Areas under Simple Curves and Regions Bounded by Curves

Introduction to Area Calculation using Integration

In calculus, integration is a powerful tool that allows us to find the area under a curve. When we talk about finding the area under a curve, we are essentially summing up an infinite number of infinitesimally small rectangles under the curve between specified limits. This concept is fundamental to understanding how to calculate the area of complex shapes and regions bounded by different curves.

The definite integral of a function $f(x)$ from $x=a$ to $x=b$, denoted as $\int_{a}^{b} f(x) \, dx$, geometrically represents the area of the region bounded by the curve $y=f(x)$, the x-axis, and the vertical lines $x=a$ and $x=b$.

For this to represent a positive area, the function $f(x)$ must be non-negative ($f(x) \ge 0$) in the interval $[a, b]$. If $f(x)$ is negative in the interval, the definite integral will yield a negative value, which represents the signed area. To find the actual geometric area when $f(x) < 0$, we take the absolute value of the integral, or equivalently, integrate the absolute value of the function: Area $= \left| \int_{a}^{b} f(x) \, dx \right|$ or Area $= \int_{a}^{b} |f(x)| \, dx$.

Area Under a Curve $y=f(x)$ Above the x-axis

Consider a function $y = f(x)$ that is continuous and non-negative on the closed interval $[a, b]$. The area of the region bounded by the curve, the x-axis, and the vertical lines $x=a$ and $x=b$ is given by the definite integral:

Area $= \int_{a}^{b} f(x) \, dx$

To visualize this, imagine dividing the interval $[a, b]$ into $n$ subintervals of equal width $\Delta x = (b-a)/n$. In each subinterval $[x_i, x_{i+1}]$, we can approximate the area by a rectangle of height $f(x_i^*)$ (where $x_i^*$ is any point in the subinterval) and width $\Delta x$. The total area is the sum of these rectangular areas: $\sum_{i=1}^{n} f(x_i^*) \Delta x$. As $n$ approaches infinity and $\Delta x$ approaches zero, this sum becomes the definite integral.

Example: Find the area under the curve $y = x^2$ from $x=1$ to $x=3$. Here, $f(x) = x^2$, which is non-negative for $x \in [1, 3]$. The limits of integration are $a=1$ and $b=3$. Area $= \int_{1}^{3} x^2 \, dx$ Using the power rule for integration ($\int x^n \, dx = \frac{x^{n+1}}{n+1}$), we get: Area $= \left[ \frac{x^3}{3} \right]_{1}^{3}$ Area $= \frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$ So, the area is $\frac{26}{3}$ square units.

Area Under a Curve $y=f(x)$ Below the x-axis

If the function $f(x)$ is non-positive ($f(x) \le 0$) on the interval $[a, b]$, the definite integral $\int_{a}^{b} f(x) \, dx$ will be negative. The geometric area is the absolute value of this integral. An alternative approach is to integrate the negative of the function, i.e., $-f(x)$, which will be non-negative.

Area $= -\int_{a}^{b} f(x) \, dx = \int_{a}^{b} (-f(x)) \, dx$

Example: Find the area of the region bounded by the curve $y = -x^2$ and the x-axis from $x=1$ to $x=3$. Here, $f(x) = -x^2$, which is non-positive for $x \in [1, 3]$. Area $= -\int_{1}^{3} (-x^2) \, dx = \int_{1}^{3} x^2 \, dx$ This is the same integral as in the previous example. Area $= \left[ \frac{x^3}{3} \right]_{1}^{3} = \frac{27}{3} - \frac{1}{3} = \frac{26}{3}$ The geometric area is $\frac{26}{3}$ square units.

Area Bounded by a Curve $y=f(x)$ and the x-axis

When the curve $y=f(x)$ crosses the x-axis within the interval $[a, b]$, we need to split the integral into parts. Let $c$ be a point in $(a, b)$ where $f(c) = 0$. If $f(x) \ge 0$ on $[a, c]$ and $f(x) \le 0$ on $[c, b]$, the total area is the sum of the areas of these two parts.

Total Area $= \int_{a}^{c} f(x) \, dx + \left| \int_{c}^{b} f(x) \, dx \right|$ or Total Area $= \int_{a}^{c} f(x) \, dx - \int_{c}^{b} f(x) \, dx$

Example: Find the area bounded by the curve $y = x^2 - 4$ and the x-axis between $x=0$ and $x=3$. First, find the x-intercepts by setting $y=0$: $x^2 - 4 = 0 \implies x = \pm 2$. The interval is $[0, 3]$. The x-intercept within this interval is $x=2$. For $x \in [0, 2]$, $x^2 - 4$ is negative (e.g., at $x=1$, $y = 1-4 = -3$). For $x \in [2, 3]$, $x^2 - 4$ is positive (e.g., at $x=2.5$, $y = 6.25-4 = 2.25$). So, we split the integral at $x=2$. Area $= \left| \int_{0}^{2} (x^2 - 4) \, dx \right| + \int_{2}^{3} (x^2 - 4) \, dx$ Calculate the first integral: $\int_{0}^{2} (x^2 - 4) \, dx = \left[ \frac{x^3}{3} - 4x \right]_{0}^{2} = \left( \frac{2^3}{3} - 4(2) \right) - \left( \frac{0^3}{3} - 4(0) \right) = \frac{8}{3} - 8 = \frac{8 - 24}{3} = -\frac{16}{3}$ The absolute value is $\frac{16}{3}$. Calculate the second integral: $\int_{2}^{3} (x^2 - 4) \, dx = \left[ \frac{x^3}{3} - 4x \right]_{2}^{3} = \left( \frac{3^3}{3} - 4(3) \right) - \left( \frac{2^3}{3} - 4(2) \right)$ $= \left( \frac{27}{3} - 12 \right) - \left( \frac{8}{3} - 8 \right) = (9 - 12) - \left( \frac{8 - 24}{3} \right) = -3 - \left(-\frac{16}{3}\right) = -3 + \frac{16}{3} = \frac{-9 + 16}{3} = \frac{7}{3}$ Total Area $= \frac{16}{3} + \frac{7}{3} = \frac{23}{3}$ square units.

Area Between Two Curves $y=f(x)$ and $y=g(x)$

To find the area of the region bounded by two curves $y = f(x)$ and $y = g(x)$, we first need to find their points of intersection. These points are found by setting $f(x) = g(x)$ and solving for $x$. Let the points of intersection be $x=a$ and $x=b$.

In the interval $[a, b]$, one function will be greater than the other. Let's assume $f(x) \ge g(x)$ for all $x \in [a, b]$. The area of the region between the curves is then given by the integral of the difference between the upper curve and the lower curve:

Area $= \int_{a}^{b} [f(x) - g(x)] \, dx$

If $g(x) \ge f(x)$ on $[a, b]$, the formula becomes:

Area $= \int_{a}^{b} [g(x) - f(x)] \, dx$

In general, the area is $\int_{a}^{b} |f(x) - g(x)| \, dx$.

Example: Find the area of the region bounded by the curves $y = x^2$ and $y = x+2$. First, find the points of intersection: $x^2 = x+2$ $x^2 - x - 2 = 0$ $(x-2)(x+1) = 0$ So, the points of intersection are $x=-1$ and $x=2$. These will be our limits of integration, $a=-1$ and $b=2$. Now, determine which function is greater on the interval $[-1, 2]$. Let's test a point, say $x=0$: For $y=x^2$, $y = 0^2 = 0$. For $y=x+2$, $y = 0+2 = 2$. Since $2 > 0$, the line $y=x+2$ is above the parabola $y=x^2$ on the interval $[-1, 2]$. So, $f(x) = x+2$ and $g(x) = x^2$. Area $= \int_{-1}^{2} [(x+2) - x^2] \, dx$ Area $= \int_{-1}^{2} (-x^2 + x + 2) \, dx$ Integrate term by term: Area $= \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^{2}$ Evaluate at the upper limit ($x=2$): $-\frac{2^3}{3} + \frac{2^2}{2} + 2(2) = -\frac{8}{3} + \frac{4}{2} + 4 = -\frac{8}{3} + 2 + 4 = 6 - \frac{8}{3} = \frac{18 - 8}{3} = \frac{10}{3}$ Evaluate at the lower limit ($x=-1$): $-\frac{(-1)^3}{3} + \frac{(-1)^2}{2} + 2(-1) = -\frac{-1}{3} + \frac{1}{2} - 2 = \frac{1}{3} + \frac{1}{2} - 2 = \frac{2 + 3 - 12}{6} = -\frac{7}{6}$ Subtract the lower limit value from the upper limit value: Area $= \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{10}{3} + \frac{7}{6} = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$ The area of the region bounded by the curves is $\frac{9}{2}$ square units.

Area Calculation Using Integration with Respect to y

Sometimes, it is easier to express the curves as $x$ in terms of $y$ (i.e., $x = g(y)$) and integrate with respect to $y$. This is particularly useful when dealing with curves that are easier to describe horizontally, such as parabolas opening sideways ($x = ay^2 + by + c$) or circles.

If we have two curves $x = f(y)$ and $x = g(y)$, and they intersect at $y=c$ and $y=d$, with $f(y) \ge g(y)$ for $y \in [c, d]$, the area between them is:

Area $= \int_{c}^{d} [f(y) - g(y)] \, dy$

Here, $f(y)$ represents the "rightmost" curve and $g(y)$ represents the "leftmost" curve.

Example: Find the area of the region bounded by the curves $x = y^2$ and $x = y+2$. First, find the points of intersection by setting $x$ values equal: $y^2 = y+2$ $y^2 - y - 2 = 0$ $(y-2)(y+1) = 0$ The points of intersection occur at $y=-1$ and $y=2$. These are our limits of integration, $c=-1$ and $d=2$. Now, determine which function gives a larger $x$ value for $y \in [-1, 2]$. Let's test $y=0$: For $x = y^2$, $x = 0^2 = 0$. For $x = y+2$, $x = 0+2 = 2$. Since $2 > 0$, the curve $x = y+2$ is to the right of $x = y^2$ on the interval $[-1, 2]$. So, $f(y) = y+2$ (right curve) and $g(y) = y^2$ (left curve). Area $= \int_{-1}^{2} [(y+2) - y^2] \, dy$ This integral is identical in form to the previous example where we integrated with respect to $x$. Area $= \left[ \frac{y^2}{2} + 2y - \frac{y^3}{3} \right]_{-1}^{2}$ Evaluate at $y=2$: $\frac{2^2}{2} + 2(2) - \frac{2^3}{3} = \frac{4}{2} + 4 - \frac{8}{3} = 2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{18-8}{3} = \frac{10}{3}$ Evaluate at $y=-1$: $\frac{(-1)^2}{2} + 2(-1) - \frac{(-1)^3}{3} = \frac{1}{2} - 2 - \frac{-1}{3} = \frac{1}{2} - 2 + \frac{1}{3} = \frac{3 - 12 + 2}{6} = -\frac{7}{6}$ Subtract: Area $= \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{10}{3} + \frac{7}{6} = \frac{20+7}{6} = \frac{27}{6} = \frac{9}{2}$ The area is $\frac{9}{2}$ square units.

Regions Bounded by Curves and Axes

Often, problems involve regions bounded by curves and the coordinate axes (x-axis and y-axis). In such cases, you need to identify the limits of integration based on the intersection points of the curves and the axes.

Example: Find the area of the region bounded by the curve $y = \sqrt{4-x^2}$, the x-axis, and the y-axis in the first quadrant. The equation $y = \sqrt{4-x^2}$ represents the upper semi-circle of a circle centered at the origin with radius 2. Squaring both sides gives $y^2 = 4-x^2$, so $x^2 + y^2 = 4$. The x-axis is $y=0$. The y-axis is $x=0$. We are in the first quadrant, so $x \ge 0$ and $y \ge 0$. The curve $y = \sqrt{4-x^2}$ intersects the x-axis ($y=0$) when $\sqrt{4-x^2} = 0 \implies 4-x^2 = 0 \implies x = \pm 2$. Since we are in the first quadrant, the x-intercept is $x=2$. The curve intersects the y-axis ($x=0$) when $y = \sqrt{4-0^2} = \sqrt{4} = 2$. So, the region is bounded by $y=\sqrt{4-x^2}$, $y=0$ (x-axis), and $x=0$ (y-axis). The limits for $x$ are from $0$ to $2$. Area $= \int_{0}^{2} \sqrt{4-x^2} \, dx$ This integral represents the area of a quarter circle of radius 2. The area of a full circle is $\pi r^2 = \pi (2^2) = 4\pi$. The area of a quarter circle is $\frac{1}{4}(4\pi) = \pi$. We can also solve this using trigonometric substitution. Let $x = 2\sin\theta$. Then $dx = 2\cos\theta \, d\theta$. When $x=0$, $0 = 2\sin\theta \implies \sin\theta = 0 \implies \theta = 0$. When $x=2$, $2 = 2\sin\theta \implies \sin\theta = 1 \implies \theta = \frac{\pi}{2}$. $\sqrt{4-x^2} = \sqrt{4 - (2\sin\theta)^2} = \sqrt{4 - 4\sin^2\theta} = \sqrt{4(1-\sin^2\theta)} = \sqrt{4\cos^2\theta} = 2|\cos\theta|$. Since $\theta \in [0, \frac{\pi}{2}]$, $\cos\theta \ge 0$, so $\sqrt{4-x^2} = 2\cos\theta$. Area $= \int_{0}^{\pi/2} (2\cos\theta) (2\cos\theta \, d\theta) = \int_{0}^{\pi/2} 4\cos^2\theta \, d\theta$ Using the identity $\cos^2\theta = \frac{1+\cos(2\theta)}{2}$: Area $= \int_{0}^{\pi/2} 4 \left(\frac{1+\cos(2\theta)}{2}\right) \, d\theta = \int_{0}^{\pi/2} 2(1+\cos(2\theta)) \, d\theta$ Area $= 2 \int_{0}^{\pi/2} (1+\cos(2\theta)) \, d\theta = 2 \left[ \theta + \frac{\sin(2\theta)}{2} \right]_{0}^{\pi/2}$ Evaluate at $\theta = \frac{\pi}{2}$: $2 \left[ \frac{\pi}{2} + \frac{\sin(\pi)}{2} \right] = 2 \left[ \frac{\pi}{2} + 0 \right] = \pi$. Evaluate at $\theta = 0$: $2 \left[ 0 + \frac{\sin(0)}{2} \right] = 2[0] = 0$. Area $= \pi - 0 = \pi$. The area is $\pi$ square units.

Composite Areas

Some regions are bounded by multiple curves and may require breaking down the area calculation into several parts. This involves identifying all intersection points and determining the upper and lower bounds for each segment of the region.

Example: Find the area of the region enclosed by the curves $y = x^2$, $y = 2-x$, and $y=0$ (the x-axis). First, find the intersection points: 1. $y = x^2$ and $y = 2-x$: $x^2 = 2-x \implies x^2 + x - 2 = 0 \implies (x+2)(x-1) = 0$. Intersections at $x=-2$ (gives $y=4$) and $x=1$ (gives $y=1$). 2. $y = x^2$ and $y = 0$: $x^2 = 0 \implies x = 0$. Intersection at $(0,0)$. 3. $y = 2-x$ and $y = 0$: $2-x = 0 \implies x = 2$. Intersection at $(2,0)$. The region is bounded by $y=x^2$ (parabola), $y=2-x$ (line), and $y=0$ (x-axis). We need to consider the area in the first quadrant where $x \ge 0$. The relevant intersection points are $(0,0)$, $(1,1)$ (between $y=x^2$ and $y=2-x$), and $(2,0)$ (between $y=2-x$ and $y=0$). The region needs to be split at $x=1$. For $x \in [0, 1]$, the region is bounded above by $y=x^2$ and below by $y=0$. For $x \in [1, 2]$, the region is bounded above by $y=2-x$ and below by $y=0$. Area $= \int_{0}^{1} x^2 \, dx + \int_{1}^{2} (2-x) \, dx$ Calculate the first integral: $\int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}$ Calculate the second integral: $\int_{1}^{2} (2-x) \, dx = \left[ 2x - \frac{x^2}{2} \right]_{1}^{2}$ $= \left( 2(2) - \frac{2^2}{2} \right) - \left( 2(1) - \frac{1^2}{2} \right)$ $= \left( 4 - \frac{4}{2} \right) - \left( 2 - \frac{1}{2} \right)$ $= (4 - 2) - \left( \frac{4-1}{2} \right) = 2 - \frac{3}{2} = \frac{4-3}{2} = \frac{1}{2}$ Total Area $= \frac{1}{3} + \frac{1}{2} = \frac{2+3}{6} = \frac{5}{6}$ square units.

Summary of Formulas

Let $A$ be the area.

• Area under $y=f(x)$ from $a$ to $b$ (where $f(x) \ge 0$): $A = \int_{a}^{b} f(x) \, dx$
• Area under $y=f(x)$ from $a$ to $b$ (where $f(x) \le 0$): $A = -\int_{a}^{b} f(x) \, dx$
• Area between $y=f(x)$ and $y=g(x)$ from $a$ to $b$ (where $f(x) \ge g(x)$): $A = \int_{a}^{b} [f(x) - g(x)] \, dx$
• Area between $x=f(y)$ and $x=g(y)$ from $c$ to $d$ (where $f(y) \ge g(y)$): $A = \int_{c}^{d} [f(y) - g(y)] \, dy$

Common Curves and Their Properties

Knowing the shapes of common curves helps in sketching and solving area problems.

Practice Problems Strategy

When faced with an area problem in an exam:

1. Read Carefully: Understand which curves define the boundary of the region.
2. Sketch: Draw a rough sketch of the curves. This is crucial.
3. Find Intersections: Solve equations to find all points where the boundary curves intersect each other and the axes.
4. Determine Integration Variable: Decide whether to integrate with respect to $x$ or $y$. Usually, integrate with respect to the variable that results in fewer functions or simpler functions.
5. Set Up Integral(s): Identify the limits of integration and the integrand (difference between the 'outer' and 'inner' functions). If the region needs to be split, set up multiple integrals.
6. Evaluate: Calculate the definite integral(s).
7. Check: Does the answer make sense geometrically? Is it positive?