Applications of Integration for O Level A Math

The definite integral serves as a powerful mathematical tool for computing areas beneath curves. Rather than manually dividing irregular shapes into countless small rectangles, the definite integral provides an elegant solution.

A definite integral operates on a closed interval with distinct starting and ending points. Unlike indefinite integrals which include a constant of integration $c$, definite integrals produce a specific numerical value.

The Fundamental Formula:

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

Where:

• $F(x)$ is the antiderivative (integral) of $f(x)$
• $a$ is the lower limit of integration
• $b$ is the upper limit of integration
• The result $F(b) - F(a)$ gives us the exact area under the curve between $x = a$ and $x = b$

How to Evaluate Definite Integrals:

1. Find the indefinite integral $F(x)$ of $f(x)$ (without the constant $c$) 2. Evaluate $F(x)$ at the upper limit: $F(b)$ 3. Evaluate $F(x)$ at the lower limit: $F(a)$ 4. Subtract: $F(b) - F(a)$

Important Notes:

One of the most powerful techniques in integration is recognizing that it's the inverse operation of differentiation. This relationship allows us to solve integration problems by working backwards from differentiation.

The Fundamental Relationship:

If $\frac{d}{dx}[F(x)] = f(x)$, then $\int f(x)\,dx = F(x) + c$

For definite integrals: If $\frac{d}{dx}[F(x)] = f(x)$, then $\int_{a}^{b} f(x)\,dx = F(b) - F(a)$

Many examination questions follow this pattern:

1. Show that the derivative of some function equals another expression 2. Hence, evaluate a definite integral of that expression

The key word "hence" tells you to use the result from part (1) to solve part (2). You don't need to integrate from scratch – you already know the antiderivative from the differentiation!

Step-by-Step Process:

1. In part (1), differentiate the given function to show it equals the required expression 2. From this, you now know: $\frac{d}{dx}[\text{given function}] = \text{expression}$ 3. Therefore: $\int \text{expression}\,dx = \text{given function} + c$ 4. For definite integrals, evaluate the given function at the limits 5. If the integral doesn't match exactly, adjust for constants (multiply/divide as needed)

Key Technique – Adjusting for Constants:

Sometimes the integral you need to find is similar but not exactly the same as the derivative you showed. You may need to:

Example Pattern:

If we showed that $\frac{d}{dx}[x\cos 3x] = -3x\sin 3x + \cos 3x$

Then to find $\int x\sin 3x\,dx$, we rearrange:

$-3x\sin 3x = \frac{d}{dx}[x\cos 3x] - \cos 3x$

$x\sin 3x = -\frac{1}{3}\frac{d}{dx}[x\cos 3x] + \frac{1}{3}\cos 3x$

$\int x\sin 3x\,dx = -\frac{1}{3}x\cos 3x + \frac{1}{9}\sin 3x + c$

This technique saves enormous time in examinations and is a favorite of examiners because it tests your understanding of the fundamental relationship between differentiation and integration.

One of the most practical applications of definite integrals is calculating the area enclosed between a curve and the $x$-axis. This technique is fundamental for solving real-world problems in physics, engineering, and economics.

The area enclosed by the curve $y = f(x)$, the $x$-axis, and the vertical lines $x = a$ and $x = b$ is given by:

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

The absolute value $|f(x)|$ is crucial because:

• When $f(x) > 0$, the curve is above the $x$-axis, and the integral is positive
• When $f(x) < 0$, the curve is below the $x$-axis, and the integral is negative
• We use absolute value to ensure we calculate the total area (always positive)

Key Concept - Net Area vs Total Area:

Step-by-Step Process:

Step 1: Find the $x$-intercepts (where the curve crosses the $x$-axis)

Set $y = 0$ and solve for $x$. These points divide the region into parts where the curve is above or below the axis.

Step 2: Determine if the curve is above or below the $x$-axis in each interval

Test a point in each interval:

• If $y > 0$, curve is above the $x$-axis → area is $\int f(x)\,dx$
• If $y < 0$, curve is below the $x$-axis → area is $-\int f(x)\,dx$ or $\int |f(x)|\,dx$

Step 3: Set up separate integrals for each region

If the curve crosses the $x$-axis within $[a, b]$, split the integral:

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

where $c$ is an $x$-intercept.

Step 4: Evaluate each integral and sum the absolute values

Important Notes:

Common Mistakes to Avoid:

1. ❌ Forgetting to use absolute value when curve is below $x$-axis 2. ❌ Not splitting the integral when curve crosses the $x$-axis 3. ❌ Mixing up $x$-intercepts with $y$-intercepts 4. ❌ Using net area instead of total area

Example Pattern:

For a curve that crosses the $x$-axis at $x = c$ between $a$ and $b$:

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

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

This technique is essential for Problems 8-11 where you need to find areas enclosed by curves and coordinate axes.

Finding the area between two curves is a powerful extension of integration techniques. This method is used extensively in economics (consumer surplus, producer surplus), physics (work done), and engineering (stress analysis).

The area between two curves $y = f(x)$ (upper curve) and $y = g(x)$ (lower curve) from $x = a$ to $x = b$ is:

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

Key Principle: Always subtract the lower curve from the upper curve.

Think of it as:

• Area under upper curve: $\int_{a}^{b} f(x)\,dx$
• Minus area under lower curve: $-\int_{a}^{b} g(x)\,dx$
• Combined: $\int_{a}^{b} [f(x) - g(x)]\,dx$

This automatically gives us the area between the curves.

Step-by-Step Process:

Set $f(x) = g(x)$ and solve for $x$. These are your integration limits (unless different limits are specified).

Test a point $x_0$ between the limits:

• If $f(x_0) > g(x_0)$, then $f(x)$ is the upper curve
• If $g(x_0) > f(x)$, then $g(x)$ is the upper curve

Important: The curves might switch positions! Check each region separately.

$\text{Area} = \int_{a}^{b} [\text{upper curve} - \text{lower curve}]\,dx$

Expand, simplify, and integrate the difference, then evaluate at the limits.

Special Cases:

Split the region into intervals where one curve is consistently above the other:

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

where the curves switch at $x = c$.

If the problem specifies $x = a$ and $x = b$ (instead of finding intersections), use these as your limits.

If it's easier to express curves as $x = f(y)$ and $x = g(y)$:

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

where $c$ and $d$ are the $y$-coordinates of intersection.

Common Mistakes to Avoid:

1. ❌ Subtracting in the wrong order (lower - upper instead of upper - lower) 2. ❌ Not finding all intersection points 3. ❌ Assuming one curve is always on top without checking 4. ❌ Using individual areas instead of the difference 5. ❌ Forgetting to split the integral when curves cross

Verification Strategy:

Practical Applications:

Example Pattern:

For curves $y = x^2$ and $y = 2x$ intersecting at $x = 0$ and $x = 2$:

Test at $x = 1$:

• Upper: $y = 2(1) = 2$
• Lower: $y = 1^2 = 1$
• So $2x$ is above $x^2$

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

This technique appears in Question 10 and is a critical skill for advanced integration problems.