You encounter ∫₀¹ ∫ᵧ¹ e^(x²) dx dy. Evaluating this directly is impossible because e^(x²) has no elementary antiderivative in x. What is the correct approach?
AApproximate numerically — no exact answer exists for this integral
BReverse the order of integration: rewrite the same region as a Type I description and integrate y first
CApply a trigonometric substitution to eliminate the x² in the exponent
DFactor e^(x²) as a product and integrate each factor separately
The region is {0 ≤ y ≤ 1, y ≤ x ≤ 1} — equivalently {0 ≤ x ≤ 1, 0 ≤ y ≤ x} as a Type I region. Reversing gives ∫₀¹ ∫₀ˣ e^(x²) dy dx = ∫₀¹ x·e^(x²) dx, which is solvable by u-substitution (u = x²). The exact answer is (e − 1)/2. This is the canonical example of why order-of-integration reversal is an essential skill: one order is intractable, the other is straightforward.
Question 2 Multiple Choice
The region D is bounded by y = 0, y = √x, and x = 4. As a Type II region (y as the outer variable), the correct bounds are:
A0 ≤ y ≤ 4, 0 ≤ x ≤ √y
B0 ≤ y ≤ 2, y² ≤ x ≤ 4
C0 ≤ y ≤ 2, 0 ≤ x ≤ y²
D0 ≤ x ≤ 4, 0 ≤ y ≤ √x
The curve y = √x has maximum y = √4 = 2, so y ranges from 0 to 2. For a fixed y, x ranges from the curve (x = y², obtained by solving y = √x) to the right boundary x = 4. Option C reverses the x-bounds: x should go from y² upward to 4, not from 0 to y². Option D describes the original Type I version, not the Type II reversal.
Question 3 True / False
In a Type I region, the inner integral limits are functions of the outer variable x, and the outer limits on x are constants.
TTrue
FFalse
Answer: True
This is the defining structure of a Type I region: x ranges between constants a and b (the outer integral), while y — for each fixed x — ranges between functions g₁(x) and g₂(x) (the inner integral). This means the inner limits depend on x, which is evaluated before x is integrated out. The outer limits must be constants so that the resulting expression after the inner integration is purely a function of x.
Question 4 True / False
To reverse the order of integration, you can simply swap x and y in the integral limits without changing anything else.
TTrue
FFalse
Answer: False
You cannot just swap the variable names in the limits — you must re-describe the same geometric region in the new order. If the original Type II setup has y from c to d and x from h₁(y) to h₂(y), reversing to Type I requires finding the x-range as constants and deriving new y-limit functions that describe the same region. Simply swapping the symbols gives mathematically different limits and integrates over a different region.
Question 5 Short Answer
Why can reversing the order of integration sometimes transform an otherwise impossible integral into a solvable one?
Think about your answer, then reveal below.
Model answer: Because the integrand may have an antiderivative with respect to one variable but not the other. The function e^(x²), for instance, has no elementary antiderivative in x, so integrating in x first is impossible. But for a fixed x, integrating e^(x²) in y is trivial (it's a constant in y). Reversing the order lets us perform the easy integration first, leaving a simpler function of one variable to integrate second.
The key insight is that the choice of integration order affects which variable is treated as the 'constant' in the inner integral. Swapping which variable is integrated first can change e^(x²) (hard) into x·e^(x²) (easy via u-substitution). The region is the same; only the order of slicing changes.