(x) = -x2 + 4x and g(x) = x2-6x+5 over the interval [0, 1] (Figure 3)

1. Calculate the points of intersection of the two curves $-x^2+4x = x^2-6x+5 ↭ 2x^2-10x +5 = 0↭ x = \frac{10±\sqrt{100-40}}{4}= \frac{5±\sqrt{15}}{2}, x = \frac{5-\sqrt{15}}{2} ∈ [0, 1]$. For the sake of brevity, i = $\frac{5-\sqrt{15}}{2}$.

2. Determine the intervals of interest: $[0, \frac{5-\sqrt{15}}{2}], [\frac{5-\sqrt{15}}{2}, 1]$

3. Take the absolute difference between the two functions and integrate over each interval separately to find the area.

$\int_{0}^{i} x^2-6x+5 -(-x^2+4x)dx + \int_{i}^{1} -x^2+4x - (x^2-6x+5)dx = \int_{0}^{i} (2x^2-10x+5)dx + \int_{i}^{1} (-2x^2+10x-5)dx = \frac{2x^3}{3}-\frac{10x^2}{2}+5x\bigg|_{0}^{i}$ +

$-\frac{2x^3}{3}+\frac{10x^2}{2}-5x\bigg|_{i}^{1} =$… = $\frac{-52}{3}+5\sqrt{15}$

• Calculate the area of the shaded region bounded by three curves, y = -x-1, y = 1 -x2, y = x -1 (Figure A)

To calculate the area of the shaded region bounded by the curves y = -x-1, y = 1 -x2, y = x -1, we need to find the points of intersection. For y = -x-1, y = 1 -x2, -x -1 = 1 -x2 ↭ x2-x -2 = 0 ↭ (x-2)(x+1) = 0 ↭ x = 2, -1. For y = 1 -x2 and y = x -1, 1 -x2 = x -1 ↭ x2 + x -2 = 0 ↭ (x+2)(x-1) = 0 ↭ x = -2, x = 1.

The shaded region is composed of two segments: The segment between y = 1 - x2 and y = x −1 from x = -1 to x = 0. The segment between 1 - x2 and y = x - 1 from x = 0 to x = 1.

$\int_{-1}^{0} ((1-x^2)-(-x-1))dx + \int_{0}^{1} ((1-x^2)-(x-1))dx =$[Because the graph is symmetric respect to the y-axix] $2\int_{0}^{1} (-x^2-x+2)dx = 2(\frac{-x^3}{3}-\frac{x^2}{2}+2x)\bigg|_{0}^{1} = 2(\frac{-1}{3} -\frac{1}{2}+2)= 2(\frac{-2}{6}-\frac{3}{6}+\frac{12}{6}) = 2·\frac{7}{6} = \frac{7}{3}.$

# Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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