Finding the Area Under a Curve Revision Notes for VCE SSCE Mathematical Methods

Finding the Area Under a Curve

Introduction to area and definite integrals

The definite integral $\int_a^b f(x) \, dx$ gives us the net signed area under a curve. This means the integral takes into account whether the curve is above or below the $x$ -axis, which affects the sign of the result.

infoNote

When finding areas, the sign of f(x) in your interval is crucial. This determines whether you get a positive or negative value from the integral, and affects how you calculate the actual area.

Finding the area of a region

When the function is above the x-axis

If your function satisfies $f(x) \geq 0$ for all $x$ in the interval $[a, b]$ , this means the curve stays above or on the x-axis throughout the entire region. In this case, finding the area is straightforward.

The area $A$ of the region between the curve, the $x$ -axis, and the vertical lines $x = a$ and $x = b$ is given by:

A = \int_a^b f(x) \, dx

When the function is below the x-axis

If your function satisfies $f(x) \leq 0$ for all $x$ in the interval $[a, b]$ , the curve stays below or on the x-axis. Here's where we need to be careful: the definite integral will give us a negative number, but area is always positive.

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Why the negative sign?

When a function is below the $x$ -axis, the integral $\int_a^b f(x) \, dx$ produces a negative value. However, area must always be positive. Therefore, you must take the negative of the integral to obtain the actual area.

To find the actual area $A$ , we need to use the negative of the integral:

A = -\int_a^b f(x) \, dx

Alternatively, you can reverse the limits of integration:

A = \int_b^a f(x) \, dx

Both approaches give the same positive area value.

When the function crosses the x-axis

Sometimes your function crosses the $x$ -axis within the interval. Suppose the function crosses at point $c$ where $a < c < b$ , so that:

$f(x) \geq 0$ for $x$ in the interval $(c, b]$ (above the axis)
$f(x) \leq 0$ for $x$ in the interval $[a, c)$ (below the axis)

chatImportant

Splitting the integral:

When the function crosses the x-axis, you cannot use a single integral. Instead, you must split your calculation at the crossing point and add the absolute values of each region together.

The area $A$ is:

A = \int_c^b f(x) \, dx + \left(-\int_a^c f(x) \, dx\right)

The first integral is positive (above axis), and we add the negative of the second integral (which was below the axis) to get the total area.

infoNote

Critical factor: The sign of f(x) in your given interval is the key determining factor when calculating area under a curve. Always check whether your function is positive, negative, or crosses the axis before setting up your integral.

Worked examples

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Worked Example: Finding area above the x-axis

Find the area of the region between the $x$ -axis, the line $y = x + 1$ and the lines $x = 2$ and $x = 4$ . Check your answer by calculating the area of the trapezium.

Solution:

First, notice that $y = x + 1$ is positive throughout the interval from $x = 2$ to $x = 4$ . When $x = 2$ , $y = 3$ , and when $x = 4$ , $y = 5$ . So the curve is above the $x$ -axis.

\text{Area} = \int_2^4 (x + 1) \, dx

= \left[\frac{x^2}{2} + x\right]_2^4

= \left(\frac{4^2}{2} + 4\right) - \left(\frac{2^2}{2} + 2\right)

= (8 + 4) - (2 + 2)

= 12 - 4 = 8

The area of the shaded region is 8 square units.

Check: We can verify this using the trapezium area formula:

\text{Area of trapezium} = \text{average height} \times \text{base} = \frac{3 + 5}{2} \times 2 = 8

This confirms our answer.

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Worked Example: Finding area below the x-axis

Find the area under the line $y = x + 1$ between $x = -4$ and $x = -2$ .

Solution:

Looking at the diagram, we can see that in this interval, the line $y = x + 1$ is below the x-axis. When $x = -4$ , $y = -3$ , and when $x = -2$ , $y = -1$ . Both values are negative.

Since the function is below the $x$ -axis, we need to use the negative of the integral:

\text{Area} = -\int_{-4}^{-2} (x + 1) \, dx

= -\left[\frac{x^2}{2} + x\right]_{-4}^{-2}

= -\left[\left(\frac{(-2)^2}{2} + (-2)\right) - \left(\frac{(-4)^2}{2} + (-4)\right)\right]

= -\left[(2 - 2) - (8 - 4)\right]

= -(0 - 4) = 4

The area of the shaded region is 4 square units.

Note: The negative sign is essential here because the integral gives us the signed area from $-4$ to $-2$ , which is negative. Taking the negative of this gives us the positive area.

lightbulbExample

Worked Example: Function crossing the x-axis

Find the exact area of the shaded region.

Solution:

The parabola $y = x^2 - 4$ crosses the $x$ -axis at $x = 2$ (since $x^2 = 4$ gives $x = \pm 2$ ). Looking at the shaded region from $x = 1$ to $x = 4$ :

From $x = 1$ to $x = 2$ : the curve is below the x-axis (since $1^2 - 4 = -3 < 0$ )
From $x = 2$ to $x = 4$ : the curve is above the x-axis (since $4^2 - 4 = 12 > 0$ )

We need to split the integral at x = 2:

\text{Area} = \int_2^4 (x^2 - 4) \, dx + \left(-\int_1^2 (x^2 - 4) \, dx\right)

Calculate the first integral:

\int_2^4 (x^2 - 4) \, dx = \left[\frac{x^3}{3} - 4x\right]_2^4

= \left(\frac{64}{3} - 16\right) - \left(\frac{8}{3} - 8\right)

= \frac{64}{3} - 16 - \frac{8}{3} + 8

= \frac{56}{3} - 8

Calculate the second integral:

-\int_1^2 (x^2 - 4) \, dx = -\left[\frac{x^3}{3} - 4x\right]_1^2

= -\left[\left(\frac{8}{3} - 8\right) - \left(\frac{1}{3} - 4\right)\right]

= -\left[\frac{8}{3} - 8 - \frac{1}{3} + 4\right]

= -\left[\frac{7}{3} - 4\right]

Add the two parts together:

\text{Area} = \frac{56}{3} - 8 - \left(\frac{7}{3} - 4\right)

= \frac{56}{3} - 8 - \frac{7}{3} + 4

= \frac{49}{3} - 4

= \frac{49 - 12}{3} = \frac{37}{3}

The area is 37/3 square units.

bookmarkSummary

Key Points to Remember:

The definite integral $\int_a^b f(x) \, dx$ gives the net signed area, taking into account whether the function is above or below the $x$ -axis
When $f(x) \geq 0$ throughout the interval, use $A = \int_a^b f(x) \, dx$ directly
When $f(x) \leq 0$ throughout the interval, use $A = -\int_a^b f(x) \, dx$ to get a positive area
When the function crosses the x-axis, split the integral at the crossing point and add the absolute values of each region
Always check which sign your function has in the given interval - this is the critical factor in determining how to calculate the area
You can verify your answers using geometric formulas (like the trapezium formula) when the shape is simple

Finding the Area Under a Curve (VCE SSCE Mathematical Methods): Revision Notes