Definite integrals and areas

A definite integral gives the signed area between a curve and the $x$-axis. It is positive where the curve is above the axis and negative where the curve is below it.

If $F'(x)=f(x)$, the fundamental theorem of calculus gives

$$\int_a^b f(x)\,dx=\left[F(x)\right]_a^b=F(b)-F(a).$$

This lesson develops the distinction between an integral and a geometric area, then applies it to regions under curves and between curves.

Prerequisites

You should be able to:

• find antiderivatives using the power rule;
• solve linear and quadratic equations;
• sketch basic graphs and identify where one graph lies above another;
• work exactly with fractions, surds, exponentials and logarithms.

Review integration basics or function graphs if these skills are uncertain.

Evaluating a definite integral

The numbers $a$ and $b$ are the limits of integration. The lower limit is $a$ and the upper limit is $b$.

To evaluate $\int_a^b f(x)\,dx$:

1. Find an antiderivative $F(x)$.
2. Substitute the upper limit to obtain $F(b)$.
3. Substitute the lower limit to obtain $F(a)$.
4. Calculate upper value minus lower value.

Worked example 1: a polynomial

Evaluate

$$\int_1^3 (2x^2-4x+5)\,dx.$$

Integrate term by term:

$$\int (2x^2-4x+5)\,dx=\frac23x^3-2x^2+5x.$$

Now apply the limits:

$$\begin{aligned} \int_1^3 (2x^2-4x+5)\,dx &=\left[\frac23x^3-2x^2+5x\right]_1^3\\ &=\left(18-18+15\right)-\left(\frac23-2+5\right)\\ &=15-\frac{11}{3}\\ &=\boxed{\frac{34}{3}}. \end{aligned}$$

There is no $+C$ in the final calculation. Any constant would occur in both $F(b)$ and $F(a)$, so it would cancel.

Worked example 2: an exact logarithmic answer

Evaluate

$$\int_1^4 \frac{3}{x}\,dx.$$

Since $\int \frac1x\,dx=\ln|x|+C$,

$$\begin{aligned} \int_1^4 \frac3x\,dx &=\left[3\ln x\right]_1^4\\ &=3\ln4-3\ln1\\ &=\boxed{3\ln4}. \end{aligned}$$

Here $x$ is positive throughout $[1,4]$, so writing $\ln x$ is sufficient. Unless a decimal is requested, $3\ln4$ is the appropriate exact answer.

Essential properties

Definite integrals obey three useful rules:

$$\int_a^a f(x)\,dx=0,$$ $$\int_b^a f(x)\,dx=-\int_a^b f(x)\,dx,$$

and, for any $c$ between $a$ and $b$,

$$\int_a^b f(x)\,dx=\int_a^c f(x)\,dx+\int_c^b f(x)\,dx.$$

The second rule explains why reversing the limits changes the sign. The third allows an interval to be split wherever the geometry changes.

Self-check 1

Given that $\int_2^7 f(x)\,dx=9$ and $\int_2^5 f(x)\,dx=4$, find:

1. $\int_7^2 f(x)\,dx$;
2. $\int_5^7 f(x)\,dx$.

Answer

Reversing the limits gives

$$\int_7^2 f(x)\,dx=-9.$$

Splitting at $x=5$ gives

$$9=4+\int_5^7 f(x)\,dx,$$

so $\int_5^7 f(x)\,dx=5$.

Signed area is not always geometric area

For a continuous function $f$:

• where $f(x)>0$, the integral contributes positively;
• where $f(x)<0$, the integral contributes negatively;
• where $f(x)=0$, the graph meets or touches the axis.

Therefore

$$\int_a^b f(x)\,dx =\text{area above the axis}-\text{area below the axis}.$$

By contrast, geometric area is never negative. The total area between $y=f(x)$ and the $x$-axis is

$$\int_a^b |f(x)|\,dx.$$

At A-level, this is usually calculated by finding every root in the interval, splitting the integral there, and changing the sign of any negative contribution.

Worked example 3: integral versus total area

The curve is

$$y=x^2-x-2=(x-2)(x+1).$$

Find both the value of $\int_{-2}^{3}y\,dx$ and the total area between the curve and the $x$-axis over this interval.

First locate the crossings:

$$x^2-x-2=0 \quad\Longrightarrow\quad x=-1\text{ or }x=2.$$

An antiderivative is

$$F(x)=\frac{x^3}{3}-\frac{x^2}{2}-2x.$$

The signed integral is

$$\begin{aligned} \int_{-2}^{3}(x^2-x-2)\,dx &=F(3)-F(-2)\\ &=-\frac32-\left(-\frac23\right)\\ &=\boxed{-\frac56}. \end{aligned}$$

This negative answer does not mean the geometric area is negative. It means the area below the axis exceeds the area above it by $\frac56$ square units.

For total area, split at both roots. The curve is above the axis on $[-2,-1]$, below it on $[-1,2]$, and above it on $[2,3]$:

$$\begin{aligned} A &=\int_{-2}^{-1}y\,dx-\int_{-1}^{2}y\,dx+\int_2^3y\,dx\\ &=\frac{11}{6}-\left(-\frac92\right)+\frac{11}{6}\\ &=\boxed{\frac{49}{6}}. \end{aligned}$$

Notice that the roots were found before integrating. A sketch or sign test then determined which pieces needed a sign change.

Area between a curve and the x-axis

If $f(x)\geq 0$ throughout $[a,b]$, then

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

The limits must match the horizontal boundaries of the region. Sometimes they are given. Sometimes they must be found from intersections.

Worked example 4: finding a boundary

The curve $y=6x-x^2$ and the positive $x$-axis enclose a finite region. Find its area.

Find the axis intersections:

$$6x-x^2=x(6-x)=0,$$

so $x=0$ or $x=6$. The downward opening parabola lies above the axis between these roots. Therefore

$$\begin{aligned} A &=\int_0^6(6x-x^2)\,dx\\ &=\left[3x^2-\frac{x^3}{3}\right]_0^6\\ &=(108-72)-0\\ &=\boxed{36\text{ square units}}. \end{aligned}$$

The equation $6x-x^2=0$ was needed to obtain the limits. Integrating first would not identify the enclosed region.

Worked example 5: a curve entirely below the axis

Find the area between $y=x^2-4x$ and the $x$-axis for $1\leq x\leq3$.

Since

$$x^2-4x=x(x-4),$$

the curve is below the axis for $0<x<4$. Thus its integral from $1$ to $3$ is negative, and the area is its negative:

$$\begin{aligned} A &=-\int_1^3(x^2-4x)\,dx\\ &=-\left[\frac{x^3}{3}-2x^2\right]_1^3\\ &=-\left(-9+\frac53\right)\\ &=\boxed{\frac{22}{3}\text{ square units}}. \end{aligned}$$

Equivalently, integrate $4x-x^2$, the vertical distance from the curve up to the axis.

Area between two curves

Suppose $y=f(x)$ is above $y=g(x)$ throughout $[a,b]$. Their vertical separation is $f(x)-g(x)$, so the enclosed area is

$$A=\int_a^b\bigl(f(x)-g(x)\bigr)\,dx.$$

The reliable rule is

$$\boxed{\text{area}=\int(\text{upper function}-\text{lower function})\,dx.}$$

Do not decide which is upper from the order in which equations are printed. Use a sketch or test a value.

Worked example 6: a line and a parabola

Find the area enclosed by

$$y=x^2 \qquad\text{and}\qquad y=2x+3.$$

First find the intersections:

$$x^2=2x+3$$ $$x^2-2x-3=0$$ $$(x-3)(x+1)=0.$$

Hence the boundaries are $x=-1$ and $x=3$.

At $x=0$, the line has value $3$ and the parabola has value $0$, so the line is above the parabola between the intersections. Therefore

$$\begin{aligned} A &=\int_{-1}^{3}\left[(2x+3)-x^2\right]dx\\ &=\left[x^2+3x-\frac{x^3}{3}\right]_{-1}^{3}\\ &=9-\left(-\frac53\right)\\ &=\boxed{\frac{32}{3}\text{ square units}}. \end{aligned}$$

A quick check is that the vertical separation is $4$ at $x=1$ and the interval has width $4$. An area of $\frac{32}{3}\approx10.7$ is plausible.

Worked example 7: the upper curve changes

Find the total area between

$$y=x \qquad\text{and}\qquad y=x^3$$

for $-1\leq x\leq1$.

The intersections satisfy

$$x=x^3 \quad\Longrightarrow\quad x(x-1)(x+1)=0,$$

so they occur at $x=-1,0,1$.

On $(-1,0)$, $x^3>x$. On $(0,1)$, $x>x^3$. The upper curve changes at $x=0$, so split the area:

$$\begin{aligned} A &=\int_{-1}^{0}(x^3-x)\,dx+\int_0^1(x-x^3)\,dx\\ &=\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^{0} +\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_0^1\\ &=\frac14+\frac14\\ &=\boxed{\frac12\text{ square unit}}. \end{aligned}$$

The symmetry gives a shorter method:

$$A=2\int_0^1(x-x^3)\,dx=\frac12.$$

Symmetry may shorten a correct setup, but it does not replace deciding which curve is upper.

Areas involving the y-axis

If a region is bounded by the $y$-axis, that boundary is $x=0$. When the curves are given as $y$ in terms of $x$, integration with respect to $x$ is usually still the direct method.

Worked example 8: a y-axis boundary

Find the area enclosed by $y=4-x^2$, $y=2x+1$, and the $y$-axis in the first quadrant.

The nonzero boundary comes from the intersection:

$$4-x^2=2x+1$$ $$x^2+2x-3=0$$ $$(x+3)(x-1)=0.$$

The first quadrant intersection is at $x=1$. At $x=0$, the parabola has value $4$ and the line has value $1$. Hence

$$\begin{aligned} A &=\int_0^1\left[(4-x^2)-(2x+1)\right]dx\\ &=\int_0^1(3-2x-x^2)\,dx\\ &=\left[3x-x^2-\frac{x^3}{3}\right]_0^1\\ &=\boxed{\frac53\text{ square units}}. \end{aligned}$$

Parameters and unknown limits

A definite integral may be given as an equation. Evaluate it in terms of the unknown, then solve while respecting any information about the region.

Worked example 9: finding an unknown boundary

The area under $y=6x-x^2$ from $x=0$ to $x=k$ is $\frac{28}{3}$ square units, where $0<k<3$. Find $k$.

Form the area equation:

$$\int_0^k(6x-x^2)\,dx=\frac{28}{3}.$$

Therefore

$$3k^2-\frac{k^3}{3}=\frac{28}{3}.$$

Multiplying by $3$ and rearranging gives

$$k^3-9k^2+28=0.$$

Since $k=2$ is a root, factorise:

$$(k-2)(k^2-7k-14)=0.$$

The quadratic factor gives

$$k=\frac{7\pm\sqrt{105}}{2}.$$

Of the three algebraic solutions, only $k=2$ satisfies $0<k<3$. Hence

$$\boxed{k=2}.$$

The interval condition is essential. Solving the integral equation can produce values that do not describe the stated region.

Units and interpretation

If both axes represent lengths, an area integral has square units. If the variables represent different quantities, multiply their units.

For example, if $v(t)$ is measured in metres per second and $t$ in seconds, then

$$\int_a^b v(t)\,dt$$

has units of metres and represents displacement, not a geometric area in square metres. If velocity changes sign, total distance requires the same splitting idea used for total area:

$$\text{distance}=\int_a^b |v(t)|\,dt.$$

Choosing the correct setup

Before integrating, ask:

1. What are the horizontal boundaries?
2. Where do the relevant graphs intersect?
3. Does any curve cross the $x$-axis inside the interval?
4. Which function is upper on each part of the interval?
5. Does the question ask for a signed integral or a geometric area?

This short analysis prevents most errors in area questions.

Common misconceptions

• A definite integral is always an area. It is a signed accumulation. Contributions below the $x$-axis are negative.
• Area can be negative. Geometric area cannot. Split at crossings and use positive vertical distances.
• The larger formula is the upper curve. Which curve is upper depends on $x$. Test a point in each interval.
• The limits can be read from the $y$-coordinates. When integrating with respect to $x$, the limits are $x$-coordinates.
• Intersection points are found after integration. They define the region, so find them first.
• Upper minus lower means upper limit minus lower limit. These are separate ideas. Upper function minus lower function forms the integrand; upper limit minus lower limit is not a valid evaluation rule. Evaluate $F(b)-F(a)$.
• Every antiderivative needs $+C$. An indefinite integral does. In a definite integral, the constant cancels.
• A decimal is as good as an exact answer. Keep fractions, surds, $\pi$, exponentials and logarithms exact unless approximation is requested.

Mixed self-check

Question 1

Evaluate

$$\int_0^2(3x^2+2x-1)\,dx.$$

Answer

$$\begin{aligned} \int_0^2(3x^2+2x-1)\,dx &=\left[x^3+x^2-x\right]_0^2\\ &=8+4-2\\ &=\boxed{10}. \end{aligned}$$

Question 2

Find the total area between $y=x-2$ and the $x$-axis for $0\leq x\leq5$.

Answer

The curve crosses the axis at $x=2$. Therefore

$$\begin{aligned} A &=-\int_0^2(x-2)\,dx+\int_2^5(x-2)\,dx\\ &=2+\frac92\\ &=\boxed{\frac{13}{2}\text{ square units}}. \end{aligned}$$

The signed integral would instead be $\frac52$.

Question 3

Find the area enclosed by $y=x^2+1$ and $y=3x+1$.

Answer

The intersections satisfy

$$x^2+1=3x+1 \quad\Longrightarrow\quad x(x-3)=0,$$

so $x=0$ and $x=3$. At $x=1$, the line is above the parabola. Thus

$$\begin{aligned} A &=\int_0^3\left[(3x+1)-(x^2+1)\right]dx\\ &=\left[\frac32x^2-\frac{x^3}{3}\right]_0^3\\ &=\boxed{\frac92\text{ square units}}. \end{aligned}$$

Question 4

Without evaluating an antiderivative, state the value of

$$\int_{-4}^{4}x^3\,dx.$$

Answer

$x^3$ is an odd function, and the interval is symmetric about zero. The positive and negative signed areas cancel, so

$$\boxed{\int_{-4}^{4}x^3\,dx=0}.$$

The total geometric area is not zero.

Summary

For an antiderivative $F$ of $f$,

$$\int_a^b f(x)\,dx=F(b)-F(a).$$

Remember the three distinct setups:

$$\begin{aligned} \text{signed area} &:\quad \int_a^b f(x)\,dx,\\ \text{total area against the axis} &:\quad \int_a^b|f(x)|\,dx,\\ \text{area between curves} &:\quad \int_a^b(\text{upper}-\text{lower})\,dx. \end{aligned}$$

Find intersections and sign changes before integrating. Split the interval whenever a curve crosses the axis or the upper curve changes.

Next steps

Study integration as the limit of a sum to understand why a definite integral represents accumulated area. Use the trapezium rule when an antiderivative is unavailable or values are given in a table. Then extend your techniques with integration by substitution and integration by parts.