Evaluating Type II Integrals With Variable Limits

Evaluating a Type II Integral

A Type II region is described by horizontal slices: the vertical coordinate $y$ ranges over a fixed interval, and for each $y$, the horizontal coordinate $x$ ranges between two boundary functions. Evaluating an integral over such a region means first integrating across each horizontal slice and then combining the results over the full range of $y$.

The order of integration determines which variable is integrated first. In a Type II iterated integral, $x$ is the inner variable and $y$ is the outer variable. This matters when finding antiderivatives: while integrating with respect to $x$, treat $y$ as fixed.

Definition: A Type II region has the form $R=\{(x,y):c\le y\le d,\ g_1(y)\le x\le g_2(y)\}$, where $g_1$ and $g_2$ are boundary functions and $g_1(y)\le g_2(y)$ on $[c,d]$. For a continuous function $f$, its iterated integral over this region is $$ \iint_R f(x,y)\,dA = \int_c^d\left(\int_{g_1(y)}^{g_2(y)} f(x,y)\,dx\right)dy. $$

The inner integral evaluates the contribution along one horizontal slice at a fixed $y$. Its limits, $g_1(y)$ and $g_2(y)$, are values of $x$. Once the inner integral is evaluated, the result is a function of $y$, which is then integrated from $c$ to $d$.

The Evaluation Procedure

Suppose $f$ is continuous on the region and the boundary functions are continuous on $[c,d]$. To evaluate the iterated integral, find an antiderivative of $f(x,y)$ with respect to $x$, keeping $y$ fixed. Evaluate that antiderivative at $x=g_2(y)$ and subtract its value at $x=g_1(y)$. The result is the integrand for the outer integral.

\int_c^d\left(\int_{g_1(y)}^{g_2(y)} f(x,y)\,dx\right)dy = \int_c^d \left[F(g_2(y),y)-F(g_1(y),y)\right]dy,

where $F$ is an antiderivative with respect to $x$, so that $\frac{\partial F}{\partial x}(x,y)=f(x,y)$. The outer integral must still be evaluated; computing only the inner integral is not the complete answer.

Integrate with respect to $x$.
Find an antiderivative, treating $y$ as a constant.

Apply the inner limits.
Substitute $g_2(y)$ and $g_1(y)$ for $x$, and subtract lower from upper.

Integrate with respect to $y$.
Evaluate the resulting one-variable integral from $c$ to $d$.

Why the Procedure Works

For a fixed value of $y$, the inner integral is an ordinary one-variable definite integral across the corresponding horizontal slice. The Fundamental Theorem of Calculus says that if $F$ is an antiderivative of $f$ with respect to $x$, then the slice contributes $F(g_2(y),y)-F(g_1(y),y)$. Integrating those slice contributions over all $y$ from $c$ to $d$ gives the iterated integral. Thus each step follows from one-variable integration together with the description of the region by its horizontal slices.

Key point: In the inner integral, the variable is $x$. The coordinate $y$ is held fixed while integrating, but it can still appear in the antiderivative and in the limits.

Worked Examples

Worked Example: A Linear Integrand on a Triangular Region

Evaluate the integral of $f(x,y)=x+2y$ over the Type II region $0\le y\le 2,\ y\le x\le 2$.

Write the iterated integral using the given limits:

\int_0^2\int_y^2 (x+2y)\,dx\,dy.

For the inner integral, $y$ is constant. An antiderivative with respect to $x$ is $\frac{x^2}{2}+2yx$, since its derivative with respect to $x$ is $x+2y$. Apply the upper and lower limits:

\begin{aligned} \int_y^2 (x+2y)\,dx &=\left[\frac{x^2}{2}+2yx\right]_y^2\\ &=\left(\frac{2^2}{2}+2y(2)\right) -\left(\frac{y^2}{2}+2y(y)\right)\\ &=2+4y-\frac{y^2}{2}-2y^2\\ &=2+4y-\frac{5}{2}y^2. \end{aligned}

Now integrate the resulting expression over $0\le y\le 2$:

\begin{aligned} \int_0^2\left(2+4y-\frac{5}{2}y^2\right)dy &=\left[2y+2y^2-\frac{5}{6}y^3\right]_0^2\\ &=\left(4+8-\frac{5}{6}(8)\right)-0\\ &=12-\frac{20}{3}\\ &=\frac{16}{3}. \end{aligned}

Check the antiderivatives: differentiating $\frac{x^2}{2}+2yx$ with respect to $x$ gives $x+2y$, and differentiating $2y+2y^2-\frac{5}{6}y^3$ with respect to $y$ gives $2+4y-\frac{5}{2}y^2$. The endpoint evaluation therefore gives $\iint_R (x+2y)\,dA=\frac{16}{3}$.

Worked Example: A Parabolic Inner Boundary

Evaluate the integral of $f(x,y)=x$ over $0\le y\le 1,\ y^2\le x\le 1$.

The corresponding iterated integral is:

\int_0^1\int_{y^2}^1 x\,dx\,dy.

An antiderivative of $x$ with respect to $x$ is $\frac{x^2}{2}$. Its derivative is $x$. Substitute both inner limits, remembering that the lower limit $y^2$ depends on $y$:

\begin{aligned} \int_{y^2}^1 x\,dx &=\left[\frac{x^2}{2}\right]_{y^2}^1\\ &=\frac{1^2}{2}-\frac{(y^2)^2}{2}\\ &=\frac{1}{2}-\frac{y^4}{2}. \end{aligned}

The outer integral is then:

\begin{aligned} \int_0^1\left(\frac{1}{2}-\frac{y^4}{2}\right)dy &=\left[\frac{y}{2}-\frac{y^5}{10}\right]_0^1\\ &=\left(\frac{1}{2}-\frac{1}{10}\right)-0\\ &=\frac{2}{5}. \end{aligned}

Differentiating $\frac{x^2}{2}$ with respect to $x$ returns $x$, and differentiating $\frac{y}{2}-\frac{y^5}{10}$ returns $\frac{1}{2}-\frac{y^4}{2}$. Thus the evaluated integral is $\frac{2}{5}$.

Worked Example: A Polynomial with Two Variable Inner Limits

Evaluate the integral of $f(x,y)=x^2$ over $0\le y\le 2,\ y\le x\le y+1$.

Start with the given Type II bounds:

\int_0^2\int_y^{y+1} x^2\,dx\,dy.

An antiderivative of $x^2$ with respect to $x$ is $\frac{x^3}{3}$. Evaluate it at both limits and expand the difference:

\begin{aligned} \int_y^{y+1}x^2\,dx &=\left[\frac{x^3}{3}\right]_y^{y+1}\\ &=\frac{(y+1)^3-y^3}{3}\\ &=\frac{y^3+3y^2+3y+1-y^3}{3}\\ &=y^2+y+\frac{1}{3}. \end{aligned}

Integrate this expression over the outer interval:

\begin{aligned} \int_0^2\left(y^2+y+\frac{1}{3}\right)dy &=\left[\frac{y^3}{3}+\frac{y^2}{2}+\frac{y}{3}\right]_0^2\\ &=\left(\frac{8}{3}+2+\frac{2}{3}\right)-0\\ &=\frac{10}{3}+2\\ &=\frac{16}{3}. \end{aligned}

Check: the derivative of $\frac{x^3}{3}$ with respect to $x$ is $x^2$, and the derivative of $\frac{y^3}{3}+\frac{y^2}{2}+\frac{y}{3}$ is $y^2+y+\frac{1}{3}$. Hence the integral evaluates to $\frac{16}{3}$.

A Common Substitution Error

When the lower inner limit depends on $y$, it is easy to substitute that limit incorrectly. For example, in the inner integral from the second worked example, the lower limit is $x=y^2$. A common error is to write:

\left[\frac{x^2}{2}\right]_{y^2}^1 \ \overset{\text{wrong}}{=}\ \frac{1}{2}-\frac{y^2}{2}.

This treats the lower limit as if its value were $y$. The antiderivative contains $x^2$, so substituting $x=y^2$ gives $(y^2)^2=y^4$, not $y^2$. The correct evaluation is:

\left[\frac{x^2}{2}\right]_{y^2}^1 = \frac{1}{2}-\frac{(y^2)^2}{2} = \frac{1}{2}-\frac{y^4}{2}.

A reliable check is to replace every $x$ in the antiderivative by the entire expression for the limit before simplifying. Do not drop the exponent or simplify the limit before substituting into the antiderivative.

What to Check Before Finishing

Several checks help keep a Type II calculation organized:

The inner limits are $x$-values, and the outer limits are $y$-values.
The inner antiderivative is taken with respect to $x$; any $y$ in the integrand is treated as constant.
The upper inner value is evaluated first, and the lower inner value is subtracted from it.
After applying the inner limits, the expression should depend only on $y$.
The outer integral is evaluated over the full interval from $c$ to $d$.

The order of the limits also determines the sign of each slice contribution. If $g_1(y)\le g_2(y)$, the inner integral runs from the left boundary to the right boundary. Reversing those limits would negate the slice integral; it would not describe the same iterated integral.

Evaluating a Type II integral is therefore a two-stage application of one-variable integration. First, the Fundamental Theorem of Calculus evaluates each horizontal slice. Then a second definite integral combines the slice contributions across the region.

Check Your Understanding

For each question, pay attention to which variable is integrated first and how the inner limits are substituted.

In $\int_1^3\int_{y}^{2y} f(x,y)\,dx\,dy$, which variable is integrated first?
Find an antiderivative of $x+y$ with respect to $x$, treating $y$ as fixed.
Evaluate $\int_{y}^{2}x\,dx$ as an expression in $y$.
For the inner antiderivative $\frac{x^3}{3}$, what expression results when the lower limit is $x=y+1$?
After evaluating the inner integral, which variable should appear in the integrand of the outer integral?