Source: Integral Calculus, University of Texas at Austin (wolesensky, 54390)
Difficulty: Intermediate | Prerequisites: Single-variable integration techniques (substitution, integration by parts, partial fractions), basic properties of logarithms and exponentials.
Once you understand what a double integral represents geometrically, the next step is computing them by hand using iterated integrals. This section covers the mechanics: integrating the inner variable while treating the outer variable as a constant, then integrating the result. The integrands here range from simple polynomials to exponentials and rational functions, each requiring you to recall and apply single-variable techniques inside a two-variable framework.
To evaluate an iterated integral, treat the outer variable as a constant while you integrate the inner one, then plug in the inner limits and integrate the result over the outer variable. The functions you will meet here include polynomials, 1/(x + y)^2, e^(x - y), (3 + x^2)/(1 + y^2), and xe^(xy), each of which calls on a specific single-variable technique.
Inner integral
The integral you evaluate first in an iterated integral. You integrate with respect to one variable while holding the other constant.
Outer integral
The integral you evaluate second, after substituting the result of the inner integral. Its variable was the constant during the inner step.
Separable integrand
A function f(x, y) that can be written as g(x) * h(y). When the integrand separates, the double integral factors into a product of two single integrals. In simple terms, you can split the work into two independent pieces.
Average value of a function of two variables
The double integral of f over a region A, divided by the area of A. Think of it as the height a uniform layer of the same total volume would have over the rectangle.
Q4: Evaluate the integral of (3 + 2xy) dx dy, with x from 1 to 3 and y from 0 to 1.
Inner (x): integrate (3 + 2xy) dx = 3x + x^2 y, evaluated from 1 to 3 = (9 + 9y) - (3 + y) = 6 + 8y.
Outer (y): integrate (6 + 8y) dy from 0 to 1 = 6 + 4 = 10.
Answer: I = 10.
Q5: Evaluate the integral of 1/(x + y)^2 dx dy, with x from 0 to 4 and y from 1 to 4.
Inner (x): integrate 1/(x + y)^2 dx = -1/(x + y), evaluated from 0 to 4 = -1/(4 + y) + 1/y.
Outer (y): integrate (1/y - 1/(4 + y)) dy from 1 to 4 = [ln y - ln(4 + y)] from 1 to 4 = (ln 4 - ln 8) - (ln 1 - ln 5) = ln 4 - ln 8 + ln 5 = ln(20/8) = ln(5/2).
But we need to check: the integral gives 1/2 * ln(5/2)? Let us recheck. Actually -1/(x+y) evaluated from x=0 to x=4: -1/(4+y) - (-1/(0+y)) = 1/y - 1/(4+y). Integrate from y=1 to y=4: [ln y - ln(y+4)] from 1 to 4 = (ln 4 - ln 8) - (0 - ln 5) = ln 4 - ln 8 + ln 5 = ln(4*5/8) = ln(5/2).
Answer: I = (1/2) ln(5/2)... No. Recheck: the integral is just ln(5/2). But the answer choices include (1/2) ln(5/2). The outer integral is straightforward: ln(5/2). Answer: I = ln(5/2).
Q6: Evaluate the integral of e^(x - y) dx dy, with x from 0 to 2 and y from 2 to 3.
The integrand separates: e^(x - y) = e^x * e^(-y).
x-integral: integrate e^x from 0 to 2 = e^2 - 1.
y-integral: integrate e^(-y) from 2 to 3 = -e^(-3) + e^(-2) = e^(-2) - e^(-3).
Product: (e^2 - 1)(e^(-2) - e^(-3)) = 1 - e^(-1) - e^(-2) + e^(-3). Rearranged: e^(-3) - e^(-2) - e^(-1) + 1.
Wait: (e^2 - 1)(e^(-2) - e^(-3)) = e^2 * e^(-2) - e^2 * e^(-3) - 1 * e^(-2) + 1 * e^(-3) = 1 - e^(-1) - e^(-2) + e^(-3).
Reordering: e^(-3) - e^(-2) - e^(-1) + 1. Answer: I = e^(-3) - e^(-2) - e^(-1) + 1.
Q7: Evaluate the integral of (3 + x^2)/(1 + y^2) over A = {0 <= x <= 1, 0 <= y <= 1}.
The integrand separates into (3 + x^2) * 1/(1 + y^2).
x-integral: integrate (3 + x^2) dx from 0 to 1 = 3 + 1/3 = 10/3.
y-integral: integrate 1/(1 + y^2) dy from 0 to 1 = arctan(1) - arctan(0) = pi/4.
Product: (10/3)(pi/4) = 10pi/12 = 5pi/6.
Answer: I = (5/6)pi.
Q8: Evaluate the integral of x * e^(xy) over A = {0 <= x <= 1, 0 <= y <= 3}.
Integrate first with respect to y (treating x as constant): the integral of x * e^(xy) dy = e^(xy), evaluated from y = 0 to y = 3 gives e^(3x) - 1.
Now integrate (e^(3x) - 1) dx from 0 to 1 = [e^(3x)/3 - x] from 0 to 1 = (e^3/3 - 1) - (1/3 - 0) = e^3/3 - 1 - 1/3 = e^3/3 - 4/3 = (1/3)(e^3 - 4).
Answer: I = (1/3)(e^3 - 4).
Q9: Find the average value of f(x, y) = 6(x + y) ln x over A = {1 <= x <= 3, 0 <= y <= 3}.
Area of A = 2 * 3 = 6.
Average value = (1/6) * double integral of 6(x + y) ln x dA = double integral of (x + y) ln x dA / 1.
Actually, average = (1/Area) * integral. So (1/6) * integral of 6(x + y) ln x = integral of (x + y) ln x over A.
Integrate (x + y) ln x dy from 0 to 3 = 3x * ln x + (9/2) ln x = (3x + 9/2) ln x.
Then integrate (3x + 9/2) ln x dx from 1 to 3. Using integration by parts for x * ln x and ln x:
The integral of 3x ln x dx = 3[x^2/2 * ln x - x^2/4] from 1 to 3 = 3[(9/2 ln 3 - 9/4) - (0 - 1/4)] = 3[9/2 ln 3 - 2] = 27/2 ln 3 - 6.
The integral of (9/2) ln x dx = 9/2 [x ln x - x] from 1 to 3 = 9/2 [(3 ln 3 - 3) - (0 - 1)] = 9/2 [3 ln 3 - 2] = 27/2 ln 3 - 9.
Total = 27/2 ln 3 - 6 + 27/2 ln 3 - 9 = 27 ln 3 - 15.
Average = (1/6) * 6 * (27 ln 3 - 15)... Let me reconsider. The average is (1/6) times the double integral of 6(x+y) ln x. So (1/6)(6)(27 ln 3 - 15) = 27 ln 3 - 15.
Checking against choices: 6(9/2 ln 3 - 5/2) = 27 ln 3 - 15. Answer: I = 6(9/2 ln 3 - 5/2).
\int_c^d \int_a^b f(x,y)\,dx\,dyIterated integral: integrate f with respect to x from a to b (treating y as constant), then integrate the result with respect to y from c to d.
\bar{f} = \frac{1}{\text{Area}(A)} \iint_A f(x,y)\,dAAverage value of f over a region A.
\int \frac{1}{(x+y)^2}\,dx = -\frac{1}{x+y} + CUseful antiderivative for rational integrands of this type.
\int x e^{xy}\,dy = e^{xy} + C \quad (\text{treating } x \text{ as constant})When the exponent is xy and you integrate with respect to y, the x in front cancels with the 1/x from the chain rule.
Iterated integrals with exponential and rational integrands show up in probability (computing expected values over joint distributions), heat transfer (integrating temperature distributions over rectangular plates), and signal processing (evaluating convolutions over rectangular windows).
Forgetting to treat the outer variable as a constant during the inner integration. If you are integrating dx first, then y is a constant, not a variable.
Misapplying the product rule instead of the chain rule when integrating xe^(xy) with respect to y. The x in front is a constant here, not a function of y.
Confusing the average value formula. The average divides by the area of the region, not the length of an interval. For a rectangle [a, b] x [c, d], that denominator is (b - a)(d - c).
Splitting a non-separable integrand into a product. The function (3 + 2xy) does not separate into g(x) * h(y), so you cannot factor the double integral into two single integrals. (3 + x^2)/(1 + y^2) does separate.
Recognise separable integrands immediately. When f(x, y) = g(x) * h(y), the double integral is the product of two single integrals, which saves substantial time.
Integration by parts appears inside iterated integrals (Q9's x * ln x). Be prepared to use it as part of the inner or outer integration step.
The average value formula is a common exam question. Remember: divide the double integral by the area of the region.
Watch for sign errors in exponential integrals. The integral of e^(-y) is -e^(-y), not e^(-y).
True or false: The integral of e^(x - y) over a rectangle can be computed as the product of two single integrals. (True. The integrand separates into e^x * e^(-y).)
Fill in the blank: To find the average value of f over a rectangle, divide the double integral of f by ______. (The area of the rectangle.)
True or false: When evaluating the inner integral of (3 + 2xy) dx, y is treated as a constant. (True.)
Fill in the blank: The integral of x * e^(xy) with respect to y equals ______ + C. (e^(xy).)
Q: Evaluate the iterated integral of (3 + 2xy) dx dy, with x from 1 to 3 and y from 0 to 1.
A: Inner integral in x gives 6 + 8y. Outer integral in y gives 6 + 4 = 10.
Q: Evaluate the double integral of e^(x - y) dx dy, with x from 0 to 2 and y from 2 to 3.
A: Separate into (e^2 - 1)(e^(-2) - e^(-3)) = e^(-3) - e^(-2) - e^(-1) + 1.
Q: Evaluate the double integral of (3 + x^2)/(1 + y^2) over A = {0 <= x <= 1, 0 <= y <= 1}.
A: Separable. x-integral = 10/3. y-integral = pi/4. Product = 5pi/6.
Q: Find the average value of f(x, y) = 6(x + y) ln x over A = {1 <= x <= 3, 0 <= y <= 3}.
A: Average = 6(9/2 ln 3 - 5/2).
These evaluation techniques feed directly into double integrals over non-rectangular regions (where the limits become functions rather than constants) and into polar-coordinate integrals. The separability idea reappears in differential equations and probability theory wherever joint distributions factor into marginals.
Iterated integral, double integral evaluation, separable integrand, Fubini's theorem, inner integral, outer integral, integration by parts, average value, mean value, exponential integrand, rational integrand, arctangent integral, multivariable calculus, integral calculus, xe^(xy) integral