Area between two curves

MATArea between Curves

This subtopic finds the area of a region enclosed between two curves by integrating the difference of their equations. The core idea is to take strips between the upper and lower (or right and left) boundaries: $\int_a^b \left(y_{\text{top}}-y_{\text{bottom}}\right)\,dx$ , where the limits are the x-coordinates of the points of intersection.

It is often dressed as a real-life modelling problem (ponds, flower beds, embankments) to test finding intersections and identifying which curve is on top.

Area between two curves (vertical strips)

A = \int_{a}^{b} \left[\, f(x) - g(x) \,\right]\,dx

f(x)\ge g(x)

[a,b]

;

f

is the upper curve,

g

the lower, and

a,b

are the x-coordinates of their intersection points.

Area between two curves (horizontal strips)

A = \int_{c}^{d} \left[\, p(y) - q(y) \,\right]\,dy

p(y)\ge q(y)

[c,d]

;

p

is the right curve,

q

the left, with

c,d

the y-coordinates of intersection.

Finding intersection limits

f(x) = g(x) \;\Rightarrow\; x = a,\; x = b

the limits of integration are obtained by solving the two curve equations simultaneously.

Area between two parabolas (worked form)

A = \int_{0}^{4}\left( 2\sqrt{x} - \frac{x^2}{4} \right)dx = \frac{16}{3}

region enclosed by

y^2=4x

(upper,

y=2\sqrt{x}

) and

x^2=4y

(lower,

y=x^2/4

), intersecting at

x=0

and

x=4

Step 1: find intersection points by solving the equations simultaneously; these give the limits.
Step 2: on the interval, decide which curve is upper (larger $y$ ) by testing an interior point.
Always integrate (top $-$ bottom); the difference is non-negative on the interval so the area comes out positive.
If the curves cross within the region, split into subintervals where the ordering of top/bottom is constant.
Use horizontal strips $\int(x_{\text{right}}-x_{\text{left}})\,dy$ when curves are more naturally written as $x=g(y)$ .
Exploit symmetry for symmetric regions (e.g. $|x|+|y|=1$ encloses a square of area $2$ , computable as $4$ times the first-quadrant triangle of area $\dfrac{1}{2}$ ).
Standard results: the region between $y^2=4x$ and $x^2=4y$ has area $\dfrac{16}{3}$ ; between $y=x^2$ and $y^2=x$ it is $\dfrac{1}{3}$ .
For modelling problems where the curve and the x-axis are the two boundaries, the 'lower curve' is simply $y=0$ .

Where the marks go

Writing (bottom $-$ top) by mistake, producing a negative answer, then ignoring the sign issue.
Skipping the simultaneous solution and guessing the limits, so the wrong interval is integrated.
Failing to split the integral when the two curves swap top/bottom positions inside the region.
Mishandling $|x|+|y|=1$ by integrating only one linear piece instead of using symmetry over all four edges.

How the board asks it

Numericalarea between a parabola and a line using vertical strips
Using integration, find the area of the region bounded by the parabola $y^2=4x$ and the line $y=2x$ .
Numericalstandard result: region between $y^2=4x$ and $x^2=4y$ has area $\dfrac{16}{3}$
Using integration, find the area enclosed between the two parabolas $y^2=4x$ and $x^2=4y$ .
Diagram / graphsketch the region, then decide which curve is upper by testing an interior point
Draw a rough sketch of the region bounded by the circle $x^2+y^2=4$ and the line $x+y=2$ in the first quadrant, and hence find its area by integration.
Numericalsplit into subintervals where the curves cross so top/bottom ordering stays constant
Find the area of the region bounded by the curves $y=x$ and $y=x^3$ between $x=-1$ and $x=1$ .
Applicationreal-life modelling between two curves; intersections give the limits
A flower bed is laid out in the region bounded by the parabola $y=x^2$ and the line $y=x+2$ . Using integration, find the area of the flower bed in square units.

Practise this topic

PreviousArea between a curve and a line NextArea bounded by standard curves (circle, parabola, ellipse)

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.