Area under a Curve

Q: How do you find the area under a curve when the function is below the x-axis?

When f(x) < 0 on an interval [a, b], the definite integral ∫ₐᵇ f(x) dx returns a negative number. To find the positive geometric area, take the absolute value of the integral, or equivalently integrate −f(x) instead. If part of the curve is above and part is below the axis, split the integral at each zero crossing.

Q: Why does the definite integral give the area under a curve?

The definite integral is defined as the limit of Riemann sums — sums of the areas of thin rectangles that approximate the region under the curve. As the rectangles become infinitely narrow, their total converges to the exact area. This connection between integration and area is formalized by the Fundamental Theorem of Calculus.

Area under a Curve

The area between the graph of y = f(x) and the x-axis is given by the definite integral below. This formula gives a positive result for a graph above the x-axis, and a negative result for a graph below the x-axis.

Note: If the graph of y = f(x) is partly above and partly below the x-axis, the formula given below generates the net area. That is, the area above the axis minus the area below the axis.

Formula:

Graph of y = f(x) above the x-axis with shaded yellow area between the curve and x-axis from x = a to x = b.

Example 1:

Find the area between y = 7 – x² and the x-axis between the values x = –1 and x = 2.

A semicircular curve from x=-2 to x=2 with maximum y=10, yellow shaded area under curve between x=0 and x=2.
Calculation showing Area = integral from -1 to 2 of (7 - x²)dx = (7x - x³/3) evaluated = 18

Example 2:

Find the net area between y = sin x and the x-axis between the values x = 0 and x = 2π.

A sine-like curve with x-axis labels 2, 4, 6; yellow shading above axis (0–~3.5) and below (~3.5–6).
Net area calculation: integral from 0 to 2π of sin x dx = (-cos x) from 0 to 2π = (-1)-(-1) = 0

Key Formula

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

Where:

$A$ = The net (signed) area between the curve and the x-axis over the interval [a, b]
$f(x)$ = The function whose graph forms the curve
$a$ = The lower limit (left boundary) of the interval
$b$ = The upper limit (right boundary) of the interval

Worked Example

Problem: Find the area between y = 7 − x² and the x-axis from x = −1 to x = 2.

Step 1: Set up the definite integral with the given function and limits.

A = \int_{-1}^{2} (7 - x^2)\, dx

Step 2: Find the antiderivative of 7 − x².

F(x) = 7x - \frac{x^3}{3}

Step 3: Evaluate the antiderivative at the upper limit x = 2.

F(2) = 7(2) - \frac{2^3}{3} = 14 - \frac{8}{3} = \frac{42 - 8}{3} = \frac{34}{3}

Step 4: Evaluate the antiderivative at the lower limit x = −1.

F(-1) = 7(-1) - \frac{(-1)^3}{3} = -7 + \frac{1}{3} = \frac{-21 + 1}{3} = \frac{-20}{3}

Step 5: Subtract the lower-limit value from the upper-limit value.

A = \frac{34}{3} - \frac{-20}{3} = \frac{34 + 20}{3} = \frac{54}{3} = 18

Answer: The area between y = 7 − x² and the x-axis from x = −1 to x = 2 is 18 square units.

Another Example

This example shows the important case where the curve crosses the x-axis within the interval. A single definite integral would give only the net area (−16/3 + 7/3 = −3), which is not the total geometric area. You must split the interval at each x-intercept and sum the absolute values.

Problem: Find the total (unsigned) area between y = x² − 4 and the x-axis from x = 0 to x = 3.

Step 1: Find where the curve crosses the x-axis by solving x² − 4 = 0. The roots are x = −2 and x = 2. Within the interval [0, 3], the crossing point is x = 2. The function is negative on [0, 2] and positive on [2, 3].

x^2 - 4 = 0 \implies x = \pm 2

Step 2: To get total area (not net area), integrate each piece separately and take absolute values. First, compute the integral from 0 to 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}

Step 3: The integral is negative because the curve is below the x-axis. Take the absolute value to get the area of this piece.

\left| -\frac{16}{3} \right| = \frac{16}{3}

Step 4: Now compute the integral from 2 to 3, where the curve is above the x-axis.

\int_{2}^{3} (x^2 - 4)\, dx = \left[\frac{x^3}{3} - 4x\right]_2^3 = \left(9 - 12\right) - \left(\frac{8}{3} - 8\right) = -3 + \frac{16}{3} = \frac{7}{3}

Step 5: Add the two pieces to get the total area.

A_{\text{total}} = \frac{16}{3} + \frac{7}{3} = \frac{23}{3}

Answer: The total area between y = x² − 4 and the x-axis from x = 0 to x = 3 is 23/3 ≈ 7.67 square units.

Frequently Asked Questions

What is the difference between net area and total area under a curve?

Net area is the value you get directly from the definite integral ∫ f(x) dx — regions below the x-axis contribute negative values that partially cancel regions above. Total area treats every region as positive, so you split the integral at each x-intercept and take the absolute value of each piece before adding. If the curve stays entirely above (or entirely below) the axis on the interval, net area and total area differ only by a sign.

How do you find the area under a curve when the function is below the x-axis?

When f(x) < 0 on an interval [a, b], the definite integral ∫ₐᵇ f(x) dx returns a negative number. To find the positive geometric area, take the absolute value of the integral, or equivalently integrate −f(x) instead. If part of the curve is above and part is below the axis, split the integral at each zero crossing.

Why does the definite integral give the area under a curve?

The definite integral is defined as the limit of Riemann sums — sums of the areas of thin rectangles that approximate the region under the curve. As the rectangles become infinitely narrow, their total converges to the exact area. This connection between integration and area is formalized by the Fundamental Theorem of Calculus.

Area under a Curve vs. Area between Curves

	Area under a Curve	Area between Curves
Definition	Region between y = f(x) and the x-axis	Region between two curves y = f(x) and y = g(x)
Formula	∫ₐᵇ f(x) dx	∫ₐᵇ \|f(x) − g(x)\| dx
When to use	One function and the x-axis as a boundary	Two functions forming the upper and lower boundaries
Sign issues	Negative when curve is below x-axis	Absolute value handles which curve is on top

Why It Matters

Area under a curve is one of the first major applications of integration you encounter in calculus. It appears in physics (work done by a variable force, displacement from a velocity-time graph), in statistics (probabilities under a probability density function), and in economics (consumer and producer surplus). Mastering this concept builds the foundation for more advanced integral applications such as volumes of revolution and arc length.

Common Mistakes

Mistake: Using a single integral over the entire interval when the curve crosses the x-axis, and reporting the result as the total area.

Correction: A single definite integral gives the net (signed) area. To find the total geometric area, identify where f(x) = 0 within [a, b], split into sub-intervals, and add the absolute value of each integral.

Mistake: Forgetting to evaluate the antiderivative at both limits and subtract correctly (especially with negative lower limits).

Correction: Always compute F(b) − F(a) carefully. When a is negative, substituting it into polynomial terms can change signs unexpectedly. Write out each substitution step to avoid arithmetic errors.

Related Terms

Definite Integral — The tool used to calculate area under a curve
Area between Curves — Extends the concept to area between two functions
Area Using Parametric Equations — Area calculation when curves are given parametrically
Area Using Polar Coordinates — Area calculation for curves in polar form
Graph of an Equation or Inequality — The visual representation of the curve
Formula — General term for the mathematical expressions used
Positive Number — Integral result when curve is above the x-axis
Negative Number — Integral result when curve is below the x-axis