# Definite integrals and area under a curve

> IB Mathematics: Applications and Interpretation SL · IB AI SL Unit 5: Calculus
> Source: https://www.owlsprep.com/study/ib-math-ai-sl-u5-definite-integrals-and-area-under/

This module covers how to use the Fundamental Theorem of Calculus to evaluate definite integrals, and calculate the area between a curve and the x-axis. You will learn the key difference between net signed area and total geometric area.

**Prerequisites:** [Indefinite integration and basic antiderivatives](https://www.owlsprep.com/study/ib-math-ai-sl-u5-indefinite-integration-antiderivatives/); [Function graphing and intercepts](https://www.owlsprep.com/study/ib-math-ai-sl-u2-functions-graphs/)

## Learning objectives

- Evaluate definite integrals using antiderivatives and the Fundamental Theorem of Calculus
- Calculate the area between a curve and the x-axis over a given interval
- Distinguish between net signed area (definite integral value) and total geometric area
- Use a GDC to evaluate definite integrals for exam-style problems

## Definite Integrals and the Fundamental Theorem of Calculus

**Definite Integral** — The net signed area between the curve $y = f(x)$, the x-axis, and vertical lines $x=a$ (lower bound) and $x=b$ (upper bound), calculated via the Fundamental Theorem of Calculus (FTC).

*Notation:* $\int_a^b f(x) \, dx$

*Example:* $\int_0^2 2x dx$ is the definite integral of $2x$ from 0 to 2.

The FTC connects differentiation and integration, eliminating the need to approximate area with rectangles for most integrable functions. The core rule is:

$$\int_a^b f(x) \, dx = F(b) - F(a) \quad \text{where} \quad F'(x) = f(x)$$

**Worked example:** Evaluate the definite integral $\int_1^3 (2x + 1) dx$

1. 1. Find the antiderivative of $(2x + 1)$. The constant of integration cancels in definite integrals, so we can ignore it.
2. $$F(x) = \int (2x + 1) dx = x^2 + x$$
3. 2. Evaluate $F(x)$ at the upper bound $x=3$ and lower bound $x=1$:
4. $$F(3) = 3^2 + 3 = 12, \quad F(1) = 1^2 + 1 = 2$$
5. 3. Subtract lower bound value from upper bound value:
6. $$F(3) - F(1) = 12 - 2 = 10$$
7. Final result: $\int_1^3 (2x + 1) dx = 10$

## Area for Curves Entirely Above the x-Axis

When $f(x) \geq 0$ for all $x$ in $[a, b]$, the entire region between the curve and the x-axis is above the axis. In this case, the value of the definite integral is exactly equal to the geometric area of the region, so no extra adjustments are needed.

> **tip**
>
> Always check if your function crosses the x-axis between the bounds before you integrate. If it crosses, the integral will not equal the total area.

**Worked example:** Find the area under the curve $y = x^2 + 1$ between $x=0$ and $x=2$

1. 1. Check the sign of the function on $[0, 2]$: $x^2$ is always non-negative, so $x^2 + 1 \geq 1 > 0$. The entire region is above the x-axis.
2. 2. Find the antiderivative:
3. $$F(x) = \frac{x^3}{3} + x$$
4. 3. Apply the FTC to get the area:
5. $$F(2) - F(0) = \left(\frac{8}{3} + 2\right) - 0 = \frac{14}{3} \approx 4.67$$
6. Area = $\frac{14}{3} \approx 4.67$ square units.

## Area When the Curve Crosses the x-Axis

When $f(x)$ is negative for any part of $[a, b]$, the definite integral calculates net signed area, where regions below the x-axis contribute negative values. This means the integral will be smaller than the total geometric area. To get total area, split the integral at each x-intercept, integrate each interval separately, take the absolute value of each result, then add them together.

**Total Geometric Area** — The sum of the absolute values of the signed areas of each region between the curve and the x-axis. It is always non-negative, unlike the definite integral which can be negative.

**Worked example:** Find the total area between $y = x$ and the x-axis between $x=-2$ and $x=3$

1. 1. Find the x-intercept: $y = x = 0$ at $x=0$. $y < 0$ for $x < 0$ and $y > 0$ for $x>0$.
2. 2. Split the integral into two intervals: $[-2, 0]$ and $[0, 3]$. Find the antiderivative $F(x) = \frac{x^2}{2}$.
3. 3. Calculate each definite integral:
4. $$\int_{-2}^0 x dx = F(0) - F(-2) = 0 - 2 = -2$$
5. $$\int_0^3 x dx = F(3) - F(0) = 4.5 - 0 = 4.5$$
6. 4. Take absolute values and add for total area:
7. $$\text{Area} = |-2| + |4.5| = 6.5 = \frac{13}{2} \text{ square units}$$
8. Note: The net definite integral is $2.5$, which is not equal to the total area.

## Evaluating Definite Integrals with a GDC

For IB AI SL, you can use your graphing display calculator (GDC) to evaluate definite integrals directly, even for complex functions like exponentials, trigonometric functions, or higher-order polynomials that are time-consuming to integrate by hand.

**Worked example:** Use your GDC to evaluate $\int_0^\pi \sin(x) dx$, correct to 3 significant figures

1. 1. Graph $y = \sin(x)$ on your GDC, confirm it is non-negative on $[0, \pi]$.
2. 2. Open the GDC's integral function, select your function, input lower bound $0$ and upper bound $\pi$.
3. 3. The GDC returns 2.00, which matches the analytical result:
4. $$\int_0^\pi \sin(x) dx = -\cos(\pi) + \cos(0) = 1 + 1 = 2$$
5. Final result: $2.00$ (3 s.f.)

> **tip**
>
> Always write the integral expression in your exam response before using your GDC. You will earn method marks for correct setup even if your final answer is wrong.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to split the integral at x-intercepts and integrating directly from $a$ to $b$.
  - Why it fails: This gives the net signed area, not the total geometric area that the question almost always asks for.
  - Correct: Check for x-intercepts between $a$ and $b$, split the integral at each intercept, then add absolute values of each integral.
- **Wrong:** Subtracting $F(b)$ (upper bound) from $F(a)$ (lower bound) instead of the reverse.
  - Why it fails: Swapping the order of bounds flips the sign of the integral, leading to an incorrect negative area value.
  - Correct: Always remember FTC: definite integral equals $F(\text{upper}) - F(\text{lower})$.
- **Wrong:** Taking the absolute value of the total integral instead of each individual split integral.
  - Why it fails: Positive and negative areas cancel out inside the absolute value, leading to an area that is too small.
  - Correct: Split the integral at every intercept, take the absolute value of each separate result, then sum them.
- **Wrong:** Only writing the final GDC result without setting up the integral first.
  - Why it fails: Examiners award method marks for correct setup, which you will lose if you only write the final number.
  - Correct: Always write the full definite integral expression explicitly in your working before stating your GDC result.

## Cheatsheet

| Scenario | Area Formula | Key Note |
| --- | --- | --- |
| $f(x) \geq 0$ on $[a,b]$ | Area = $\int_a^b f(x) dx$ | No adjustment needed |
| $f(x) \leq 0$ on $[a,b]$ | Area = $\left\|\int_a^b f(x) dx\right\|$ | Integral is negative, flip sign for area |
| Crosses x-axis once at $c \in (a,b)$ | Area = $\left\|\int_a^c f(x) dx\right\| + \left\|\int_c^b f(x) dx\right\|$ | Split at intercept, add absolute values |
| FTC Rule | $\int_a^b f(x) dx = F(b) - F(a), F'(x)=f(x)$ | Constant of integration cancels out |

## What's next

Now that you understand definite integrals and area under a single curve, you can extend this knowledge to more advanced integration problems common in IB AI SL exams. The next step is learning how to calculate the area between two different curves, which uses the same core idea of splitting integrals and absolute values, just with an adjusted integrand. You can also build on this foundation to learn about applications of integration to kinematics, where definite integrals are used to calculate displacement and total distance traveled from velocity functions. This topic is also the base for learning integration by substitution, a method to integrate more complex functions that appear frequently on Paper 1.

- [Kinematic applications of calculus](https://www.owlsprep.com/study/ib-math-ai-sl-u5-kinematic-applications-of-calculus/)

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