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IB Mathematics AI SL Integration as anti-differentiation MAI Study Notes - New Syllabus

IB Mathematics AI SL Integration as anti-differentiation MAI Study Notes

LEARNING OBJECTIVE

  • Introduction to integration as anti-differentiation of functions

Key Concepts: 

  • Anti-differentiation of functions
  • Definite integrals using technology. 
  • Area of a region enclosed by a curve

MAI HL and SL Notes – All topics

INTEGRATION AS ANTI-DIFFERENTIATION

Integration as Anti-Differentiation

Anti-differentiation (or indefinite integration) is the reverse process of differentiation. If:

\( \frac{d}{dx} F(x) = f(x) \)

Then we say:

\( \int f(x)\,dx = F(x) + C \), where \( C \) is the constant of integration.

Example: Polynomial Anti-Differentiation

Find the general solution of the differential equation:
\( \frac{dy}{dx} = 5x^4 – 3x^3 + 2x^{-2} – \frac{1}{x^3} + 7 \)

▶️ Answer/Explanation
  • Use anti-differentiation (power rule) term-by-term:
  • \( \int 5x^4\,dx = x^5 \)
  • \( \int -3x^3\,dx = -\frac{3}{4}x^4 \)
  • \( \int 2x^{-2}\,dx = -2x^{-1} \)
  • \( \int -x^{-3}\,dx = \frac{1}{2}x^{-2} \)
  • \( \int 7\,dx = 7x \)
  • Final Answer:
    \( \displaystyle y = x^5 – \frac{3}{4}x^4 – \frac{2}{x} + \frac{1}{2x^2} + 7x + C \)

 Polynomial Anti-Differentiation

Rule: For any term of the form \( ax^n+bx^{n-1} \), where \( n \neq -1 \), the anti-derivative is:

\( \int ax^n dx = \frac{a}{n+1}x^{n+1} + C \)

Note: Do not forget to add the constant of integration \( C \).

Example : Polynomial Anti-Differentiation

Find \( \int (3x^2 – 4x + 7)\,dx \)

▶️ Answer/Explanation

Solution:

  • \( \int 3x^2\,dx = x^3 \)
  • \( \int -4x\,dx = -2x^2 \)
  • \( \int 7\,dx = 7x \)
  • Final Answer: \( x^3 – 2x^2 + 7x + C \)

BOUNDARY CONDITIONS AND THE CONSTANT TERM

Boundary Conditions and the Constant Term

When given a specific point on the graph, we can find the constant \( C \).

Key Idea: If \( F(x) \) is the anti-derivative of \( f(x) \), and we know \( F(x_0) = y_0 \), then substitute into the equation to find \( C \).

Example : Boundary Conditions

Given \( \frac{dy}{dx} = 2x + 1 \), and \( y = 5 \) when \( x = 2 \), find the original function.

▶️ Answer/Explanation
  • \( \int (2x + 1)dx = x^2 + x + C \)
  • Use the point \( (2, 5) \):
  • \( 5 = 4 + 2 + C \Rightarrow C = -1 \)
  • Final Answer: \( y = x^2 + x – 1 \)

DEFINITE INTEGRALS AND TECHNOLOGY

Definite Integrals and Technology

A definite integral calculates the net area under the curve between two bounds \( a \) and \( b \):

\( \int_a^b f(x)\,dx = F(b) – F(a) \)

Use a GDC (Graphing Display Calculator) or software like Desmos to evaluate definite integrals when the expression is complex.

Steps to Use a GDC (TI-84 or Desmos):

  • On a TI-84, press `MATH` → select `fnInt(` or `∫(`.
  • On Desmos, just type the integral: `int(f(x), a, b)`.
  • Make sure the function is correctly typed (use parentheses if needed).
  • Lower limit: $a$ and Upper limit: $b$
  • Press `ENTER` (on GDC) or view the result (in Desmos).

Example

Evaluate the definite integral:

$\int_0^2 (x^2 + 1) \, dx$

▶️Answer/Explanation

Solution:

  • Open Desmos or a graphing calculator.
  • Type: ∫(x^2 + 1, x, 0, 2)
  • The result is: \(\boxed{4.67}\) (rounded to 2 decimal places)

So, the area under the curve from \(x = 0\) to \(x = 2\) is approximately 4.67.

Link Between Anti-Derivatives, Definite Integrals, and Area

The definite integral \( \int_a^b f(x)\,dx \) gives the signed area under the curve between \( x = a \) and \( x = b \).

If \( f(x) \ge 0 \), the integral gives the area above the x-axis.

If \( f(x) \le 0 \), the integral is negative, corresponding to area below the x-axis.

Example : Definite Integral

Evaluate \( \int_1^3 (x^2 – 2x + 1)\,dx \)

▶️ Answer/Explanation
  • Antiderivative: \( F(x) = \frac{1}{3}x^3 – x^2 + x \)
  • \( F(3) = 9 – 9 + 3 = 3 \)
  • \( F(1) = \frac{1}{3} – 1 + 1 = \frac{1}{3} \)
  • Answer: \( 3 – \frac{1}{3} = \frac{8}{3} \)

AREA BETWEEN A CURVE AND THE X-AXIS

Area Between a Curve and the x-axis

If a function crosses the x-axis between $a$ and $c$, the definite integral must be split at the point of intersection $b$, and the absolute value of each integral must be taken to find the total area.

That is, the total area is $A_1 + A_2 = \left| \int_a^b f(x) \, dx \right| + \left| \int_b^c f(x) \, dx \right|$.

$A_1$ is below the x-axis (negative integral),
$A_2$ is above the x-axis (positive integral),
and $f(x)$ crosses the x-axis at $b$.

Example : Area Between Curve and x-axis

Find the area between the curve \( f(x) = x^2 – 4 \) and the x-axis over \( [-3, 3] \).

▶️ Answer/Explanation
  • The function crosses x-axis at \( x = -2 \) and \( x = 2 \).
  • Break into 3 intervals and evaluate:
  • \( A_1 = \int_{-3}^{-2}(x^2 – 4)\,dx \), \( A_2 = \left| \int_{-2}^{2}(x^2 – 4)\,dx \right| \), \( A_3 = \int_2^3(x^2 – 4)\,dx \)
  • Using calculator, total area ≈ 8.67
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