Multivariable-Calculus · Unit 14 · Video 4 · Interactive Practice

Exchanging the Order of Integration: Redraw the Region

IKey Formulas

Formula	Name	What it says
$\int_0^1\!\int_0^2 dx\,dy = \int_0^2\!\int_0^1 dy\,dx = 2$	Rectangle	Every bound is a constant, so the swap is immediate
$\int_0^1\!\int_x^{\sqrt{x}} \frac{e^y}{y}\,dy\,dx = \int_0^1\!\int_{y^2}^{y} \frac{e^y}{y}\,dx\,dy$	Sliver, order exchanged	Same integrand; new bounds read from horizontal slices
$\int_{y^2}^{y} \frac{e^y}{y}\,dx = \frac{e^y}{y}(y - y^2) = e^y - y\,e^y$	Inner integral	$\frac{e^y}{y}$ is constant in $x$ ; the factor $\frac{1}{y}$ cancels
$\int_0^1 (e^y - y\,e^y)\,dy = \big[-y\,e^y + 2e^y\big]_0^1 = e - 2$	Outer integral	Antiderivative by guess-and-adjust (or by parts)

Key Insight: Outer bounds are numbers — the first and last slice. Inner bounds are where one slice, at a fixed value of the outer variable, enters and leaves the region.

IIRead the Bounds Off the Slices

For each order, a slice's entry and exit give the inner bounds; the first and last slices give the outer bounds.

Region Order Slice at x = 0.40

💡 Swapping the quarter disk's bounds without redrawing puts $\sqrt{1-x^2}$ on the outer integral — after $x$ has been integrated out, so the result could never be a number.

IIIAdding Up the Horizontal Slices

In the order $dx\,dy$ , the slice at height $y$ contributes $e^y(1-y)$ , and the slices add up to $e - 2$ .

Height y 0.60

IVQuiz Questions

Problem 1 · Exchange the Order on a Triangle

Given: $\displaystyle\int_0^2\!\int_0^{x} f(x,y)\,dy\,dx$ — rewrite it in the order $dx\,dy$ .

✅ Correct! A horizontal line at height

y

enters the triangle at the line

x = y

and leaves at

x = 2

; heights run from

0

2

❌ Wrong side of the line.

0 \le x \le y

is the triangle to the left of

y = x

. The region

0 \le y \le x

lies to the right of that line, so

x

runs from

y

2

❌ That only swaps the bounds. The outer upper limit is

x

, which has already been integrated out — outer bounds must be numbers.

❌ Check the top of the region. The triangle reaches the point

(2, 2)

, so the highest slice is at

y = 2

, not

y = 1

Show solution

Draw the region. For each $x$ from $0$ to $2$ , $y$ runs from $0$ up to the line $y = x$ : the triangle with corners $(0,0)$ , $(2,0)$ , $(2,2)$ .

Fix $y$ . A horizontal line at height $y$ enters on the left at the line $y = x$ , that is $x = y$ , and leaves on the right at $x = 2$ . So $y \le x \le 2$ .

Outer bounds. The lowest slice is at $y = 0$ and the highest at $y = 2$ .

\int_0^2\!\int_0^{x} f(x,y)\,dy\,dx = \int_0^2\!\int_y^{2} f(x,y)\,dx\,dy

Problem 2 · Which Border Is on the Left?

Given: $\displaystyle\int_0^1\!\int_{x^2}^{x} f(x,y)\,dy\,dx$ — rewrite it in the order $dx\,dy$ .

✅ Correct! For

0 < y < 1

y < \sqrt{y}

: the line

x = y

is the left border and the parabola

x = \sqrt{y}

the right one.

❌ Left and right are reversed. For

0 < y < 1

\sqrt{y} > y

, so this inner integral runs right to left and comes out with the wrong sign.

❌ That is the video's sliver, not this one. Here the region lies between the parabola

y = x^2

and the line

y = x

; solving

y = x^2

for

x

gives

x = \sqrt{y}

, not

x = y^2

❌ The bounds were only swapped. The outer limits

x^2

and

x

contain

x

, which is gone after the inner integration.

Show solution

Draw the region. For $0 \le x \le 1$ , $y$ runs from the parabola $y = x^2$ up to the line $y = x$ ; the two meet at $(0,0)$ and $(1,1)$ .

Fix $y$ . Solve each border for $x$ : the line gives $x = y$ , the parabola gives $x = \sqrt{y}$ . For $0 < y < 1$ we have $y < \sqrt{y}$ (e.g. $y = \tfrac14$ : $\tfrac14 < \tfrac12$ ), so a horizontal line enters at $x = y$ and leaves at $x = \sqrt{y}$ .

Outer bounds. $y$ runs from $0$ to $1$ .

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

The parabola was the lower border in the original order and is the right border now.

Problem 3 · An Integral You Cannot Start

Given: $\displaystyle\int_0^1\!\int_x^{1} e^{y^2}\,dy\,dx$ . The inner integral has no elementary antiderivative, so exchange the order and evaluate.

After exchanging, what does the inner integral $\int (\ldots)\,dx$ evaluate to?

What is the value of the integral?

✅ Correct! In the order

dx\,dy

the slice width

y

supplies exactly the factor that

e^{y^2}

needs:

\int_0^1 y\,e^{y^2}\,dy = \tfrac{e-1}{2}

❌ Wrong side of the line. The region is

x \le y \le 1

, i.e. below

y = 1

and above

y = x

. At height

y

x

runs from

0

y

, a width of

y

, not

1 - y

❌ The slice width is missing.

e^{y^2}

is constant in

x

, so integrating over

0 \le x \le y

multiplies it by the width

y

❌ The bounds were never substituted. The antiderivative is

x\,e^{y^2}

, but it must be evaluated from

x = 0

x = y

; no

x

can survive the inner integral.

❌ Missing a factor of $\tfrac12$ .

\frac{d}{dy}e^{y^2} = 2y\,e^{y^2}

, so the antiderivative of

y\,e^{y^2}

\tfrac12 e^{y^2}

❌ The lower limit was dropped.

\tfrac12 e^{y^2}

y = 0

\tfrac12

, not

0

, so subtract it.

❌ That is the video's answer for a different integral. Here the outer integral is

\int_0^1 y\,e^{y^2}\,dy

Show solution

Draw the region. $0 \le x \le 1$ and $x \le y \le 1$ : the triangle above the line $y = x$ and below $y = 1$ , with corners $(0,0)$ , $(0,1)$ , $(1,1)$ .

Fix $y$ . A horizontal line at height $y$ enters at $x = 0$ and leaves at the line $x = y$ . Heights run from $0$ to $1$ .

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

Inner integral. $e^{y^2}$ does not contain $x$ :

\int_0^{y} e^{y^2}\,dx = \Big[x\,e^{y^2}\Big]_{x=0}^{x=y} = y\,e^{y^2}

Outer integral. Since $\frac{d}{dy}\big(\tfrac12 e^{y^2}\big) = y\,e^{y^2}$ :

\int_0^1 y\,e^{y^2}\,dy = \Big[\tfrac12 e^{y^2}\Big]_0^1 = \tfrac12 e - \tfrac12 = \frac{e-1}{2} \approx 0.859

Problem 4 · Transfer: An Inner Integral in $x$

Given: $\displaystyle\int_0^1\!\int_y^{1} \frac{\sin x}{x}\,dx\,dy$ — $\frac{\sin x}{x}$ has no elementary antiderivative. Exchange the order and evaluate.

✅ Correct! The slice length

x

cancels the

\frac{1}{x}

, leaving

\int_0^1 \sin x\,dx = 1 - \cos 1 \approx 0.460

❌ Sign of the antiderivative. The antiderivative of

\sin x

-\cos x

, so

\big[-\cos x\big]_0^1 = -\cos 1 + 1

❌ The outer integral was not carried out. After the inner integral you have

\int_0^1 \sin x\,dx

;

\sin 1

is the integrand's value at

x = 1

, not the integral.

❌ The lower limit was dropped.

-\cos x

x = 0

-1

, so

\big[-\cos x\big]_0^1 = -\cos 1 - (-1)

Show solution

Draw the region. $0 \le y \le 1$ and $y \le x \le 1$ : the triangle below the line $y = x$ , right of it up to $x = 1$ , with corners $(0,0)$ , $(1,0)$ , $(1,1)$ .

Fix $x$ . A vertical line at $x$ enters at $y = 0$ and leaves at the line $y = x$ . The outer variable $x$ runs from $0$ to $1$ .

\int_0^1\!\int_y^{1} \frac{\sin x}{x}\,dx\,dy = \int_0^1\!\int_0^{x} \frac{\sin x}{x}\,dy\,dx

Inner integral. $\frac{\sin x}{x}$ is constant in $y$ :

\int_0^{x} \frac{\sin x}{x}\,dy = \frac{\sin x}{x}\cdot x = \sin x

Outer integral.

\int_0^1 \sin x\,dx = \big[-\cos x\big]_0^1 = 1 - \cos 1 \approx 0.460

Same move as in the video: the new inner integral multiplies the integrand by a slice length that cancels the troublesome denominator.

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