When one boundary of a region is a straight line, its contribution to the area is exact geometry — a trapezium — so you never need to integrate it at all. Spotting this shortcut turns a two-integral problem into one.
10 min read3 worked examples
The straight-line shortcut
The area between a curve $y=f(x)$ and a straight line $y=mx+c$, between their two points of intersection $x=a$ and $x=b$, can always be found the standard way: integrate the difference $f(x)-(mx+c)$ (or its negative, whichever is positive) between $a$ and $b$. But the straight line's own contribution to that integral never needs an integration step at all, because a straight line is exactly the shape a trapezium formula already describes:
Here $y(a)$ and $y(b)$ are the line's (signed) heights at the two boundary values — this is exactly the "average of the parallel sides times the distance between them" formula for the area of a trapezium, applied to the two vertical segments of height $y(a)$ and $y(b)$ standing on the $x$-axis. It holds even when a $y$-value is negative, in which case that side of the trapezium is measured below the axis rather than above it.
Why this isn't an approximation
This might look similar to the Trapezium Rule met earlier in this topic — but it is not the same idea. The (numerical) Trapezium Rule approximates the area under a curved function by pretending short sections of it are straight. Here, the boundary genuinely is a straight line, so the trapezium formula gives the exact area under it, not an estimate. No approximation, and no integration of $mx+c$, is needed at all.
Strategy — using the shortcut
(1) Find the intersection points of the curve and the line to get the limits $a$ and $b$. (2) Test a value of $x$ between $a$ and $b$ to see which of the curve or the line is on top. (3) Integrate the curve normally between $a$ and $b$. (4) Find the line's exact contribution using the trapezium formula — no integration needed. (5) Subtract (top minus bottom) to get the enclosed area.
Worked example 1
Area between a curve and a line · curve on top
Find the area of the region enclosed between the curve $y=8-x^2$ and the line $y=x+2$.
MethodIntegrate the curve, find the line's area by the trapezium formula, then subtract
1
Intersections: $8-x^2=x+2 \Rightarrow x^2+x-6=0 \Rightarrow (x+3)(x-2)=0$, so $x=-3$ or $x=2$.
Set the two expressions equal to find the limits of integration.
2
At $x=0$: curve $=8$, line $=2$ — the curve lies above the line throughout $-3<x<2$.
A single test value decides which function is "on top".
Line's area by the trapezium formula: $y(-3)=-1$, $y(2)=4$, so $\dfrac12(-1+4)(2-(-3)) = \dfrac12(3)(5) = \dfrac{15}{2}$.
No integration needed for the straight-line boundary.
5
Area $=\dfrac{85}{3}-\dfrac{15}{2} = \dfrac{170-45}{6} = \dfrac{125}{6}$.
Curve's integral minus the line's trapezium area.
▪ solved
Cross-check against the full-integral method
Integrating the difference directly gives the same answer: $\int_{-3}^{2}\big[(8-x^2)-(x+2)\big]dx = \int_{-3}^{2}(6-x-x^2)\,dx = \Big[6x-\dfrac{x^2}{2}-\dfrac{x^3}{3}\Big]_{-3}^{2} = \dfrac{22}{3}-(-13.5) = \dfrac{125}{6}$ — identical to the shortcut, but this route needed the line integrated as well as the curve.
Worked example 2
Area between a curve and a line · line on top, signed areas
Find the area of the region enclosed between the curve $y=x^2-1$ and the line $y=x+1$.
MethodFind the line's area by the trapezium formula, integrate the curve, then subtract
1
Intersections: $x^2-1=x+1 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0$, so $x=-1$ or $x=2$.
Set the two expressions equal.
2
At $x=0$: curve $=-1$, line $=1$ — the line lies above the curve throughout $-1<x<2$.
Test value shows the line is on top this time.
3
Line's area by the trapezium formula: $y(-1)=0$, $y(2)=3$, so $\dfrac12(0+3)(2-(-1)) = \dfrac12(3)(3) = \dfrac92$.
Why the curve's integral being zero doesn't mean zero area
The curve dips below the $x$-axis for part of $-1<x<2$ and rises above it for the rest, and these signed contributions happen to cancel — $\int_{-1}^{2}(x^2-1)\,dx=0$ does not mean the curve coincides with the $x$-axis, only that the signed area under it nets to zero. The enclosed region between curve and line is unaffected by this coincidence: subtracting the curve's (net-zero) signed integral from the line's trapezium area still gives the correct enclosed area, $\frac92$.
Worked example 3
Area between a curve and a chord · finding the line first
The points $A(1,1)$ and $B(3,9)$ both lie on the curve $y=x^2$. Find the area of the region enclosed between the curve and the chord $AB$.
MethodFind the equation of the chord, then apply the shortcut
1
Gradient of $AB$: $\dfrac{9-1}{3-1}=4$. Equation: $y-1=4(x-1) \Rightarrow y=4x-3$.
The chord is just the straight line through the two given points.
2
At $x=2$ (between 1 and 3): curve $=4$, chord $=5$ — the chord lies above the curve throughout $1<x<3$.
A chord to a convex curve always lies above it between the two points.
3
Chord's area by the trapezium formula: $y(1)=1$, $y(3)=9$, so $\dfrac12(1+9)(3-1) = \dfrac12(10)(2) = 10$.
Area $=10-\dfrac{26}{3} = \dfrac{30-26}{3} = \dfrac43$.
Chord's trapezium area minus the curve's integral.
▪ solved
Examiner's eye
Nothing here required finding intersection points by solving an equation — $A$ and $B$ were given directly, and their $x$-coordinates became the limits of integration immediately. This is a common exam variant: the "curve meets line" setup is replaced by "chord joining two named points", but the shortcut applies exactly the same way once the chord's equation is found.
Common mistake
Subtracting in the wrong order — always check which function is on top with a test value first, since "curve minus line" and "line minus curve" differ only in sign but a negative "area" signals the subtraction was done backwards. Also, don't confuse this exact trapezium formula with the approximate numerical Trapezium Rule from earlier in this topic: here the boundary really is straight, so there is no approximation error to worry about.
Check yourself
Find the area of the region enclosed between the curve $y=x^2+1$ and the line $y=3x-1$. (Hint: find the intersection points, then use the trapezium formula for the line's area and integrate the curve directly.)
Method
1
Intersections: $x^2+1=3x-1 \Rightarrow x^2-3x+2=0 \Rightarrow (x-1)(x-2)=0$, so $x=1,2$.
Set the expressions equal.
2
At $x=1.5$: curve $=3.25$, line $=3.5$ — the line lies above the curve.
Test value between the limits.
3
Line's area: $y(1)=2$, $y(2)=5$, so $\dfrac12(2+5)(2-1)=\dfrac72$.