1.08PureStretch

Area Under a Curve

A definite integral is a $\textit{signed}$ area: regions above the $x$ -axis count positively, regions below count negatively. Ask for "the area" of a region that crosses the axis and simply integrate across it, and the two parts cancel — sometimes to zero. This lesson turns definite integrals into reliable area calculations, including areas between two curves.

45 min Integration

The big picture

Finding an area always starts with a sketch, because the sketch tells you where the region is, where it crosses the $x$ -axis, and where curves intersect — and those points become the limits. From there the rules are few. Above the axis, the area is the integral. Below the axis, the area is minus the integral. Between two curves, integrate top minus bottom, which works regardless of the axis. When a curve cannot be integrated exactly, the trapezium rule estimates the area numerically, and the curve’s convexity tells you whether that estimate is too big or too small. The same ideas extend to areas with respect to the $y$ -axis and, via parametric integration, to curves like ellipses.

What you'll be able to do

Find the area between a curve and the $x$ -axis, including regions below the axis
Find the area between two curves, or a curve and a line
Use the trapezium rule and state whether it over- or under-estimates
Find areas with respect to the $y$ -axis

Areas above and below the axis

For a curve above the $x$ -axis between $x = a$ and $x = b$ , the area is $\int_{a}^{b} y\,\mathrm{d}x$ .

If the curve is below the axis, the integral is negative, and the area is its absolute value.

If the region crosses the axis, find where $y = 0$ , integrate each part separately, and add the absolute values.

Example 1Find the total area between

y = x(x - 2)

and the

x

-axis from

x = 0

x = 3

1Roots at

x = 0

and

x = 2

: the curve is below the axis on

(0, 2)

and above on

(2, 3)

\int_{0}^{2} (x^{2} - 2x)\,\mathrm{d}x = \big[\tfrac13x^{3} - x^{2}\big]_{0}^{2} = \tfrac83 - 4 = -\tfrac43

\int_{2}^{3} (x^{2} - 2x)\,\mathrm{d}x = (9 - 9) - \left(\tfrac83 - 4\right) = \tfrac43

4Total area

= \tfrac43 + \tfrac43

Answer

\tfrac83

(whereas

\int_{0}^{3}

gives

0

)

Why this works

The integral from 0 to 3 is exactly zero here: the negative area below the axis cancels the positive area above it. Without a sketch, you would report an area of zero for a region you can plainly see.

Area between two curves

If $y = \mathrm{f}(x)$ lies above $y = \mathrm{g}(x)$ between $x = a$ and $x = b$ , the area between them is $\int_{a}^{b} \big(\mathrm{f}(x) - \mathrm{g}(x)\big)\,\mathrm{d}x$ .

This works even if part of the region is below the $x$ -axis — the subtraction takes care of it.

The limits are usually the $x$ -coordinates of the intersection points, found by solving $\mathrm{f}(x) = \mathrm{g}(x)$ .

\text{Area} = \int_{a}^{b} (y_{\text{top}} - y_{\text{bottom}})\,\mathrm{d}x

Area between two curves.

Example 1Find the area enclosed between

y = 4x - x^{2}

and

y = x

1Intersections:

4x - x^{2} = x

, so

x(3 - x) = 0

, giving

x = 0

and

x = 3

2Between them the parabola is on top (at

x = 1

3 > 1

\int_{0}^{3} (3x - x^{2})\,\mathrm{d}x = \big[\tfrac32x^{2} - \tfrac13x^{3}\big]_{0}^{3} = \tfrac{27}{2} - 9

Answer

\tfrac92

Tip — If you are unsure which curve is on top, substitute a value between the limits. Getting it backwards gives the right size with a negative sign — a clue to check.

The trapezium rule

When a function cannot be integrated exactly, split the interval into $n$ strips of equal width $h = \dfrac{b - a}{n}$ and replace the curve on each strip by a straight chord, forming trapezia.

With ordinates $y_{0}, y_{1}, \dots, y_{n}$ , the total is $\tfrac12 h\big(y_{0} + y_{n} + 2(y_{1} + \dots + y_{n-1})\big)$ .

If the curve is $\textbf{convex}$ (bending upwards), the chords lie above it, so the rule $\textbf{overestimates}$ . If $\textbf{concave}$ , it underestimates. More strips give a better estimate.

\int_{a}^{b} y\,\mathrm{d}x \approx \tfrac12 h\left[y_{0} + y_{n} + 2(y_{1} + y_{2} + \dots + y_{n-1})\right]

Trapezium rule,

h = \tfrac{b-a}{n}

Example 1Use the trapezium rule with 4 strips to estimate

\int_{0}^{2} \sqrt{1 + x^{2}}\,\mathrm{d}x

, and state whether it is an over- or under-estimate.

h = 0.5

. Ordinates at

x = 0, 0.5, 1, 1.5, 2

1, 1.1180, 1.4142, 1.8028, 2.2361

2Estimate:

0.25\big(1 + 2.2361 + 2(1.1180 + 1.4142 + 1.8028)\big) = 0.25 \times 11.9061

3The curve

y = \sqrt{1 + x^{2}}

is convex, so the chords lie above it.

Answer

2.977

(3 d.p.), an overestimate.

Areas with respect to the y-axis

The area between a curve and the $y$ -axis, from $y = c$ to $y = d$ , is $\int_{c}^{d} x\,\mathrm{d}y$ . Rearrange the curve to give $x$ in terms of $y$ first.

Sometimes it is easier to find an area by subtracting from a rectangle, or by integrating in $y$ when the region is naturally bounded horizontally.

Example 1Find the area between

y = x^{2}

(

x \ge 0

), the

y

-axis and the lines

y = 1

and

y = 4

x = \sqrt{y} = y^{1/2}

\int_{1}^{4} y^{1/2}\,\mathrm{d}y = \big[\tfrac23 y^{3/2}\big]_{1}^{4} = \tfrac23(8 - 1)

Answer

\tfrac{14}{3}

Think like an examiner

Sketch the region and mark intersections and axis crossings before integrating.
Treat parts below the axis separately and give area as positive.
Integrate top minus bottom for areas between curves.
For the trapezium rule, show the table of ordinates and justify over- or under-estimation by convexity.

Common misconceptions

Integrating straight across an axis crossing.

Split at the root and add absolute values.

Giving a negative area.

Area is positive; take the modulus of a negative integral.

Using

n

ordinates for

n

strips.

n

strips need

n + 1

ordinates.

Saying the trapezium rule always overestimates.

Only for convex curves; it underestimates for concave ones.

Areas

A = \int_{a}^{b} y\,\mathrm{d}x

Above the

x

-axis.

A = \int_{a}^{b} (y_{1} - y_{2})\,\mathrm{d}x

Between curves.

A = \int_{c}^{d} x\,\mathrm{d}y

With the

y

-axis.

\tfrac12 h[y_{0} + y_{n} + 2(y_{1} + \dots + y_{n-1})]

Trapezium rule.

Stretch yourself

The curve $y = x^{3} - 4x$ and the line $y = 5x$ enclose two regions. Find the total area of the two regions.

Hint — Solve $x^{3} - 4x = 5x$ to find three intersections. Work out which graph is on top in each region — the answer is not the same in both.

Reveal solution

SolutionIntersections:

x^{3} - 9x = 0

, so

x(x - 3)(x + 3) = 0

, giving

x = -3, 0, 3

. On

(-3, 0)

, test

x = -1

: curve

3

, line

-5

, so the curve is on top. On

(0, 3)

, test

x = 1

: curve

-3

, line

5

, so the line is on top. First region:

\int_{-3}^{0} (x^{3} - 9x)\,\mathrm{d}x = \big[\tfrac14x^{4} - \tfrac92x^{2}\big]_{-3}^{0} = 0 - \left(\tfrac{81}{4} - \tfrac{81}{2}\right) = \tfrac{81}{4}

. Second region:

\int_{0}^{3} (9x - x^{3})\,\mathrm{d}x = \big[\tfrac92x^{2} - \tfrac14x^{4}\big]_{0}^{3} = \tfrac{81}{2} - \tfrac{81}{4} = \tfrac{81}{4}

. Total area

= \tfrac{81}{2}

. (The two regions are equal by the rotational symmetry of both graphs about the origin.)

\blacksquare

Questions students ask

Why does an integral sometimes give zero for a region with obvious area?

Parts below the axis contribute negatively and can cancel parts above. Split at the roots.

Do I need to split when finding the area between two curves?

Only if the curves cross inside the interval, so that which is on top changes. The

x

-axis itself does not matter.

How do I know if the trapezium rule over- or under-estimates?

Look at the shape: convex curves give overestimates, concave curves underestimates. The second derivative can confirm the shape.

How can I improve a trapezium rule estimate?

Use more strips. Halving the strip width typically reduces the error by about a factor of four.

Key takeaways

Always sketch: find roots and intersections for limits.
Area below the axis is the modulus of a negative integral; split regions that cross the axis.
Between curves: integrate top minus bottom.
The trapezium rule estimates areas; convex gives an overestimate, concave an underestimate.
Areas with the $y$ -axis use $\int x\,\mathrm{d}y$ .