# Integration by substitution and area

> IB Mathematics AA SL · IB AA SL
> Source: https://www.owlsprep.com/study/ib-math-aa-sl-u5-integration-by-substitution-area-and/

This sub-topic covers integration by substitution (the reverse chain rule) and calculating the total area bounded by curves and axes for IB AA SL.

**Prerequisites:** [Basic indefinite and definite integration](https://www.owlsprep.com/study/ib-math-aa-sl-u5-basic-integration-rules/); [Chain rule for differentiation](https://www.owlsprep.com/study/ib-math-aa-sl-u5-differentiation-rules/)

## Learning objectives

- Perform indefinite and definite integration using substitution
- Calculate total area bounded by curves and coordinate axes
- Identify and avoid common substitution and area exam traps

## Integration by Substitution

**Integration by substitution** — A technique to integrate composite functions of the form $
int f(g(x))g'(x)dx$. Substitute $u = g(x)$ so $du = g'(x)dx$, transforming the integral into the simpler $
int f(u)du$.

*Notation:* $u = g(x)$

*Example:* For $
int 2x\cos(x^2)dx$, set $u=x^2$ to simplify the integral to $
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 $du$ and $dx$:
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.

## Area Bounded by Curves and Axes

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 $A = \int_a^b |f(x)| dx$.

**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.

## Common pitfalls

- **Wrong:** Forgetting to change limits when doing definite substitution, and evaluating the $u$ integral with original $x$ limits
  - Why it fails: Variables are mismatched, leading to completely incorrect results
  - Correct: Either change limits to match $u$, or substitute back to $x$ before evaluating with original limits
- **Wrong:** Leaving negative areas as negative when calculating total area of a region that goes below the x-axis
  - Why it fails: Area is always positive; exam questions asking for area expect total area, not net signed area
  - Correct: Find all intercepts in the interval, split the integral, and take the absolute value of each segment
- **Wrong:** Choosing $u = f(x)$ instead of the inner composite function $g(x)$ for $
int f(g(x))g'(x)dx$
  - Why it fails: This leaves an integral that is harder or impossible to evaluate, wasting valuable exam time
  - Correct: Always pick $u$ as the inner function of the composite term, and confirm its derivative is present (up to a constant) in the integrand

## Cheatsheet

| Concept | Formula | Key Note |
| --- | --- | --- |
| Indefinite substitution | $\int f(g(x))g'(x)dx = \int f(u)du, u=g(x)$ | Differentiate back to check |
| Definite substitution | $\int_{x=a}^{x=b} ...dx = \int_{u=g(a)}^{u=g(b)}...du$ | No need to substitute back if changing limits |
| Total area (a to b, x-axis) | $A = \sum \|\int_{x_i}^{x_{i+1}} f(x)dx\|$ | Split at all x-intercepts |

## 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 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.

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