Area under a curve (between curve and an axis)

MATArea under a Curve

This subtopic uses the definite integral to compute the area of a region bounded by a single curve, one coordinate axis and a pair of bounding lines (ordinates or abscissae). The core idea is that $\int_a^b y\,dx$ measures the signed area of vertical strips between the curve $y=f(x)$ and the x-axis, while $\int_c^d x\,dy$ measures horizontal strips against the y-axis.

It is the foundational skill on which every other area problem in the chapter is built.

Area against the x-axis

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

y=f(x)

lies above the x-axis on

[a,b]

, with

a<b

;

A

is the area between the curve and the x-axis from

x=a

x=b

Area against the y-axis

A = \int_{c}^{d} x\,dy = \int_{c}^{d} g(y)\,dy

x=g(y)

is the curve expressed as a function of

y

, lying to the right of the y-axis on

[c,d]

, with

c<d

Area when the curve dips below the x-axis

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

used when

f(x)\le 0

[a,b]

, so the raw integral is negative; the magnitude gives the physical area.

Splitting at a sign change

A = \left| \int_{a}^{c} f(x)\,dx \right| + \left| \int_{c}^{b} f(x)\,dx \right|

x=c

is the point in

(a,b)

where

f(x)

changes sign (crosses the x-axis); each piece is integrated separately and the magnitudes are added.

Decide the strip direction first: use $\int y\,dx$ for regions bounded by the x-axis and ordinates, $\int x\,dy$ for regions bounded by the y-axis and horizontal lines.
Always sketch the curve and shade the region; the sketch tells you the limits and whether the area lies above or below the axis.
Area is intrinsically positive: if the region is below the x-axis the integral is negative, so report $|A|$ .
If the curve crosses the axis inside $[a,b]$ , split the integral at the crossing point and add the absolute values of the parts; do NOT integrate straight through.
For $\int x\,dy$ rewrite the curve as $x=g(y)$ (e.g. $y^2=x \Rightarrow x=y^2$ ) and use the y-limits.
The bounding lines $x=a,\,x=b$ (or $y=c,\,y=d$ ) supply the limits of integration directly.
Standard results worth recalling: $\int_0^\pi \sin x\,dx = 2$ and $\int_0^2 x^2\,dx = \dfrac{8}{3}$ .
Units of area follow from the context (e.g. square units); leave a pure number if no units are given.

Where the marks go

Computing $\int_0^{2\pi}\sin x\,dx = 0$ and reporting zero area, instead of splitting at $x=\pi$ to get total area $4$ .
Forgetting to take the absolute value when the region lies below the x-axis (e.g. $y=\cos x$ on $\left[\dfrac{\pi}{2},\pi\right]$ ), giving a negative 'area'.
Using $\int y\,dx$ when the region is bounded by the y-axis and lines $y=c,\,y=d$ , where $\int x\,dy$ is required.
Swapping or mislabelling the limits so that $a>b$ , which flips the sign of the answer.

How the board asks it

Numericalvertical strips against the x-axis using $\int_a^b y\,dx$
Find the area of the region bounded by the curve $y = x^2 + 2$ , the x-axis and the ordinates $x = 1$ and $x = 3$ .
Diagram / graphthe sketch fixes the limits and tells you if the region is above or below the axis
Draw a rough sketch of the curve $y = \sin x$ for $0 \le x \le 2\pi$ and hence find the total area enclosed between the curve and the x-axis.
Numericalsplitting at the crossing point and adding the absolute values of the parts
Find the area bounded by the curve $y = x^3$ , the x-axis and the lines $x = -2$ and $x = 1$ , taking into account the portion lying below the x-axis.
Numericalhorizontal strips against the y-axis using $\int_c^d x\,dy$ with $x = g(y)$
Find the area of the region bounded by the parabola $y^2 = 4x$ , the y-axis and the lines $y = 1$ and $y = 3$ .
Give reasonsa below-axis integral is negative while area is intrinsically positive
Explain, with the help of a sketch, why $\int_0^{2\pi} \sin x\,dx = 0$ does not give the area enclosed between $y = \sin x$ and the x-axis on $[0, 2\pi]$ , and state the correct area.

Practise this topic

NextArea between a curve and a line

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.