Mathematics 9709 · Pure Mathematics 1

Area under a curve

A definite integral gives the area between a curve and the $x$ -axis.
This is the most common way Paper 1 tests integration. So it is worth getting really good at it.

Area under a curve is $\displaystyle A=\int_a^b y\,dx$ when $y\ge 0$ . Below the axis the integral is negative. So take its size. If the region crosses the axis, split at the crossing and add the sizes.

A definite integral is an area

For a region above the $x$ -axis, between $x=a$ and $x=b$ , the area is just the definite integral of $y$ :

$A=\int_a^b y\,dx\qquad(\text{for }y\ge 0)$

The area from a to b is the integral of y. One thin gold strip has height y and width dx; the integral adds them all up.

Integrate the curve, then do top limit minus bottom limit as before. The number you get is the area.
The integral and the area are the same number. Drag the two limits below. The value of the shaded region matches $F(b)-F(a)$ .

limit alimit b

∫ from 1.0 to 4.0·area 8.25·F(b) − F(a)

Worked example

Find the area under $y=3x^2$ between $x=4$ and $x=6$

The region sits above the axis, so the area is just the integral:

$A=\int_4^6 3x^2\,dx=\big[x^3\big]_4^6$

$=6^3-4^3=216-64$

Answer

152\ \text{units}^2

Why integrating gives the area (the strip picture)

Cut the region into thin vertical strips. Each strip has height $y$ and a tiny width $dx$ . So it is almost a rectangle of area $y\,dx$ . (That is the one gold strip in the picture above.)

Add up very many of these strips as their width gets smaller and smaller. That total is what the long $\int$ sign means: a sum of $y\,dx$ from $a$ to $b$ . So integrating the height of the curve adds up the area under it.

When the curve dips below the axis

Below the axis $y$ is negative. So the integral is negative. This is called the signed area.
A real area is never negative. So take its size (drop the minus sign).
The trap is a region partly above and partly below the axis. If you integrate all the way across, the positive and negative parts cancel. You get a number that is too small (sometimes even zero).
The fix: split at the crossing point. Work out each piece on its own. Then add the sizes.

y = x(x − 2)(x − 4): the lobe on [0, 2] gives +4, the lobe on [2, 4] gives −4. Split at x = 2 and add the sizes; never let them cancel.

Worked example

Find the area between $y=x(x-5)$ and the $x$ -axis, from $x=0$ to $5$

This curve is below the axis between its roots. So expect a negative integral. Expand first, then integrate:

$\int_0^5\big(x^2-5x\big)\,dx=\left[\tfrac{x^3}{3}-\tfrac{5x^2}{2}\right]_0^5$

$=\left(\tfrac{125}{3}-\tfrac{125}{2}\right)-0=-\tfrac{125}{6}$

The integral is negative because the whole region is below the axis. The area is its size:

Answer

\tfrac{125}{6}\ \text{units}^2

Finding the limits: they are usually the roots

A question may say “the area bounded by the curve and the $x$ -axis” but not give you the limits. Then the limits are the curve’s roots: where it meets the axis. Find them first.
To find roots, solve $y=0$ . Let your calculator give them (the Quadratics tips work here too).
The roots become your $a$ and $b$ . Three roots also tell you where to split.

Bounded by lines parallel to the axes

The syllabus also asks for a region bounded by a curve and lines parallel to the axes. A vertical line like $x=4$ just gives a limit. A horizontal line like $y=5$ closes the region at the top.
For a horizontal top line, the area is the rectangle minus the area under the curve. That is $\int(\,y_{\text{line}}-y_{\text{curve}}\,)\,dx$ .

Worked example

Find the area bounded by the curve $y=x^2+1$ , the $y$ -axis, and the line $y=5$ .

The horizontal line $y=5$ meets the curve where $x^2+1=5$ , so $x=2$ (take the right-hand side). The region runs from $x=0$ (the $y$ -axis) to $x=2$ , closed at the top by $y=5$ . Integrate the gap, line minus curve:

$A=\int_0^2\big(5-(x^2+1)\big)\,dx=\int_0^2\big(4-x^2\big)\,dx=\left[4x-\tfrac{x^3}{3}\right]_0^2$

$=\Big(8-\tfrac{8}{3}\Big)-0=\tfrac{16}{3}$

Answer

\tfrac{16}{3}\ \text{units}^2

When the curve is a linear bracket, the two vertical lines just become your limits. Then you use the $(ax+b)^n$ rule.
This exact shape is a real Paper 1 question:

Worked example

Find the area bounded by $y=\dfrac{3}{\sqrt{2x+1}}$ , the lines $x=0$ , $x=7$ and $y=0$ .

The region sits above the axis between $x=0$ and $x=7$ , so the area is the integral. Rewrite the root as a power: $\dfrac{3}{\sqrt{2x+1}}=3(2x+1)^{-\frac12}$ .

Integrate with the linear-bracket rule: new power $\tfrac12$ , divide by it, then divide by $a=2$ :

$\int_0^7 3(2x+1)^{-\frac12}\,dx=\left[\dfrac{3(2x+1)^{\frac12}}{2\left(\frac12\right)}\right]_0^7=\Big[3\sqrt{2x+1}\,\Big]_0^7$

At $x=7$ the bracket is $15$ ; at $x=0$ it is $1$ . So top minus bottom: $3\sqrt{15}-3\sqrt{1}$ :

Answer

3\sqrt{15}-3\approx 8.62\ \text{units}^2

(9709/11 Oct/Nov 2024 Q7)

Where the marks go

Common mistake

A region below the axis gives a negative integral. Take its size for the area. If the region is partly above and partly below, split at the crossing points and add the sizes. Never let a positive and a negative part cancel.

Common mistake

For an area “bounded by the curve and the

x

-axis” the limits are the roots. You usually have to find them first by setting

y=0

. Do not guess them from the diagram.

Now you try

Find the total area bounded by $y=x(x-2)(x-4)$ and the $x$ -axis. (9709-style)

Your turn— tap to reveal the worked answer (split at x = 2)

Roots at $0,\ 2,\ 4$ . So integrate over $[0,2]$ and $[2,4]$ on their own. Expand: $x^3-6x^2+8x$ . The piece from $0$ to $2$ is above the axis. The piece from $2$ to $4$ is below.

$\int_0^2(x^3-6x^2+8x)\,dx=4$ and $\int_2^4(x^3-6x^2+8x)\,dx=-4$ .

Add the sizes: $4+4$ .

Answer

8\ \text{units}^2

PreviousDefinite integrals NextArea between curves & volumes of revolution

Area under a curve

A definite integral gives the area between a curve and the $x$ -axis.
This is the most common way Paper 1 tests integration. So it is worth getting really good at it.

A definite integral is an area

For a region above the $x$ -axis, between $x=a$ and $x=b$ , the area is just the definite integral of $y$ :

$A=\int_a^b y\,dx\qquad(\text{for }y\ge 0)$

The area from a to b is the integral of y. One thin gold strip has height y and width dx; the integral adds them all up.

Integrate the curve, then do top limit minus bottom limit as before. The number you get is the area.
The integral and the area are the same number. Drag the two limits below. The value of the shaded region matches $F(b)-F(a)$ .

limit alimit b

∫ from 1.0 to 4.0·area 8.25·F(b) − F(a)

Worked example

Find the area under $y=3x^2$ between $x=4$ and $x=6$

The region sits above the axis, so the area is just the integral:

$A=\int_4^6 3x^2\,dx=\big[x^3\big]_4^6$

$=6^3-4^3=216-64$

Answer

152\ \text{units}^2

Why integrating gives the area (the strip picture)

Cut the region into thin vertical strips. Each strip has height $y$ and a tiny width $dx$ . So it is almost a rectangle of area $y\,dx$ . (That is the one gold strip in the picture above.)

When the curve dips below the axis

Below the axis $y$ is negative. So the integral is negative. This is called the signed area.
A real area is never negative. So take its size (drop the minus sign).
The trap is a region partly above and partly below the axis. If you integrate all the way across, the positive and negative parts cancel. You get a number that is too small (sometimes even zero).
The fix: split at the crossing point. Work out each piece on its own. Then add the sizes.

y = x(x − 2)(x − 4): the lobe on [0, 2] gives +4, the lobe on [2, 4] gives −4. Split at x = 2 and add the sizes; never let them cancel.

Worked example

Find the area between $y=x(x-5)$ and the $x$ -axis, from $x=0$ to $5$

This curve is below the axis between its roots. So expect a negative integral. Expand first, then integrate:

$\int_0^5\big(x^2-5x\big)\,dx=\left[\tfrac{x^3}{3}-\tfrac{5x^2}{2}\right]_0^5$

$=\left(\tfrac{125}{3}-\tfrac{125}{2}\right)-0=-\tfrac{125}{6}$

The integral is negative because the whole region is below the axis. The area is its size:

Answer

\tfrac{125}{6}\ \text{units}^2

Finding the limits: they are usually the roots

A question may say “the area bounded by the curve and the $x$ -axis” but not give you the limits. Then the limits are the curve’s roots: where it meets the axis. Find them first.
To find roots, solve $y=0$ . Let your calculator give them (the Quadratics tips work here too).
The roots become your $a$ and $b$ . Three roots also tell you where to split.

Bounded by lines parallel to the axes

The syllabus also asks for a region bounded by a curve and lines parallel to the axes. A vertical line like $x=4$ just gives a limit. A horizontal line like $y=5$ closes the region at the top.
For a horizontal top line, the area is the rectangle minus the area under the curve. That is $\int(\,y_{\text{line}}-y_{\text{curve}}\,)\,dx$ .

Worked example

Find the area bounded by the curve $y=x^2+1$ , the $y$ -axis, and the line $y=5$ .

$A=\int_0^2\big(5-(x^2+1)\big)\,dx=\int_0^2\big(4-x^2\big)\,dx=\left[4x-\tfrac{x^3}{3}\right]_0^2$

$=\Big(8-\tfrac{8}{3}\Big)-0=\tfrac{16}{3}$

Answer

\tfrac{16}{3}\ \text{units}^2

When the curve is a linear bracket, the two vertical lines just become your limits. Then you use the $(ax+b)^n$ rule.
This exact shape is a real Paper 1 question:

Worked example

Find the area bounded by $y=\dfrac{3}{\sqrt{2x+1}}$ , the lines $x=0$ , $x=7$ and $y=0$ .

The region sits above the axis between $x=0$ and $x=7$ , so the area is the integral. Rewrite the root as a power: $\dfrac{3}{\sqrt{2x+1}}=3(2x+1)^{-\frac12}$ .

Integrate with the linear-bracket rule: new power $\tfrac12$ , divide by it, then divide by $a=2$ :

$\int_0^7 3(2x+1)^{-\frac12}\,dx=\left[\dfrac{3(2x+1)^{\frac12}}{2\left(\frac12\right)}\right]_0^7=\Big[3\sqrt{2x+1}\,\Big]_0^7$

At $x=7$ the bracket is $15$ ; at $x=0$ it is $1$ . So top minus bottom: $3\sqrt{15}-3\sqrt{1}$ :

Answer

3\sqrt{15}-3\approx 8.62\ \text{units}^2

(9709/11 Oct/Nov 2024 Q7)

Where the marks go

Common mistake

For an area “bounded by the curve and the

x

-axis” the limits are the roots. You usually have to find them first by setting

y=0

. Do not guess them from the diagram.

Now you try

Find the total area bounded by $y=x(x-2)(x-4)$ and the $x$ -axis. (9709-style)

Your turn— tap to reveal the worked answer (split at x = 2)

Roots at $0,\ 2,\ 4$ . So integrate over $[0,2]$ and $[2,4]$ on their own. Expand: $x^3-6x^2+8x$ . The piece from $0$ to $2$ is above the axis. The piece from $2$ to $4$ is below.

$\int_0^2(x^3-6x^2+8x)\,dx=4$ and $\int_2^4(x^3-6x^2+8x)\,dx=-4$ .

Add the sizes: $4+4$ .

Answer

8\ \text{units}^2