Build the integral for a region between curves, integrals from easy to advanced lesson

The line $y = 2x$ and the parabola $y = x^2$ meet at the origin and once more to the right, and between those points they enclose a lens-shaped region. No formula hands us its area, so the integral has to be built from the picture. This page finds the intersection points that give the limits, shows that the line stays on top, turns one thin strip of height top minus bottom into an integral and evaluates it to $\tfrac43$ . It also shows how to choose the direction of the strips and what changes when two curves cross inside the interval.

Key terms

Intersection points

Definition The curves $y = f(x)$ and $y = g(x)$ meet at the points $(c, f(c))$ with $f(c) = g(c)$ , so the $x$ -coordinates of their intersection points are the solutions of the equation $f(x) - g(x) = 0$ . When the two curves alone enclose a region, the region starts and ends at intersection points, and their $x$ -coordinates are the limits of integration for vertical strips. To find every solution, move all terms to one side and factor. Dividing both sides by an expression that can be zero loses the solutions where it vanishes.

In plain words Two curves meet where their formulas give the same height. Set the formulas equal, collect everything on one side, factor and keep every root.

Example For the line and the parabola, setting the heights equal gives

\begin{aligned} x^2 &= 2x, \\[0.4em] x^2 - 2x &= 0, \\[0.4em] x(x - 2) &= 0, \end{aligned}

so $x = 0$ or $x = 2$ , and the curves meet at $(0, 0)$ and $(2, 4)$ . Dividing $x^2 = 2x$ by $x$ would leave only $x = 2$ and lose the origin, which is the left end of the region.

Top minus bottom

Definition Suppose that for every $x$ in $[a, b]$ the vertical line through $x$ meets a region in one segment, from a lower boundary $y = g(x)$ up to an upper boundary $y = f(x)$ . The height of the strip at $x$ is top minus bottom, $f(x) - g(x)$ , and a strip of width $\Delta x$ has area approximately $\bigl(f(x) - g(x)\bigr)\,\Delta x$ . The order of the two curves has to be checked on the whole interval, so show that $f(x) - g(x) \ge 0$ for every $x$ in $[a, b]$ , for example by factoring. A single test point settles the order when $f - g$ is continuous and has no zero strictly between $a$ and $b$ , because by the intermediate value theorem it then keeps one sign there.

In plain words A strip reaches from the lower curve up to the upper curve, so its height is the upper $y$ minus the lower $y$ . Check the order across the whole interval and not only at one convenient point, because curves that cross can swap places.

Example On $[0, 2]$ the difference of the line and the parabola factors as

2x - x^2 = x(2 - x),

and both factors are nonnegative when $0 \le x \le 2$ . So the line is the upper boundary and the parabola the lower one across the whole region. The strip at $x = \tfrac12$ runs from $y = \tfrac14$ up to $y = 1$ , and its height $1 - \tfrac14 = \tfrac34$ is the value of $2x - x^2$ at $\tfrac12$ .

Area between two curves

Definition Let $f$ and $g$ be continuous on $[a, b]$ with $g(x) \le f(x)$ for every $x$ in $[a, b]$ . The region of points $(x, y)$ with $a \le x \le b$ and $g(x) \le y \le f(x)$ has area

A = \int_a^b \bigl(f(x) - g(x)\bigr)\,dx.

The integral is the limit of the sums of strip areas $\bigl(f(x_i^{*}) - g(x_i^{*})\bigr)\,\Delta x_i$ as the widest strip shrinks to width $0$ .

In plain words Cut the region into thin vertical strips, measure each one as top minus bottom times its width, and let the integral add them. The subtraction follows from the region. When both curves lie above the horizontal axis, integrating $f$ alone would also count the area under the lower curve, which is outside the region, and subtracting the integral of $g$ removes exactly that part.

Example For the region enclosed by $y = 2x$ and $y = x^2$ , the line is on top on $[0, 2]$ , so

\begin{aligned} A &= \int_0^2 (2x - x^2)\,dx \\[0.4em] &= \left[x^2 - \frac{x^3}{3}\right]_0^2 \\[0.4em] &= 4 - \frac83 = \frac43. \end{aligned}

The area under the line over $[0, 2]$ is a triangle of area $\int_0^2 2x\,dx = 4$ , the area under the parabola is $\int_0^2 x^2\,dx = \tfrac83$ , and their difference is the same $\tfrac43$ . A shape check confirms the size. Since $2x - x^2 = 1 - (x - 1)^2$ , the strip height is $0$ at both ends and reaches its largest value $1$ at $x = 1$ . Sliding every strip straight down onto the horizontal axis gives the region under the graph of $2x - x^2$ , which has the same area. That region fits inside a rectangle of width $2$ and height $1$ , so $A < 2$ . It contains the triangle with vertices $(0, 0)$ , $(1, 1)$ and $(2, 0)$ , which has area $1$ , because the graph lies above both slanted sides, so $A > 1$ .

Vertical and horizontal strips

Definition Vertical strips have width $\Delta x$ and give an integral in $x$ . They need each vertical line to meet the region in one segment whose bottom and top each follow one formula. Horizontal strips have thickness $\Delta y$ , run from a left boundary $x = v(y)$ to a right boundary $x = u(y)$ , and give an integral in $y$ . If $u$ and $v$ are continuous on $[c, d]$ with $v(y) \le u(y)$ for every $y$ in $[c, d]$ , the region of points with $c \le y \le d$ and $v(y) \le x \le u(y)$ has area

A = \int_c^d \bigl(u(y) - v(y)\bigr)\,dy.

When an end of the strips changes formula partway across, the integral splits at that point. The other direction sometimes avoids the split, and both directions give the same area.

In plain words Before calculating, slide a strip through the region. Choose the direction in which both ends of the strip keep one manageable formula, and write those formulas in the variable of integration.

Example The same lens can be cut into horizontal strips for $0 \le y \le 4$ . Solved for $x$ , the line is $x = y/2$ and the right half of the parabola is $x = \sqrt{y}$ . Their difference $\sqrt{y} - \tfrac{y}{2} = \tfrac12\sqrt{y}\,(2 - \sqrt{y})$ is nonnegative on $[0, 4]$ , so the parabola is on the right and

\begin{aligned} A &= \int_0^4 \left(\sqrt{y} - \frac{y}{2}\right)\,dy \\[0.4em] &= \left[\frac23 y^{3/2} - \frac{y^2}{4}\right]_0^4 \\[0.4em] &= \frac{16}{3} - 4 = \frac43. \end{aligned}

Both directions give one integral here, and the vertical one is simpler because it needs no square roots.

Absolute difference

Definition If $f$ and $g$ are continuous on $[a, b]$ , the total area between their graphs over $[a, b]$ is

A = \int_a^b \lvert f(x) - g(x) \rvert\,dx.

When $f - g$ has finitely many zeros in $[a, b]$ , list them together with $a$ and $b$ in increasing order. Between neighboring points of the list the difference keeps one sign, by the intermediate value theorem, so integrate whichever of $f - g$ and $g - f$ is nonnegative on each piece and add the results. The signed integral of $f - g$ equals this area exactly when $f(x) \ge g(x)$ on all of $[a, b]$ .

In plain words Area is never negative, so every strip must be counted with a nonnegative height. Where the curves swap order, split the interval and subtract in the right order on each piece. Reversing the order on the lens gives $\int_0^2 (x^2 - 2x)\,dx = -\tfrac43$ , which is a signed difference and not an area.

Example Extend the line and the parabola to $[0, 3]$ . They cross at $x = 2$ , with the line on top before the crossing and the parabola on top after it, since $x^2 - 2x = x(x - 2) > 0$ for $2 < x \le 3$ . So

\begin{aligned} A &= \int_0^2 (2x - x^2)\,dx \\[0.4em] &\quad + \int_2^3 (x^2 - 2x)\,dx \\[0.4em] &= \frac43 + \frac43 = \frac83. \end{aligned}

The signed integral $\int_0^3 (2x - x^2)\,dx = 9 - 9 = 0$ misses all of this area, because the two pieces cancel.

Check yourself

Find the area enclosed by the parabolas $y = x^2$ and $y = 2 - x^2$ .

Show the answer

The curves meet where $x^2 = 2 - x^2$ , so $x^2 = 1$ and $x = -1$ or $x = 1$ . The difference $(2 - x^2) - x^2 = 2(1 - x^2)$ is nonnegative on $[-1, 1]$ , so $y = 2 - x^2$ is on top and

\begin{aligned} A &= \int_{-1}^1 (2 - 2x^2)\,dx \\[0.4em] &= \left[2x - \frac{2x^3}{3}\right]_{-1}^1 \\[0.4em] &= \frac43 + \frac43 = \frac83. \end{aligned}

Find the area enclosed by the parabola $x = 4 - y^2$ and the line $x = y + 2$ . Which strip direction needs only one integral?

Show the answer

Horizontal strips need only one. Setting $4 - y^2 = y + 2$ gives $y^2 + y - 2 = 0$ , so $(y + 2)(y - 1) = 0$ and the curves meet at $(0, -2)$ and $(3, 1)$ . For $-2 \le y \le 1$ the difference $(4 - y^2) - (y + 2) = (1 - y)(y + 2)$ is nonnegative, so the parabola is on the right and

\begin{aligned} A &= \int_{-2}^1 (2 - y - y^2)\,dy \\[0.4em] &= \left[2y - \frac{y^2}{2} - \frac{y^3}{3}\right]_{-2}^1 \\[0.4em] &= \frac76 + \frac{10}{3} = \frac92. \end{aligned}

Vertical strips need two integrals. For $0 \le x \le 3$ a vertical strip runs from the lower half of the parabola, $y = -\sqrt{4 - x}$ , up to the line $y = x - 2$ , but for $3 \le x \le 4$ it runs from the lower half of the parabola to the upper half. The two pieces give $\tfrac{19}{6} + \tfrac43 = \tfrac92$ , the same area.

A classmate wants the area between $y = 4x$ and $y = x^3$ for $0 \le x \le 3$ . The classmate checks that the line is higher at $x = 1$ and then computes

\left\lvert \int_0^3 (4x - x^3)\,dx \right\rvert = \frac94.

Find the correct area and explain what went wrong.

Show the answer

The difference $4x - x^3 = x(2 - x)(2 + x)$ is positive on $(0, 2)$ and negative on $(2, 3)$ , so the curves cross at $x = 2$ and the cubic is on top after the crossing. One test point cannot detect that. Splitting at the crossing gives

\begin{aligned} A &= \int_0^2 (4x - x^3)\,dx \\[0.4em] &\quad + \int_2^3 (x^3 - 4x)\,dx \\[0.4em] &= 4 + \frac{25}{4} = \frac{41}{4}. \end{aligned}

The signed integral is $4 - \tfrac{25}{4} = -\tfrac94$ , because the piece where the cubic is higher counts negatively and cancels part of the first piece. Taking the absolute value after integrating cannot undo that cancellation. The absolute value belongs inside the integral, on the height of every strip.