Integration by substitution, area and volume of revolution

IB Mathematics AA SL· IB AA SL Topic 5: Calculus, 5.9, 5.10· 25 min read

1. Integration by Substitution★★★☆☆⏱ 10 min

📘 Definition

Integration by substitution

$u = g (x)$

A technique to integrate composite functions of the form $ int f(g(x))g'(x)dx $. S u b s t i t u t e$ u = g(x) $so$ du = g'(x)dx $, t r an s f or min g t h e in t e g r a l in t o t h es im pl er$ int f(u)du$.

Example:

For $ int 2x\cos(x^2)dx $, se t$ u=x^2 $t os im pl i f y t h e in t e g r a l t o$ int \cos u du$.

Substitution works for both indefinite and definite integrals. For definite integrals, you have two valid options: (1) change the limits to match $u$ and evaluate directly in terms of $u$ , or (2) integrate in terms of $u$ , substitute back to $x$ , then evaluate with original limits. Both are accepted in IB exams if working is clear.

📐 Worked Example

Evaluate $ int_0^2 2x e^{x^2} dx$

1
Choose $u$ as the inner function of the composite exponential:
2
$u = x^2$
3
Differentiate $u$ to find the relationship between $d u$ and $d x$ :
4
$\frac{du}{dx} = 2x \implies du = 2x dx$
5
Change the $x$ limits to match $u$ :
6
When $x = 0$ , $u = 0^{2} = 0$ ; when $x = 2$ , $u = 2^{2} = 4$
7
Substitute into the original integral to simplify:
8
$\int_{0}^{4} e^u du$
9
Integrate and evaluate:
10
$\left[ e^u \right]_0^4 = e^4 - e^0 = e^4 - 1 \approx 54.6$

Exam tip:

Always check your result by differentiating back. This takes 10 seconds and catches most substitution errors.

2. Area Bounded by Curves and Axes★★★☆☆⏱ 8 min

A standard definite integral gives net signed area, where regions below the x-axis contribute negative values. To find actual total area, you must split the integral at any x-intercepts in the interval and take the absolute value of each segment. The total area between $y = f (x)$ , the x-axis, $x = a$ and $x = b$ is $A = \int_{a}^{b} ∣ f (x) ∣ d x$ .

📘 Definition

Total area between a curve and the x-axis

The positive sum of areas of all regions between the curve and the axis, regardless of whether the curve is above or below the axis.

📐 Worked Example

Find the total area between $y = x^{3} - x$ and the x-axis from $x = 0$ to $x = 2$ .

1
Find x-intercepts in the interval to split the integral:
2
$x^3 - x = x(x^2 - 1) = 0 \implies x=0, x=1$
3
The function is negative between $0 < x < 1$ and positive between $1 < x < 2$ , so area is:
4
$A = \left| \int_0^1 (x^3 - x) dx \right| + \int_1^2 (x^3 - x) dx$
5
Evaluate the first integral:
6
$\int_0^1 (x^3 - x) dx = \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_0^1 = -\frac{1}{4}$
7
Take absolute value and evaluate the second integral:
8
$A = \left| -\frac{1}{4} \right| + \left[ \frac{x^4}{4} - \frac{x^2}{2} \right]_1^2 = \frac{1}{4} + \left(2 + \frac{1}{4}\right) = 2.5$

Exam tip:

If a question asks for area, it always means total area, not net signed area. Always check if the curve crosses the x-axis in the interval.

3. Volume of Revolution Around the x-axis★★★★☆⏱ 7 min

When you rotate the region bounded by $y = f (x)$ , the x-axis, $x = a$ and $x = b$ 360° around the x-axis, the volume of the resulting solid is given by the formula:

V = \pi \int_a^b [f(x)]^2 dx

📐 Worked Example

Find the volume of revolution formed by rotating $y = x + 1$ around the x-axis from $x = 0$ to $x = 2$ , in terms of $ pi$.

1
Write down the volume formula and substitute the known values:
2
$V = \pi \int_0^2 (x+1)^2 dx$
3
Use substitution to evaluate the integral, set $u = x + 1$ , $d u = d x$ :
4
Change limits: when $x = 0$ , $u = 1$ ; when $x = 2$ , $u = 3$
5
$V = \pi \int_1^3 u^2 du$
6
Integrate and evaluate:
7
$V = \pi \left[ \frac{u^3}{3} \right]_1^3 = \pi \left(9 - \frac{1}{3}\right) = \frac{26}{3}\pi$

Exam tip:

Don't forget the $π$ at the end of your volume calculation! Examiners frequently penalize this common mistake.

4. Common Pitfalls

Wrong move:

Forgetting to change limits when doing definite substitution, and evaluating the $u$ integral with original $x$ limits

Why:

Variables are mismatched, leading to completely incorrect results

Correct move:

Either change limits to match $u$ , or substitute back to $x$ before evaluating with original limits

Wrong move:

Leaving negative areas as negative when calculating total area of a region that goes below the x-axis

Why:

Area is always positive; exam questions asking for area expect total area, not net signed area

Correct move:

Find all intercepts in the interval, split the integral, and take the absolute value of each segment

Wrong move:

Forgetting to square $f (x)$ when calculating volume of revolution

Why:

The formula uses the area of the circular cross-section, which requires squaring the radius $y = f (x)$

Correct move:

Always write the formula first, and confirm you have squared $f (x)$ before integrating

Wrong move:

Choosing $u = f (x)$ instead of the inner composite function $g (x)$ for $ int f(g(x))g'(x)dx$

Why:

This leaves an integral that is harder or impossible to evaluate, wasting valuable exam time

Correct move:

Always pick $u$ as the inner function of the composite term, and confirm its derivative is present (up to a constant) in the integrand

5. Quick Reference Cheatsheet

Concept	Formula	Key Note
Indefinite substitution	$\int f (g (x)) g^{'} (x) d x = \int f (u) d u, u = g (x)$	Differentiate back to check
Definite substitution	$\int_{x = a}^{x = b} ... d x = \int_{u = g (a)}^{u = g (b)} ... d u$	No need to substitute back if changing limits
Total area (a to b, x-axis)	$A = \sum ∣ \int_{x_{i}}^{x_{i + 1}} f (x) d x ∣$	Split at all x-intercepts
Volume of revolution (x-axis)	$V = π \int_{a}^{b} [f (x)]^{2} d x$	Don't forget the $π$

6. Frequently Asked

Do I need to change limits of integration for substitution?

Yes for definite integrals. You can either change limits to match your new variable, or substitute back to the original variable before evaluating at the original limits. Both are acceptable.

Is volume of revolution around the y-axis required for AA SL?

No, AA SL only examines volumes of revolution around the x-axis. You do not need to practice rotation around the y-axis.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2021 · 1
Indefinite integration by substitution
2022 · 2
Total area between curve and x-axis
2023 · 1
Volume of revolution about x-axis

Going deeper

formula_referenceIB AA SL Formula BookletVolume of revolution formula is provided here

What's Next

Integration by substitution is the foundation for all advanced integration techniques, and is a prerequisite for topics like integration by parts if you continue to HL Mathematics. Area and volume of revolution problems are very common in IB AA SL Paper 2, often combined with substitution, so consistent practice of varied problems will help you score full marks on these frequently high-mark questions. Mastery of this topic also builds intuition for integral calculus that is useful for university studies in STEM and economics.