Evaluating Type I Integrals With Variable Limits

Evaluating a Type I Integral

A Type I region is described by vertical slices: the outer variable ranges over a fixed interval, and for each value of that variable, the vertical variable ranges between two boundary curves. Once the limits are written in that form, evaluating the iterated integral is a matter of carrying out the inner integral first and then the outer integral.

The order matters. In the integral below, $y$ is the inner variable, so the first antiderivative is with respect to $y$. During that step, $x$ is held fixed. Only after the inner limits have been applied do we integrate with respect to $x$.

Definition: A Type I region has the form $D=\{(x,y):a\le x\le b,\ g_1(x)\le y\le g_2(x)\}$, where $g_1$ and $g_2$ are continuous on $[a,b]$ and $g_1(x)\le g_2(x)$ throughout that interval. Its iterated integral is written $\int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx$.

For a continuous integrand $f$ on such a region, the inner limits describe the bottom and top of each vertical slice. The outer limits describe the full interval of $x$-values. This tutorial focuses on evaluating an integral once those limits are known.

The Evaluation Rule

To evaluate a Type I iterated integral, find an antiderivative of the integrand with respect to $y$, treating $x$ as a constant. Substitute the upper $y$-limit and subtract the value at the lower $y$-limit. The result is a function of $x$; integrate that function from $a$ to $b$.

Theorem: Suppose $f$ is continuous on $D$, and $F(x,y)$ is an antiderivative of $f(x,y)$ with respect to $y$, so that $\frac{\partial F}{\partial y}(x,y)=f(x,y)$. Then $$ \int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx = \int_a^b \bigl(F(x,g_2(x))-F(x,g_1(x))\bigr)\,dx. $$ If the resulting function of $x$ has an antiderivative $G(x)$, the value is $G(b)-G(a)$.

This rule follows from the fundamental theorem of calculus applied to the inner integral for each fixed $x$. The endpoints $g_1(x)$ and $g_2(x)$ are numbers during that inner integration, even though their values depend on $x$. Once the endpoint difference has been formed, it is integrated over the outer interval.

A Reliable Calculation Sequence

Identify the inner variable.
In $dy\,dx$, integrate with respect to $y$ first; treat $x$ as fixed.

Find the inner antiderivative.
Use the usual antiderivative rules, keeping any $x$-dependent factors in the expression.

Apply both inner limits.
Evaluate at the upper boundary and subtract the evaluation at the lower boundary.

Integrate over the outer interval.
Simplify the endpoint difference, then integrate it with respect to $x$ from $a$ to $b$.

The subtraction in Step 3 is essential. The inner integral measures accumulation from the lower boundary to the upper boundary, not from zero unless the lower boundary is zero.

Worked Example: A Triangular Type I Region

Evaluate the integral over $0\le x\le 2$, $0\le y\le x$:

\int_0^2\int_0^x (2x+3y)\,dy\,dx.

First integrate with respect to $y$. Here $x$ is constant, so an antiderivative is $2xy+\frac{3}{2}y^2$. Apply the inner limits $0$ and $x$:

\left[2xy+\frac{3}{2}y^2\right]_{0}^{x} = \left(2x(x)+\frac{3}{2}x^2\right) - \left(2x(0)+\frac{3}{2}(0)^2\right) = \frac{7}{2}x^2.

Now integrate this result over $0\le x\le2$:

\int_0^2\frac{7}{2}x^2\,dx = \frac{7}{2}\left[\frac{x^3}{3}\right]_0^2 = \frac{7}{6}(8-0) = \frac{28}{3}.

Check: Differentiating $2xy+\frac{3}{2}y^2$ with respect to $y$ gives $2x+3y$. Differentiating $\frac{7}{6}x^3$, the outer antiderivative, gives $\frac{7}{2}x^2$, the simplified inner result. The endpoint evaluations produce $\frac{28}{3}$.

Variable Lower and Upper Boundaries

When both inner boundaries depend on $x$, evaluate the antiderivative at each boundary separately. In particular, a term involving $y$ may produce different powers of $x$ at the two endpoints. Keeping the two evaluations in parentheses until they are subtracted helps prevent sign errors.

Worked Example: A Curved Lower Boundary

Evaluate the integral over $0\le x\le1$, $x^2\le y\le x$:

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

The bounds are correctly ordered on this interval: $x^2\le x$ for $0\le x\le1$. With respect to $y$, an antiderivative of $x+2y$ is $xy+y^2$. Substituting the upper limit first gives $x(x)+x^2=2x^2$. Substituting the lower limit gives $x(x^2)+(x^2)^2=x^3+x^4$. Thus the inner integral is

\left[xy+y^2\right]_{x^2}^{x} = 2x^2-(x^3+x^4) = 2x^2-x^3-x^4.

Integrate each term over the outer interval:

\begin{aligned} \int_0^1(2x^2-x^3-x^4)\,dx &= \left[\frac{2}{3}x^3-\frac{1}{4}x^4-\frac{1}{5}x^5\right]_0^1\\ &= \frac{2}{3}-\frac{1}{4}-\frac{1}{5}\\ &= \frac{40}{60}-\frac{15}{60}-\frac{12}{60} = \frac{13}{60}. \end{aligned}

Check: Differentiating $xy+y^2$ with respect to $y$ gives $x+2y$. Differentiating $\frac{2}{3}x^3-\frac{1}{4}x^4-\frac{1}{5}x^5$ gives $2x^2-x^3-x^4$, as required. Its value at $1$ is $\frac{13}{60}$, and its value at $0$ is $0$.

When the Inner Antiderivative Uses a Boundary Function

The inner antiderivative may contain an exponential, logarithm, or another function evaluated at an $x$-dependent endpoint. The procedure remains the same: substitute the endpoint into the antiderivative, simplify the resulting expression in $x$, and then evaluate the outer integral. The outer integration is performed after this simplification; the inner endpoint itself is not differentiated at that stage.

Worked Example: An Exponential Integrand

Evaluate the integral over $0\le x\le3$, $0\le y\le\ln(1+x)$:

\int_0^3\int_0^{\ln(1+x)} e^y\,dy\,dx.

An antiderivative of $e^y$ with respect to $y$ is $e^y$. Apply both limits and use $e^{\ln(1+x)}=1+x$, which is valid here because $1+x>0$:

\left[e^y\right]_0^{\ln(1+x)} = e^{\ln(1+x)}-e^0 = (1+x)-1 = x.

The remaining integral is therefore

\int_0^3 x\,dx = \left[\frac{x^2}{2}\right]_0^3 = \frac{9}{2}-0 = \frac{9}{2}.

Check: The derivative of $e^y$ with respect to $y$ is $e^y$. The derivative of $x^2/2$ is $x$, which is the simplified inner integral. Evaluating at $x=3$ and $x=0$ gives $\frac{9}{2}$.

Why the Two Integrations Give the Result

For a fixed value of $x$, the fundamental theorem of calculus states that integrating $f(x,y)$ from $g_1(x)$ to $g_2(x)$ gives the difference of the antiderivative at those endpoints. In symbols, if $\frac{\partial F}{\partial y}=f$, then

\int_{g_1(x)}^{g_2(x)}f(x,y)\,dy = F(x,g_2(x))-F(x,g_1(x)).

This equation holds for each $x$ in $[a,b]$. Integrating both sides with respect to $x$ over that interval gives

\int_a^b\left(\int_{g_1(x)}^{g_2(x)}f(x,y)\,dy\right)dx = \int_a^b\bigl(F(x,g_2(x))-F(x,g_1(x))\bigr)\,dx.

If $G'(x)=F(x,g_2(x))-F(x,g_1(x))$, the fundamental theorem of calculus evaluates the right-hand side as $G(b)-G(a)$. This explains both why the inner endpoint difference is formed first and why the outer limits are applied last. The continuity assumptions ensure the integrals in this rule are well-defined on the bounded region.

A Common Mistake: Forgetting the Lower Endpoint

A frequent error is to evaluate the inner antiderivative at the upper endpoint and stop, even when the lower endpoint is not zero. For the second example, the incorrect step would be

\left[xy+y^2\right]_{x^2}^{x} \ \text{incorrectly replaced by}\ xy\big|_{y=x}+y^2\big|_{y=x} = 2x^2.

This omits the value at $y=x^2$. It would lead to the incorrect outer integral $\int_0^1 2x^2\,dx=\frac{2}{3}$. The correction is to subtract the full lower-endpoint value:

2x^2-\left(x(x^2)+(x^2)^2\right) = 2x^2-x^3-x^4,

so the correct value is $\frac{13}{60}$, not $\frac{2}{3}$. The square brackets with upper and lower limits always mean “upper value minus lower value.”

Practical Checks and Interpretation

After evaluating a Type I integral, check the calculation at both stages. Differentiate the inner antiderivative with respect to $y$ to recover the integrand, then differentiate the outer antiderivative with respect to $x$ to recover the simplified inner result. Finally, verify the endpoint subtraction. These checks catch many errors in signs, powers, and constants.

If $f(x,y)\ge0$ throughout the region, the value of the integral cannot be negative. This provides a useful reasonableness check, though it does not by itself verify an exact value. For example, all three integrands above are nonnegative on their stated regions, and all three resulting integrals are positive.

Evaluating the integral correctly depends on keeping the roles of the variables clear. The order $dy\,dx$ prescribes the calculation: integrate in $y$, use the two vertical boundary functions, simplify, and then integrate in $x$. Changing that sequence without changing the limits would generally calculate a different quantity.

Check Your Understanding

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

In $\int_0^2\int_1^{x+1} f(x,y)\,dy\,dx$, which variable is held fixed during the inner integration?
Find an antiderivative with respect to $y$ of $3x+y^2$, treating $x$ as constant.
For inner bounds $y=x^2$ to $y=x$, what is the correct order for subtracting the endpoint values?
Evaluate $\int_0^1\int_0^x 2y\,dy\,dx$.
What two derivative checks can help verify an evaluated Type I integral?