FAD1014 L11-L12: Area Under Curves

Lecture notes from Week 6 covering the definite integral, Riemann sums, and their applications to calculating areas under and between curves.

1. From Riemann Sums to the Definite Integral

1.1 Approximating Area with Rectangles

To find the area under a curve $y = f(x)$, we approximate it by adding up the areas of rectangles:

Riemann sum: the result of adding rectangles to approximate area under a curve.
Subinterval: the width of one rectangle.
Partition: the entire interval $[a, b]$ divided into subintervals.
Subintervals do not all have to be the same size.

If the partition is denoted by $P$, the length of the longest subinterval is called the norm of $P$, denoted $|P|$.

As $|P|$ gets smaller, the approximation for the area gets better.

1.2 The Limit Definition

For a partition $P$ of $[a, b]$:

$$\text{Area} = \lim_{|P| \to 0} \sum_{k=1}^{n} f(c_k),\Delta x_k$$

This limit is called the definite integral of $f$ over $[a, b]$.

1.3 Equal Subintervals

If we use $n$ subintervals of equal length:

$$\Delta x = \frac{b - a}{n}$$

The definite integral is then:

$$\lim_{n \to \infty} \sum_{k=1}^{n} f(c_k),\Delta x = \int_a^b f(x),dx$$

Leibnitz introduced the integral symbol $\int$ as a stylized "S" for sum. Note that the very small change in $x$ becomes $dx$.

1.4 Anatomy of the Definite Integral

$$\int_a^b f(x),dx$$

Part	Meaning
$\int$	Integration symbol
$a$	Lower limit of integration
$b$	Upper limit of integration
$f(x)$	Integrand
$dx$	Variable of integration (dummy variable)

The variable of integration is called a dummy variable because the answer does not depend on the variable chosen.

1.5 The Area Function and FTC

Let $A_a(x)$ = area under the curve from $a$ to $x$ (where $a$ is constant).

Then $A_a(x+h) - A_a(x)$ is the area of a thin strip from $x$ to $x+h$.

By the squeeze theorem using min and max heights:

$$h \cdot \min f \le A_a(x+h) - A_a(x) \le h \cdot \max f$$

Dividing by $h$ and taking the limit as $h \to 0$:

$$\frac{d}{dx} A_a(x) = f(x)$$

Since $A_a(x) = F(x) + c$ and $A_a(a) = 0$:

$$A_a(x) = F(x) - F(a)$$

Area under curve from $a$ to $x$ = antiderivative at $x$ minus antiderivative at $a$.

2. Summary Chain

$$\text{Area} = \lim_{|P| \to 0} \sum_{k=1}^{n} f(c_k),\Delta x_k = \int_a^b f(x),dx = F(x) - F(a)$$

3. Area Under a Single Curve

3.1 Fundamental Theorem of Calculus (Evaluation Form)

If $f$ is continuous on $[a, b]$ and $F$ is an antiderivative of $f$:

$$\int_a^b f(x),dx = F(b) - F(a) = \bigl[F(x)\bigr]_a^b$$

Example: Find the area under $y = x^2$ from $x = 1$ to $x = 2$.

$$\int_1^2 x^2,dx = \left[\frac{x^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$$

3.2 Four-Step Procedure for Area Under a Curve

Sketch a graph of $f(x)$ over $[a, b]$.
Find any $x$-intercepts of $f(x)$ in $[a, b]$. These divide the total region into subregions.
The definite integral will be positive for subregions above the $x$-axis and negative for subregions below the $x$-axis. Use separate integrals to find the (positive) areas of the subregions.
The total area is the sum of the areas of all of the subregions.

3.3 Area When the Curve Crosses the x-axis

If the curve dips below the $x$-axis, the definite integral gives a negative value for that portion. Since area is always positive, take the absolute value or negate the integral for regions below the axis.

Example 6.2.3: Find the area bounded by $f(x) = x^2 - 4$, the $x$-axis, and $x = 0, x = 2$.

$$\int_0^2 (x^2 - 4),dx = \left[\frac{x^3}{3} - 4x\right]_0^2 = \left(\frac{8}{3} - 8\right) - 0 = -\frac{16}{3}$$

The answer is negative because the area is below the $x$-axis. Since area must be positive:

$$\text{Area} = \left|-\frac{16}{3}\right| = \frac{16}{3}$$

Example 6.2.4: Find the area between the $x$-axis and $f(x) = x^2 - 4$ from $x = 0$ to $x = 4$.

The curve crosses the $x$-axis at $x = 2$. Split into two integrals:

$$\text{Area} = \left|\int_0^2 (x^2 - 4),dx\right| + \left|\int_2^4 (x^2 - 4),dx\right|$$

$$= \left|-\frac{16}{3}\right| + \left|\frac{16}{3}\right| = \frac{16}{3} + \frac{32}{3} = 16$$

Example 6.2.2: Find the area between the $x$-axis and $y = \cos x$ from $x = 0$ to $x = \frac{3\pi}{2}$.

$$\int_0^{\pi/2} \cos x,dx - \int_{\pi/2}^{3\pi/2} \cos x,dx = \bigl[\sin x\bigr]0^{\pi/2} - \bigl[\sin x\bigr]{\pi/2}^{3\pi/2}$$

$$= (1 - 0) - (-1 - 1) = 1 + 2 = 3$$

3.4 Key Notes on Sign

An area is always positive.
The definite integral is positive for areas above the $x$-axis but negative for areas below the axis.
To find an area, we need to know whether the curve crosses the $x$-axis between the boundaries.
For areas above the axis, the definite integral gives the area directly.
For areas below the axis, we need to change the sign of the definite integral to find the area.

4. Area Between Two Curves

4.1 Vertical Strips

The area $A$ of the region bounded by $y = f(x)$ (top), $y = g(x)$ (bottom), and the lines $x = a, x = b$, where $f(x) \ge g(x)$ on $[a, b]$:

$$A = \int_a^b \bigl[f(x) - g(x)\bigr],dx$$

Or using the lecture notation:

$$\text{Area} = \int_a^b \bigl[f_1(x) - f_2(x)\bigr],dx$$

Example: Find the area between $y_1 = 2 - x^2$ and $y_2 = -x$.

Intersections: $2 - x^2 = -x \Rightarrow x^2 - x - 2 = 0 \Rightarrow x = -1, 2$.

$$\text{Area} = \int_{-1}^{2} \bigl[(2 - x^2) - (-x)\bigr],dx = \int_{-1}^{2} (2 - x^2 + x),dx$$

$$= \left[2x - \frac{x^3}{3} + \frac{x^2}{2}\right]_{-1}^{2} = \left(4 - \frac{8}{3} + 2\right) - \left(-2 + \frac{1}{3} + \frac{1}{2}\right) = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{27}{6} = \frac{9}{2}$$

4.2 Horizontal Strips

When vertical strips would require splitting the integral into multiple parts, it may be easier to use horizontal strips (integrate with respect to $y$).

Solve both curves for $x$ in terms of $y$.
The width of the strip is $dy$.
The length of the strip is $(\text{right curve} - \text{left curve})$.

Example: Area between $y = \sqrt{x}$ and $y = x - 2$.

With vertical strips: $$\int_0^2 \sqrt{x},dx + \int_2^4 \bigl[\sqrt{x} - (x - 2)\bigr],dx \quad \text{(two integrals)}$$

With horizontal strips: rewrite as $x = y^2$ and $x = y + 2$.

$$\text{Area} = \int_0^2 \bigl[(y + 2) - y^2\bigr],dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_0^2 = 2 + 4 - \frac{8}{3} = \frac{10}{3}$$

4.3 General Strategy for Area Between Curves

Sketch the curves.
Decide on vertical or horizontal strips. Pick whichever is easier to write formulas for the length of the strip, and/or whichever will let you integrate fewer times.
Write an expression for the area of the strip. If the width is $dx$, the length must be in terms of $x$. If the width is $dy$, the length must be in terms of $y$.
Find the limits of integration. If using $dx$, the limits are $x$ values; if using $dy$, the limits are $y$ values.
Integrate to find area.

5. Area with Respect to the y-axis

The area bounded by a curve, the $y$-axis, and the lines $y = c$ and $y = d$ is found by switching the $x$s and $y$s in the formula:

$$\int_a^b y,dx \quad \text{becomes} \quad \int_c^d x,dy$$

Example: Find the area between $y = \sqrt{x}$, the $y$-axis, and $y = 1$ to $y = 2$.

Rewrite $y = \sqrt{x}$ as $x = y^2$:

$$\int_1^2 y^2,dy = \left[\frac{y^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$$

6. Harder Areas: Two Methods

Method 1: Subtract Areas Separately

Find the area under the upper curve, subtract the area under the lower curve (or geometric shape).

Example: Area enclosed by $y = 2x - x^2$ and $y = x$.

Intersections: $x = 2x - x^2 \Rightarrow x^2 - x = 0 \Rightarrow x = 0, 1$.

$$\text{Area under curve} = \int_0^1 (2x - x^2),dx = \left[x^2 - \frac{x^3}{3}\right]_0^1 = \frac{2}{3}$$

$$\text{Area of triangle} = \frac{1}{2}(1)(1) = \frac{1}{2}$$

$$\text{Required area} = \frac{2}{3} - \frac{1}{2} = \frac{1}{6}$$

Method 2: Integrate the Difference

Subtract the functions first, then integrate:

$$(2x - x^2) - x = x - x^2$$

$$\text{Area} = \int_0^1 (x - x^2),dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$$

Example (a): $y = x^2 + 2$ and $y = 6$.

Intersections: $x^2 + 2 = 6 \Rightarrow x = \pm 2$.

$$\text{Area under curve} = \int_{-2}^{2} (x^2 + 2),dx = \left[\frac{x^3}{3} + 2x\right]_{-2}^{2} = \frac{16}{3}$$

$$\text{Shaded area} = \text{area of rectangle} - \text{area under curve} = 24 - \frac{16}{3} = 18\frac{2}{3}$$

Example (b): $y = 4 - x^2$ and $y = x + 2$.

Intersections: $4 - x^2 = x + 2 \Rightarrow x^2 + x - 2 = 0 \Rightarrow x = -2, 1$.

Points: $(-2, 0)$ and $(1, 3)$.

7. Area of a Symmetrical Curve

If a curve crosses the $x$-axis between the limits of integration, part of the area will be above the axis and part below.

Example: $y = x^3$ between $-1$ and $+1$.

The symmetry of the curve means that the integral from $-1$ to $+1$ is $0$.

To find the area, we could integrate from $0$ to $1$ and, because of the symmetry, double the answer.

For a curve which wasn't symmetrical, we could find the two areas separately and then add.

Properties of Definite Integrals

$\displaystyle\int_a^a f(x),dx = 0$
$\displaystyle\int_a^b f(x),dx = -\int_b^a f(x),dx$
$\displaystyle\int_a^b \bigl[f(x) + g(x)\bigr],dx = \int_a^b f(x),dx + \int_a^b g(x),dx$
$\displaystyle\int_a^b c \cdot f(x),dx = c\int_a^b f(x),dx$
$\displaystyle\int_a^b f(x),dx + \int_b^c f(x),dx = \int_a^c f(x),dx$

FAD1014 L11-L12 — Area Under Curves