Change the order, reveal the missing factor, integrals from easy to advanced lesson

The iterated integral

I = \int_0^1 \int_x^1 e^{y^2}\,dy\,dx

is the volume of a solid over a triangle, under a roof whose height $e^{y^2}$ stays between $1$ and $e$ , so it is a perfectly finite number. Yet its inner integral stalls at once, because $e^{y^2}$ has no elementary antiderivative. This lesson reads the same triangle with horizontal strips instead of vertical ones, explains why Fubini’s theorem allows the switch, and shows how the new order supplies the factor $y$ that a substitution needs, which gives the exact value $I = \tfrac{e - 1}{2}$ .

Key terms

Triangular region

Definition A triangular region is a closed triangle in the plane, made of three noncollinear vertices, the three sides joining them and the points inside. A triangle is convex, so every vertical line that meets it does so in a single segment or point, and so does every horizontal line. It can therefore be described in either order, by $a \le x \le b$ and $g_1(x) \le y \le g_2(x)$ , or by $c \le y \le d$ and $h_1(y) \le x \le h_2(y)$ . The boundary functions are continuous, and each one follows a single side or two sides that meet at a vertex.

In plain words A triangle is the simplest region that is not a rectangle. The range of one variable depends on the other, and which sides bound a strip depends on the direction of the strip.

Example The inequalities $0 \le x \le y \le 1$ describe the triangle $T$ with vertices $(0, 0)$ , $(0, 1)$ and $(1, 1)$ , the half of the unit square on or above the diagonal $y = x$ . A vertical strip at a given $x$ runs from the diagonal $y = x$ up to the top side $y = 1$ . A horizontal strip at a given $y$ runs from the left side $x = 0$ to the diagonal $x = y$ . A test point settles which half of the square is meant. The point $(\tfrac14, \tfrac12)$ satisfies $0 \le \tfrac14 \le \tfrac12 \le 1$ and lies in $T$ , while $(\tfrac12, \tfrac14)$ has $y < x$ and does not. Both readings give the area $\tfrac12$ ,

\begin{aligned} \int_0^1 (1 - x)\,dx &= \frac12, \\[0.4em] \int_0^1 y\,dy &= \frac12. \end{aligned}

Bounds read from a sketch

Definition To find the limits of an iterated integral over a region $D$ in the other order, work from a picture of $D$ rather than from the old limits alone.

Write $D$ as inequalities. The outer limits are constants, and the inner limits are functions of the outer variable.
Sketch the boundary curves, shade $D$ and confirm the shading with a test point.
Draw a typical strip in the new inner direction. The curves where it enters and leaves $D$ , solved for the new inner variable, give the new inner limits.
Take the smallest and largest values of the new outer variable on $D$ as the new outer limits.
If the entering or leaving curve changes partway along, split the outer range at that value and add the pieces.

In plain words The limits are a description of the region, so a new order needs a new description of the same region. The picture is what guarantees that the region has not changed.

Example Take $\int_0^1 \int_y^{2 - y} f(x, y)\,dx\,dy$ for a continuous $f$ . The limits say $0 \le y \le 1$ and $y \le x \le 2 - y$ , so the region is the triangle with vertices $(0, 0)$ , $(2, 0)$ and $(1, 1)$ . A vertical strip always starts on the bottom side $y = 0$ , but it ends on the line $y = x$ when $x \le 1$ and on the line $y = 2 - x$ when $x \ge 1$ . So the order $dy\,dx$ splits at $x = 1$ into

\begin{aligned} &\int_0^1 \int_0^x f\,dy\,dx \\[0.4em] &\quad + \int_1^2 \int_0^{2 - x} f\,dy\,dx. \end{aligned}

With $f = 1$ the single integral is $\int_0^1 (2 - 2y)\,dy = 1$ and the two pieces give $\tfrac12 + \tfrac12 = 1$ , the area of the triangle. One region can need one integral in one order and two in the other.

Fubini’s theorem

Definition Let $D$ be the set of points with $a \le x \le b$ and $g_1(x) \le y \le g_2(x)$ , where $g_1$ and $g_2$ are continuous on $[a, b]$ and $g_1 \le g_2$ , and let $f$ be continuous on $D$ . Then

\begin{aligned} &\iint_D f\,dA \\[0.4em] &= \int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y)\,dy\,dx. \end{aligned}

The same holds with the roles of $x$ and $y$ exchanged, for a region given by $c \le y \le d$ and $h_1(y) \le x \le h_2(y)$ with continuous $h_1 \le h_2$ , in the order $dx\,dy$ . So when $f$ is continuous on a region that can be described both ways, the two iterated integrals are equal. Here the double integral over $D$ means the double integral, over a rectangle containing $D$ , of the function equal to $f$ on $D$ and to $0$ outside $D$ . For an improper integral, where $f$ or $D$ is unbounded, the condition is absolute integrability, a finite double integral of $\lvert f \rvert$ . Without it the two orders can give different values.

In plain words For a continuous integrand on such a bounded region, adding the same columns in either order gives the same total, so you may choose whichever order is easier to compute. The theorem is about the same region described two ways, never about swapping symbols while keeping the limits.

Example On the triangle $0 \le x \le y \le 1$ , the integrand $e^{y^2}$ is continuous and lies between $1$ and $e$ . The triangle is described by $g_1(x) = x$ and $g_2(x) = 1$ for $0 \le x \le 1$ , and by $h_1(y) = 0$ and $h_2(y) = y$ for $0 \le y \le 1$ , all continuous. So the two iterated integrals of $e^{y^2}$ over it are equal. The condition matters for improper integrals. On the square of points with $0 < x \le 1$ and $0 < y \le 1$ , take

f(x, y) = \frac{x - y}{(x + y)^3}.

Since $f$ is the $y$ derivative of $\frac{y}{(x + y)^2}$ , the inner integral in $y$ is $\frac{1}{(1 + x)^2}$ , and the order $dy\,dx$ gives $\tfrac12$ . Exchanging $x$ and $y$ changes the sign of $f$ , so the order $dx\,dy$ gives $-\tfrac12$ . The theorem does not apply, because $f$ is unbounded near the origin and the double integral of $\lvert f \rvert$ is infinite.

Reversing the order of integration

Definition Reversing the order of integration rewrites an iterated integral over a region $D$ as an iterated integral over the same $D$ in the other order,

\begin{aligned} &\int_a^b \int_{g_1(x)}^{g_2(x)} f\,dy\,dx \\[0.4em] &\quad = \int_c^d \int_{h_1(y)}^{h_2(y)} f\,dx\,dy. \end{aligned}

The integrand stays the same, while the differentials and all four limits change. The new limits come from a sketch of $D$ . The equality holds under the conditions of Fubini’s theorem, for example when $f$ is continuous on a bounded region $D$ described both ways by continuous boundary functions. If the new order needs a split, the right side becomes a sum of such integrals.

In plain words Same region, same integrand, different slicing. The reason to do it is that an inner integral that is hard in one direction can be easy in the other.

Example In the integral

I = \int_0^1 \int_x^1 e^{y^2}\,dy\,dx,

the limits $0 \le x \le 1$ and $x \le y \le 1$ describe the triangle $0 \le x \le y \le 1$ . A horizontal strip at height $y$ runs from $x = 0$ to $x = y$ , and $y$ runs from $0$ to $1$ . The integrand is continuous on the closed triangle, so Fubini’s theorem gives

\begin{aligned} I &= \int_0^1 \int_0^y e^{y^2}\,dx\,dy \\[0.4em] &= \int_0^1 y e^{y^2}\,dy. \end{aligned}

The new inner integral is easy because $e^{y^2}$ does not change as $x$ moves along the strip, so it equals $e^{y^2}$ times the strip length $y$ . The substitution $u = y^2$ , with $du = 2y\,dy$ and new limits $u = 0$ and $u = 1$ , finishes the work,

\begin{aligned} I &= \frac12 \int_0^1 e^u\,du \\[0.4em] &= \frac{e - 1}{2} \approx 0.859. \end{aligned}

The size is sensible. The triangle has area $\tfrac12$ and the height lies between $1$ and $e$ , so $I$ must lie between $\tfrac12$ and $\tfrac{e}{2} \approx 1.359$ .

Missing factor for substitution

Definition The substitution rule says that if $g$ is continuously differentiable on $[a, b]$ and $h$ is continuous on the interval of values that $g$ takes there, then

\begin{aligned} &\int_a^b h(g(y))\,g'(y)\,dy \\[0.4em] &\quad = \int_{g(a)}^{g(b)} h(u)\,du. \end{aligned}

The factor $g'(y)$ beside the composite $h(g(y))$ is what the rule needs. A factor that differs from $g'(y)$ by a nonzero constant can be adjusted by multiplying and dividing by that constant, but a missing factor that depends on $y$ cannot be supplied by algebra. In a double integral, an inner integral in $x$ of a function $\varphi(y)$ over $h_1(y) \le x \le h_2(y)$ equals $\bigl(h_2(y) - h_1(y)\bigr)\varphi(y)$ , the strip length times $\varphi(y)$ . When that strip length is a constant multiple of $g'(y)$ , the outer integral is ready for a substitution.

In plain words Substitution needs the derivative of the inside function as a factor, so $e^{y^2}$ needs a $y$ next to it. A missing constant is easy to repair, but a missing $y$ cannot be created by rearranging. In the triangle, the horizontal strip at height $y$ has length $y$ , and reversing the order puts exactly that factor into the integrand.

Example In $\int_0^1 y e^{y^2}\,dy$ the inside function is $y^2$ , whose derivative $2y$ is twice the factor present. With $u = y^2$ and $du = 2y\,dy$ , the factor becomes $y\,dy = \tfrac12\,du$ , the limits $y = 0$ and $y = 1$ become $u = 0$ and $u = 1$ , and the integral is $\tfrac12 \int_0^1 e^u\,du = \tfrac{e - 1}{2}$ . The limits happen to keep their values, but they are still the new limits for $u$ . Without the factor, the same substitution in $\int_0^1 e^{y^2}\,dy$ needs $dy = \frac{du}{2\sqrt{u}}$ and leads to

\int_0^1 \frac{e^u}{2\sqrt{u}}\,du,

an improper integral that has no elementary antiderivative either. Its value, about $1.4627$ , comes from numerical methods or a power series, not from an antiderivative formula.

Check yourself

Reverse the order of integration in

\int_0^2 \int_{x/2}^1 f(x, y)\,dy\,dx

for a continuous $f$ , and check the new limits by computing the area of the region in both orders.

Show the answer

The limits say $0 \le x \le 2$ and $\tfrac{x}{2} \le y \le 1$ . The region is the triangle with vertices $(0, 0)$ , $(0, 1)$ and $(2, 1)$ , above the line $y = \tfrac{x}{2}$ and below the line $y = 1$ . The test point $(\tfrac12, \tfrac12)$ satisfies $\tfrac14 \le \tfrac12 \le 1$ , so it lies inside. A horizontal strip at height $y$ runs from the $y$ axis to the line $x = 2y$ , and $y$ runs from $0$ to $1$ , so the reversed integral is

\int_0^1 \int_0^{2y} f(x, y)\,dx\,dy.

With $f = 1$ the two orders give $\int_0^2 (1 - \tfrac{x}{2})\,dx = 1$ and $\int_0^1 2y\,dy = 1$ . Swapping only the differentials, as in $\int_0^1 \int_{x/2}^1 f\,dx\,dy$ , would not even make sense, because the inner limits would contain $x$ , the very variable being integrated.

Evaluate

\int_0^1 \int_{\sqrt{x}}^1 e^{y^3}\,dy\,dx.

Show the answer

The function $e^{y^3}$ has no elementary antiderivative, so reverse the order. The limits say $0 \le x \le 1$ and $\sqrt{x} \le y \le 1$ . For nonnegative $x$ and $y$ the condition $\sqrt{x} \le y$ is the same as $x \le y^2$ , so the region is bounded by the $y$ axis, the line $y = 1$ and the curve $x = y^2$ , and it is also described by $0 \le y \le 1$ and $0 \le x \le y^2$ . The integrand is continuous on this closed bounded region, so Fubini’s theorem allows the switch,

\begin{aligned} &\int_0^1 \int_0^{y^2} e^{y^3}\,dx\,dy \\[0.4em] &\quad = \int_0^1 y^2 e^{y^3}\,dy. \end{aligned}

The strip length $y^2$ is one third of the derivative $3y^2$ of the exponent. With $u = y^3$ , $du = 3y^2\,dy$ and new limits $u = 0$ and $u = 1$ , the value is

\begin{aligned} \frac13 \int_0^1 e^u\,du &= \frac{e - 1}{3} \\[0.4em] &\approx 0.573. \end{aligned}

As a check, the region has area $\int_0^1 y^2\,dy = \tfrac13$ and the height lies between $1$ and $e$ , so the value must lie between $\tfrac13$ and $\tfrac{e}{3}$ , and it does.

Evaluate the integral below. The formula $\frac{\sin y}{y}$ is undefined at $y = 0$ , so explain why reversing the order is still allowed.

\int_0^\pi \int_x^\pi \frac{\sin y}{y}\,dy\,dx

Show the answer

The region is the triangle $0 \le x \le y \le \pi$ , and its only point with $y = 0$ is the origin. Since $\frac{\sin y}{y} \to 1$ as $y \to 0$ , giving the integrand the value $1$ at $y = 0$ makes it continuous on the closed triangle, and a value at a single point does not change any of the integrals involved. So Fubini’s theorem applies. A horizontal strip at height $y$ runs from $x = 0$ to $x = y$ , and

\begin{aligned} &\int_0^\pi \int_0^y \frac{\sin y}{y}\,dx\,dy \\[0.4em] &\quad = \int_0^\pi \sin y\,dy = 2. \end{aligned}

The strip length $y$ cancels the denominator. In the original order the inner integral would need an antiderivative of $\frac{\sin y}{y}$ , and there is no elementary one.

Change the order, reveal the missing factor

Key terms

Triangular region

Bounds read from a sketch

Fubini’s theorem

Reversing the order of integration

Missing factor for substitution

Check yourself

How this lesson connects

Builds on

Used later in