Area between curves

A region trapped between two curves is not measured by a single integral of one function — it needs the difference of two. When curve $f$ rides above curve $g$ across an interval, the sliver of area squeezed between them is what is left after you take the area under $f$ and cut away the area under $g$ . This single idea — subtract, then integrate — extends the familiar area-under-a-curve method to enclosed regions bounded by a parabola and a line, two intersecting curves, or a curve and a horizontal boundary. The only new skills are finding where the curves meet (to fix the limits) and checking, with a test point, which curve is genuinely on top before you subtract — get either wrong and the "area" comes out negative or the wrong shape entirely.

The core insight

The area enclosed between two curves $y=f(x)$ and $y=g(x)$ over $[a,b]$ , with $f(x)\ge g(x)$ throughout, is $A=\int_a^b[f(x)-g(x)]\,dx$ — upper curve minus lower curve, integrated between their points of intersection.

The core idea

Area-between-curves is area-under-a-curve applied twice. The area under $f$ from $a$ to $b$ minus the area under $g$ over the same interval leaves exactly the sandwiched strip between them, because the parts of each area lying below the other curve's boundary cancel algebraically in the subtraction. If $f(x)\ge g(x)$ for every $x$ in $[a,b]$ — curve $f$ lies above curve $g$ throughout — the enclosed area is:

A = \int_a^b \big[f(x)-g(x)\big]\,dx

where: $f(x)$ is the upper curve, $g(x)$ is the lower curve, and $a,b$ are the $x$ -values bounding the region.

Finding the limits automatically. When no vertical lines are given, the region is bounded by the curves' own intersection points: solve $f(x)=g(x)$ for $x$ , and the two roots become $a$ and $b$ .

When the curves cross inside the interval. If "upper" and "lower" swap partway through — the curves cross each other somewhere inside $[a,b]$ — split the interval at every crossing point. On each piece, subtract in the order (upper $-$ lower) for that piece, then add the resulting positive areas together, exactly as with a sign change in a single-curve area problem.

Why this generalises so cleanly. Nothing about the reasoning depends on $f$ and $g$ being any particular kind of function — a parabola against a line, two trigonometric curves, or a curve against a constant horizontal line $y=k$ (treat $k$ as the second "curve," $g(x)=k$ ) all fall under the same formula. The only real work in any problem is algebraic: locate the intersection points correctly, and settle which function is larger on each resulting sub-interval. Once both of those are pinned down, the computation is ordinary definite-integral evaluation via the fundamental theorem of calculus, identical in spirit to a single-curve area problem — properties such as even/odd symmetry of the integrand about the origin can shortcut the arithmetic exactly as they do there.

Figure

Shaded region between the line y equals x plus 2 above and the parabola y equals x squared below, from x equals negative 1 to x equals 2, with the two intersection points marked

Derivation

Derivation AArea between two curves from the fundamental theorem of calculus

The goal is to show precisely why $A=\int_a^b[f(x)-g(x)]\,dx$ gives the true enclosed area rather than some approximation, by building it from the single-curve area formula that the fundamental theorem of calculus already justifies, so that subtracting two such areas is legitimate term by term.

Step 1. Recall the single-curve result. For a curve $y=h(x)$ that stays non-negative on $[a,b]$ , the area between the curve and the $x$ -axis is $\int_a^b h(x)\,dx$ , a consequence of the fundamental theorem of calculus applied to thin vertical strips of width $dx$ and height $h(x)$ .

Step 2. Apply this twice, to $f$ and to $g$ separately, treating both as curves above the $x$ -axis for a moment (shift the whole picture up if either dips below the axis — this does not change the enclosed area between them). The area under $f$ from $a$ to $b$ is $\int_a^b f(x)\,dx$ , and the area under $g$ over the same interval is $\int_a^b g(x)\,dx$ .

Step 3. Subtract the two areas: $\int_a^b f(x)\,dx - \int_a^b g(x)\,dx$ . Because both integrals share the same limits, linearity of the definite integral lets us combine them into a single integral of the difference: $\int_a^b f(x)\,dx - \int_a^b g(x)\,dx = \int_a^b[f(x)-g(x)]\,dx$ .

Step 4. Interpret the cancellation. Picture a thin vertical strip at some $x$ inside $[a,b]$ : its height, from the lower curve up to the upper curve, is exactly $f(x)-g(x)$ , and its width is $dx$ . Summing these strips over $[a,b]$ is precisely what the integral $\int_a^b[f(x)-g(x)]\,dx$ computes, so the formula is not a shortcut but the direct strip-by-strip area of the enclosed region, valid as long as $f(x)\ge g(x)$ everywhere on $[a,b]$ (if not, split at the crossing points first).

Worked example

Find the area of the region enclosed between the parabola $y=x^2$ and the line $y=x+2$ .

Step 1. Find intersection points: set $x^2 = x+2 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0 \Rightarrow x=-1,\,2$ . These fix the limits $a=-1$ , $b=2$ .

Step 2. Identify which curve is on top: test $x=0$ , inside $[-1,2]$ . The line gives $y=2$ ; the parabola gives $y=0$ . So the line $y=x+2$ is above the parabola $y=x^2$ throughout the interval.

Step 3. Integrate the difference:

A = \int_{-1}^{2} \left[(x+2) - x^2\right] dx = \left[-\frac{x^3}{3}+\frac{x^2}{2}+2x\right]_{-1}^{2}

At $x=2$ : $-\dfrac{8}{3}+2+4 = \dfrac{10}{3}$ . At $x=-1$ : $\dfrac{1}{3}+\dfrac{1}{2}-2 = -\dfrac{7}{6}$ .

Step 4. Combine: $A = \dfrac{10}{3}-\left(-\dfrac{7}{6}\right) = \dfrac{20}{6}+\dfrac{7}{6} = \dfrac{27}{6} = \boxed{\dfrac{9}{2}\ \text{sq. units}}$

Competitive stretch

The same region can be checked by symmetry logic when curves are simpler. For $y=x^2$ and the line $y=4$ : intersections at $x=\pm2$ , line on top throughout, and since $4-x^2$ is even, $A=2\int_0^2(4-x^2)\,dx = 2\left(8-\dfrac{8}{3}\right)=\dfrac{32}{3}$ sq. units — the even/odd symmetry property of definite integrals shortcuts the work whenever the integrand and interval are both symmetric about the origin.

Common traps

Assuming one curve is always on top without checking — always test a point inside the interval; curves can swap order partway through.
Not splitting at interior intersection points — integrating across a crossing without splitting lets the two regions wrongly cancel instead of add.
Reading limits wrong — check whether $a,b$ are given directly as vertical lines or must be found by solving $f(x)=g(x)$ .
Subtracting lower minus upper instead of upper minus lower — this flips the sign; the correct order is always (top curve $-$ bottom curve) on each piece.

Ask the tutor: Not sure which curve is on top over a given interval, or where a region needs splitting because the curves cross — ask the AI tutor to walk through the test-point check with your specific functions.

How do I know which curve is upper without sketching the whole graph

Pick any convenient test value of x strictly inside the interval and evaluate both functions there. Whichever gives the larger y-value is the upper curve on that piece of the interval. This is faster and safer than trying to visualize the full graph, especially under exam time pressure, and it still works even if you cannot picture the shapes at all.

What if the intersection equation gives more than two roots

Then the curves cross more than twice, and the enclosed region is made of several separate pieces. Order all the roots along the x-axis, treat each consecutive pair as its own sub-interval, determine which curve is on top on that piece with a test point, integrate the difference there, and add all the positive sub-areas together. The method is identical to the two-root case, just repeated more times.

Do I need to know which region the question means if it does not say enclosed by intersections, and can the answer come out negative

Read carefully whether the boundaries are the curves' own crossing points or externally given vertical lines like x equals 1 and x equals 3; if vertical lines are stated, use those directly even if the curves also intersect elsewhere outside that range. Separately, a genuine enclosed area is always positive, so a negative result signals an error, almost always the subtraction order — if you subtracted lower minus upper instead of upper minus lower, the integral evaluates to the exact negative of the correct area, and flipping the order at your test point fixes it without redoing the integration.