Area between a curve and a line

MATArea between Curves

This subtopic computes the area of the region trapped between a curve (typically a parabola or modulus graph) and a straight line, a special and very common case of the area-between-two-curves method. The line and curve are intersected to fix the limits, then the area is $\int (\text{upper}-\text{lower})\,dx$ or $\int (\text{right}-\text{left})\,dy$ .

ISC also examines finding the area of a triangle by integration, where each side is treated as a line.

Area between curve and line

A = \int_{a}^{b} \left[\, y_{\text{line}} - y_{\text{curve}} \,\right]\,dx

valid where the line lies above the curve on

[a,b]

;

a,b

are the x-coordinates of their intersection (reverse the bracket where the curve is on top).

Parabola and line (worked form)

A = \int_{0}^{4}\left( 2\sqrt{x} - x \right)dx = \frac{8}{3}

region bounded by

y^2=4x

(so

y=2\sqrt{x}

) and the line

y=x

, meeting at

(0,0)

and

(4,4)

Modulus curve and horizontal line

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

region between

y=|x|

and

y=4

; symmetry about the y-axis lets you compute the right half (

y=x

) and double.

Triangle area by integration

A = \int_{x_1}^{x_2} \left(\text{upper edge} - \text{lower edge}\right)\,dx

the triangle's three sides are written as line equations

y=mx+c

; integrate the top boundary minus the bottom boundary, splitting at the middle vertex's x-coordinate.

Find intersections of the curve and line first; these give the limits of integration.
Test an interior point to determine whether the line or the curve is the upper boundary.
For a parabola $y^2=4ax$ , take the relevant branch $y=2\sqrt{ax}$ (upper) and integrate against the line.
Horizontal strips $\int(x_{\text{right}}-x_{\text{left}})\,dy$ are often cleaner for $x^2=4y$ with a horizontal line $y=k$ .
For $y=|x|$ split at $x=0$ into $y=x$ ( $x\ge0$ ) and $y=-x$ ( $x<0$ ), or use symmetry and double one side.
For a triangle with vertices $A,B,C$ , write each side as a line, split the x-range at the middle vertex, and sum the strip areas; cross-check with $\dfrac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$ .
Standard results: $y^2=4x$ with $y=x$ gives $\dfrac{8}{3}$ ; $y=x^2$ with $y=x$ gives $\dfrac{1}{6}$ .
Keep the area positive by always subtracting the lower boundary from the upper.

Where the marks go

Reversing upper and lower boundaries, yielding a negative result for what must be a positive area.
Using only the positive branch of $y^2=4x$ when the enclosed region actually spans both branches, or vice versa.
Forgetting to split the triangle's x-interval at the middle vertex, so the wrong pair of sides bounds part of the strip.
Treating $y=|x|$ as the single line $y=x$ over all $x$ , halving the true area instead of accounting for both arms.

How the board asks it

Numericalparabola and line
Using integration, find the area of the region bounded by the parabola $y^2 = 4x$ and the line $y = x$ .
Diagram / graphintersections fix the limits
Draw a rough sketch of the region enclosed by the curve $x^2 = 4y$ and the line $y = x$ , and find its area by integration.
Numericalmodulus curve and horizontal line
Find, by integration, the area of the region bounded by the curve $y = |x|$ and the line $y = 4$ .
Numericalvertical strips, upper minus lower
Using integration, find the area of the region $\{(x, y) : x^2 \le y \le x + 2\}$ .
Numericaleach side treated as a line
Using integration, find the area of the triangle whose vertices are $A(2, 0)$ , $B(4, 5)$ and $C(6, 3)$ .

Practise this topic

PreviousArea under a curve (between curve and an axis)NextArea between two curves

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.