Area Under a Curve

What You’ll Learn

In this lesson you’ll learn how to use definite integrals to calculate the area under a curve, how to handle regions below the x-axis, and how to find the area between two curves.

The Concept

The definite integral gives you the net signed area between a curve and the x-axis. “Net signed” means area above the x-axis counts as positive and area below counts as negative.

For a function that stays above the x-axis on [a, b], the area is just the integral:

$$\text{Area} = \int_a^b f(x) \, dx$$

But if the curve dips below the x-axis, the integral subtracts that region. To get the total geometric area (always positive), you need to split the integral at the x-intercepts and take absolute values of the negative pieces.

Area Between Two Curves

When you want the area trapped between two curves f(x) and g(x) from a to b, where f(x) is on top:

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

Top minus bottom, integrated. If the curves cross, you split at the crossing points and swap which is on top.

Worked Example

Example 1: Simple area above the x-axis

Find the area under f(x) = 2x from x = 1 to x = 4.

$$\int_1^4 2x \, dx = \left[x^2\right]_1^4 = 16 - 1 = 15$$

Since 2x is positive on [1, 4], the integral gives the area directly: 15 square units.

Example 2: Area between two curves

Find the area between y = x and y = x² from x = 0 to x = 1.

On [0, 1], x is greater than or equal to x² (the line is above the parabola). They meet at x = 0 and x = 1.

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

y = x (upper) y = x² (lower) area = 1/6

The green line is y = x (the upper curve) and the blue curve is y = x² (the lower curve). The shaded region between them has area 1/6. The orange dots mark where the curves intersect at (0, 0) and (1, 1).

Example 3: Handling negative regions

Find the total area between f(x) = x² - 1 and the x-axis from x = -2 to x = 2.

The function crosses the x-axis at x = -1 and x = 1. It’s negative between -1 and 1, positive outside.

$$\text{Total area} = \int_{-2}^{-1}(x^2 - 1)\,dx + \left|\int_{-1}^{1}(x^2 - 1)\,dx\right| + \int_{1}^{2}(x^2 - 1)\,dx$$

Each piece is computed separately. The middle integral gives a negative number (the curve is below the axis), so you take its absolute value before adding.

Real-World Application

Area calculations show up everywhere:

• Physics: the area under a velocity-time graph is the total distance traveled
• Economics: the area between supply and demand curves is consumer or producer surplus
• Medicine: the area under a drug concentration curve (AUC) measures total drug exposure
• Engineering: the area under a force-displacement curve is the work done

You’ve Got This

Area under a curve is one of the most intuitive applications of integration. For simple cases where the function stays positive, just integrate. When the curve crosses the x-axis, split at the roots. For area between two curves, subtract bottom from top. The visual intuition is straightforward: you’re measuring the size of a region, and the integral is the tool that computes it exactly.

Quiz

The definite integral gives:

• A.Only positive area
• B.The slope at a point
• C.Net signed area
• D.The derivative

To find total geometric area when the curve crosses the x-axis:

• A.Integrate the whole interval as one piece
• B.Split at the roots and take absolute values of negative parts
• C.Only integrate the positive parts
• D.Ignore the negative regions

The area between y = x and y = x² from 0 to 1 is:

• A.1/2
• B.1/3
• C.1/6
• D.1

For area between two curves, you integrate:

• A.The product of the two functions
• B.Top function minus bottom function
• C.Bottom minus top
• D.The sum of both functions

In physics, the area under a velocity-time graph gives:

• A.Acceleration
• B.Maximum speed
• C.Total distance traveled
• D.The slope of position