Area Under a Curve: Definite Integral & Examples

What Is The Area Under A Curve?

The area under a curve is the region trapped between the graph of a function $y = f(x)$ and the x-axis, across an interval from $x = a$ to $x = b$. When $f(x) \ge 0$ on that interval, this area is given exactly by the definite integral:

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

Here $a$ and $b$ are the limits of integration: $a$ is where the region starts and $b$ is where it ends. The symbol $\int$ is a stretched "S" for "sum," a reminder that an integral adds up infinitely many pieces. The height of the region at any point is $f(x)$, and the integral sums those heights across the whole width.

Two ideas sit underneath that one line, and the rest of this article unpacks them:

• Area as a limit. The integral is defined as the limit of Riemann sums, the combined area of rectangles that approximate the region and get thinner without end.

• Area with a sign. If the curve drops below the x-axis, the integral counts that part as negative. That gives net area, which is not always the total area you want.

For the wider map of the subject, see the calculus overview; for the meaning of area itself, see area.

How Do You Calculate The Area Under A Curve?

To calculate the area under a curve, find an antiderivative of $f$ and subtract its values at the two limits. This is the evaluation step of the Fundamental Theorem of Calculus:

$$\int_a^b f(x),dx = F(b) - F(a), \qquad \text{where } F'(x) = f(x)$$

The recipe is three steps: find $F$, plug in the top limit $b$, subtract the value at the bottom limit $a$.

Example 1: Find the area under $y = x^2$ from $x = 0$ to $x = 2$.

The curve $y = x^2$ sits on or above the x-axis, so the integral gives the area directly. An antiderivative of $x^2$ is $F(x) = \dfrac{x^3}{3}$:

$$\int_0^2 x^2,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{2^3}{3} - \frac{0^3}{3} = \frac{8}{3} - 0 = \frac{8}{3}$$

Check by differentiating the antiderivative back: $\dfrac{d}{dx}\left(\dfrac{x^3}{3}\right) = x^2$, the original integrand, so $F$ is correct.

Final answer: the area is $\dfrac{8}{3} \approx 2.6667$ square units.

Every definite integral in this article is checked the same way, by differentiating the antiderivative back to the function you started with.

How Do Riemann Sums Build The Area Under A Curve?

Before the integral gives an exact number, it is built from an approximation. Slice the interval $[a, b]$ into $n$ strips of equal width $\Delta x = \dfrac{b - a}{n}$, and on each strip draw a rectangle whose height is the value of $f$ at a sample point. Adding the rectangle areas gives a Riemann sum:

$$S_n = \sum_{i=1}^{n} f(x_i),\Delta x$$

Geometrically, the rectangles trace the region with a staircase. As $n$ grows the staircase hugs the curve more tightly, and the area under a curve is defined as the limit of that sum:

$$\int_a^b f(x),dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i),\Delta x$$

That limit is not a guess; it can be computed. Take $f(x) = x^2$ on $[0, 2]$ with right-endpoint samples $x_i = \dfrac{2i}{n}$ and $\Delta x = \dfrac{2}{n}$:

$$S_n = \sum_{i=1}^{n} \left(\frac{2i}{n}\right)^2 \cdot \frac{2}{n} = \frac{8}{n^3}\sum_{i=1}^{n} i^2 = \frac{8}{n^3}\cdot\frac{n(n+1)(2n+1)}{6}$$

Using the identity $\displaystyle\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ and letting $n \to \infty$, the fraction tends to $\dfrac{8}{6}\cdot 2 = \dfrac{8}{3}$, the same $\dfrac{8}{3}$ the antiderivative gave in Example 1. The limit of rectangles and the antiderivative agree, which is the whole reason integration measures area.

Example 2: Approximate the same area with four rectangles, then compare.

With $n = 4$ on $[0, 2]$, each strip has width $\Delta x = 0.5$, and the right-endpoint sample points are $x = 0.5, 1, 1.5, 2$:

$$S_4 = \big(0.5^2 + 1^2 + 1.5^2 + 2^2\big)(0.5) = (0.25 + 1 + 2.25 + 4)(0.5) = 3.7500$$

Final answer: four rectangles give $3.7500$, an overestimate of the exact $\dfrac{8}{3} \approx 2.6667$. The gap is expected, since $x^2$ is increasing and right-endpoint rectangles poke above the curve. More rectangles shrink the gap toward zero.

What Is Signed Area Versus Total Area?

A definite integral counts area above the x-axis as positive and area below it as negative. So the bare integral gives signed (net) area, the area above minus the area below. Often that is exactly what a problem wants, because "net" tracks direction. Other times a problem wants total area, the raw space between curve and axis with every piece counted positive.

The two split whenever the curve crosses the x-axis inside the interval. The rule is to find those crossing points, integrate over each piece, and combine.

• Signed (net) area: $\displaystyle\int_a^b f(x),dx$, one integral, signs kept.

• Total area: $\displaystyle\int_a^b \lvert f(x)\rvert,dx$, split at the zeros of $f$, each piece made positive before adding.

Example 3: Find both the net and the total area for $y = \sin x$ on $[0, 2\pi]$.

The curve is positive on $[0, \pi]$ and negative on $[\pi, 2\pi]$, crossing zero at $x = \pi$. An antiderivative of $\sin x$ is $F(x) = -\cos x$. Over the first arch:

$$\int_0^{\pi} \sin x,dx = \big[-\cos x\big]_0^{\pi} = (-\cos\pi) - (-\cos 0) = (1) - (-1) = 2$$

Over the second arch:

$$\int_{\pi}^{2\pi} \sin x,dx = \big[-\cos x\big]_{\pi}^{2\pi} = (-\cos 2\pi) - (-\cos\pi) = (-1) - (1) = -2$$

Adding the two gives the net area: $2 + (-2) = 0$. The trip above the axis exactly cancels the trip below. For the total area, make each piece positive first: $\lvert 2\rvert + \lvert -2\rvert = 4$.

Final answer: net (signed) area $= 0$; total area $= 4$ square units. A curve can enclose real space and still integrate to zero, which is the single idea behind the most common mistake on this topic.

How Do You Find The Area Between Two Curves?

The same tool measures the area trapped between two curves, not just between one curve and the axis. If $y = f(x)$ stays above $y = g(x)$ on $[a, b]$, the enclosed area is the integral of the gap between them:

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

You subtract the lower curve from the upper curve, then integrate. The axis version is just the special case $g(x) = 0$.

Example 4: Find the area between $y = x$ and $y = x^2$ from $x = 0$ to $x = 1$.

On $[0, 1]$ the line $y = x$ sits above the parabola $y = x^2$, so $f(x) = x$ is the top curve and $g(x) = x^2$ the bottom:

$$\int_0^1 \big(x - x^2\big),dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \left(\frac{1}{2} - \frac{1}{3}\right) - 0 = \frac{1}{6}$$

Final answer: the enclosed area is $\dfrac{1}{6} \approx 0.1667$ square units. For the full method, including curves that swap which one is on top, see area between two curves.

What Are The Standard Area-Under-A-Curve Setups?

Most first-year problems are one of a few shapes. This table names each and gives the setup, so you can match a question to its integral before computing.

Table: How to set up the area under a curve for the common cases.

Reading the situation before writing the integral is what separates a correct setup from an "answer" for the wrong quantity.

Why Does The Area Under A Curve Equal The Integral?

The equality feels like a coincidence the first time, because "area" sounds geometric while an "integral" looks algebraic. The bridge is the rectangle staircase, and the reason it works is geometric all the way down.

• A rectangle is area you already know. Each Riemann rectangle has area height times width, $f(x_i),\Delta x$. Summing them approximates the region with shapes whose area needs no calculus at all.

• Thinner strips mean less error. The only error is the sliver between each rectangle top and the curve. As the strips narrow, those slivers shrink faster than the strips multiply, so the total error tends to zero.

• The limit is the exact area. Because the approximation improves without bound, its limit is the true area, and that limit is precisely what the symbol $\int_a^b f(x),dx$ is defined to mean. The definition of the definite integral is the area, not a formula that happens to match it.

Seen this way, the definite integral is not a trick for finding area. It is area, written as the limit of a sum you can actually compute once you know an antiderivative.

Who Discovered The Area Under A Curve?

Measuring curved area is one of the oldest problems in mathematics, solved in special cases almost two thousand years before calculus gave a general method.

Two later figures turned the idea into a tool:

• Bonaventura Cavalieri (1598–1647, Italy) treated an area as built from infinitely many thin "indivisible" strips, an intuitive forerunner of the integral that let mathematicians compute areas the Greeks could not.

• Bernhard Riemann (1826–1866, Germany) gave the sum-of-rectangles idea its rigorous modern definition, so "the limit of the rectangles" finally rested on precise footing rather than intuition. The sums now carry his name.

Where Is The Area Under A Curve Used In The Real World?

The phrase "add up a changing quantity over an interval" describes a huge range of measurements, and each one is an area under a curve.

• Physics and motion: the area under a speed-time graph is the distance travelled, and the area under a force-distance graph is the work done by a variable force.

• Probability and statistics: the area under a probability density curve between two values is the probability of landing in that range, which is why the total area under any density is exactly one.

• Economics: the area between a demand curve and the market price is consumer surplus, a total benefit summed across every unit sold.

• Engineering: the volume of water delivered by a pipe is the area under its flow-rate curve over time, evaluated by a definite integral.

• Medicine and pharmacology: the "area under the curve" of a drug's concentration over time measures total exposure, and doctors dose by it directly.

One picture, the region beneath a graph, turns "how much accumulated" into a single number across physics, economics, and medicine. For more, see applications of integration.

What Are The Most Common Mistakes With The Area Under A Curve?

These four errors account for most lost marks on area problems, and each matches a question real students ask on r/calculus, Quora, and course help pages.

Reporting net area when the question wants total area.

Where it slips in:

A student integrates straight across an interval where the curve crosses the axis, gets a small number or zero, and writes it as the area.

Don't do this:

Do not trust one integral across a sign change. For $\int_0^{2\pi}\sin x,dx = 0$, the answer to "total area" is not zero.

The correct way:

Find the zeros of $f$, split the interval there, and add the absolute value of each piece: total area of $\sin x$ on $[0, 2\pi]$ is $\lvert 2\rvert + \lvert -2\rvert = 4$.

Never checking where the curve crosses the x-axis.

Where it slips in:

A student sets up a single integral for total area without testing whether $f$ changes sign inside the limits.

Don't do this:

Do not assume the curve stays on one side of the axis across the whole interval.

The correct way:

Solve $f(x) = 0$ first to locate every crossing in $[a, b]$, then integrate piece by piece. The crossings are the boundaries between positive and negative area.

Swapping or misreading the limits of integration.

Where it slips in:

A student writes the bounds in the wrong order, or reads $a$ and $b$ off the graph the wrong way round, flipping the sign of the whole answer.

Don't do this:

Do not reverse the limits for free. Swapping them negates the integral: $\int_a^b f = -\int_b^a f$.

The correct way:

Put the smaller x-value on the bottom and the larger on top for a left-to-right region, and double-check both against the picture before evaluating.

Treating the Riemann-sum approximation as the exact area.

Where it slips in:

A student computes a few rectangles and reports that finite sum as the area, or the reverse, doubts that the integral is exact because it "comes from rectangles."

Don't do this:

Do not confuse a finite sum with its limit. Four rectangles gave $3.7500$ for a region whose exact area is $\dfrac{8}{3} \approx 2.6667$.

The correct way:

Use a finite Riemann sum only as an estimate. The exact area is the limit as the number of rectangles goes to infinity, which the definite integral delivers in one evaluation.

Practice Problems On The Area Under A Curve

Work each one, then check against the answer. Every answer is verified.

1. Find the area under $y = x^2$ from $x = 0$ to $x = 3$.
   (Answer: $\left[\frac{x^3}{3}\right]_0^3 = 9$ square units.)

2. Find the area under $y = 2x$ from $x = 1$ to $x = 2$.
   (Answer: $\left[x^2\right]_1^2 = 4 - 1 = 3$ square units.)

3. Find the net (signed) area for $y = \cos x$ on $[0, 2\pi]$.
   (Answer: $\left[\sin x\right]_0^{2\pi} = 0$.)

4. Find the total area between $y = x^2 - 1$ and the x-axis on $[0, 2]$.
   (Answer: crosses at $x = 1$; $\left\lvert -\tfrac{2}{3}\right\rvert + \tfrac{4}{3} = 2$ square units.)

5. Find the area under $y = 4 - x^2$ from $x = -2$ to $x = 2$.
   (Answer: $\left[4x - \frac{x^3}{3}\right]_{-2}^{2} = \frac{32}{3} \approx 10.6667$ square units.)

6. Find the area between $y = x$ and $y = x^2$ on $[0, 1]$.
   (Answer: $\int_0^1 (x - x^2),dx = \frac{1}{6} \approx 0.1667$ square units.)

Where Should You Go Next After Area Under A Curve?

Area under a curve is the gateway to every geometric use of the integral, and several natural doors open from here.

1. Fundamental Theorem of Calculus. The result that lets you evaluate every area with $F(b) - F(a)$ instead of summing rectangles by hand.

2. Properties of definite integrals. The rules for splitting intervals and flipping limits that the signed-versus-total method leans on.

3. Area between two curves. The direct next step, replacing the x-axis with a second function.

4. Average value of a function and the broader applications of integration, where the same area idea gives averages, volumes, and work.

If your child is meeting the area under a curve for the first time, a live Bhanzu trainer teaches it from the rectangle picture up, so the integral feels like measured area rather than a memorised symbol, in the Bhanzu math program.