IB Mathematics AA SL 5.5 Introduction to integration as anti differentiation Study Notes- New Syllabus
IB Mathematics AA SL 5.5 Introduction to integration as anti differentiation Study Notes
IB Mathematics AA SL 5.5 Introduction to integration as anti differentiation Study Notes Offer a clear explanation of Use of Tangents and normals at a given points , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Tangents and normals at a given points.
Introduction to Integration as Anti-differentiation
Introduction to Integration as Anti-differentiation
Integration is the reverse of differentiation. For functions of the form: \( f(x) = ax^n + bx^{n-1} + \cdots \) where \( n \in \mathbb{Z} \) and \( n \neq -1 \), the anti-derivative is:
\( \int f(x)\,dx = \frac{a}{n+1} x^{n+1} + \frac{b}{n} x^{n} + \cdots + C \)
For each term
\( kx^m \): \( \int kx^m dx = \frac{k}{m+1} x^{m+1} + C \quad (m \neq -1) \)
The constant of integration \( C \) represents the family of functions.
Example:
Find the indefinite integral: \( \int \left( 3x^4 – 2x^{-2} + 5 \right) dx \)
▶️ Answer/Explanation
Integrate term by term:
- \( \int 3x^4 dx = 3 \cdot \frac{1}{5} x^5 = \frac{3}{5} x^5 \)
- \( \int (-2x^{-2}) dx = -2 \cdot \frac{1}{-1} x^{-1} = 2x^{-1} \)
- \( \int 5 dx = 5x \)
Final answer:
\( \int \left( 3x^4 – 2x^{-2} + 5 \right) dx = \frac{3}{5} x^5 + 2x^{-1} + 5x + C \)
Anti-differentiation with Boundary Condition
Anti-differentiation with Boundary Condition
When you integrate a function, the result includes a constant of integration \( C \). A boundary condition (or initial condition) is used to find \( C \) by substituting a known point \((x_0, y_0)\).
Process:
- Integrate the function (indefinite integral).
- Substitute the known \( x_0 \) and \( y_0 \) into the integrated expression.
- Solve for \( C \).
General form: If \( \frac{dy}{dx} = f(x) \) and \( y(x_0) = y_0 \), then:
\( y = \int f(x) dx + C \) Find \( C \) using \( y_0 = y(x_0) \).
Example:
A function satisfies \( \frac{dy}{dx} = 6x^2 – 4x \). Given that \( y = 3 \) when \( x = 1 \), find \( y \).
▶️ Answer/Explanation
\( y = \int (6x^2 – 4x) dx \) \( = 6 \cdot \frac{x^3}{3} – 4 \cdot \frac{x^2}{2} + C \) \( = 2x^3 – 2x^2 + C \)
Use boundary condition
\( y = 3 \) when \( x = 1 \):
\( 3 = 2(1)^3 – 2(1)^2 + C \)
\( 3 = 2 – 2 + C \)
\( 3 = 0 + C \)
\( C = 3 \)
Final solution:
\( y = 2x^3 – 2x^2 + 3 \)
Definite Integrals Using Technology
Definite Integrals Using Technology
A definite integral calculates the area under the curve of \( f(x) \) between two bounds \( a \) and \( b \):
\( \int_a^b f(x)\, dx \)
How to compute using technology:
- Enter the function into the calculator or graphing tool.
- Access the integral or calculation menu.
- TI-Nspire: Menu → Calculus → Integral → enter bounds and function.
- Casio: Interactive → Calculate → ∫ → enter bounds and function.
- Desmos: Type
integrate(f(x), a, b)directly.
- The tool computes and displays the numerical value of the integral.
Technology will handle even complicated integrals where manual work would be slow or prone to error.
Example:
Use technology to compute: \( \int_1^3 \left( 2x^2 – x + 1 \right) dx \)
▶️ Answer/Explanation
- Enter: \( 2x^2 – x + 1 \) into the graphing calculator or Desmos.
- Choose integral calculation:
- TI-Nspire: Menu → Calculus → Integral → enter 1 as lower limit, 3 as upper limit → select function.
- Casio: Interactive → Calculate → ∫ → enter 1, 3, and function.
- Desmos: Type:
integrate(2x^2 - x + 1, 1, 3)
- The calculator shows: \( \int_1^3 (2x^2 – x + 1) dx = 14 \)
Area Under a Curve \( y = f(x) \) and the x-axis (where \( f(x) > 0 \))
Area Under a Curve \( y = f(x) \) and the x-axis (where \( f(x) > 0 \))
If \( f(x) \ge 0 \) on an interval \([a, b]\), the area between the curve and the x-axis is:
\( \text{Area} = \int_a^b f(x) \, dx \)
The integral gives the exact area beneath the curve from \( x = a \) to \( x = b \).
When \( f(x) > 0 \), the integral directly gives the positive area.
How to compute:
- Identify the interval \([a, b]\).
- Set up and evaluate the definite integral: \( \int_a^b f(x) dx \)
- Use technology or manual integration.
Example:
Find the area enclosed by the curve \( y = x^2 + 1 \) and the x-axis from \( x = 0 \) to \( x = 2 \).
▶️ Answer/Explanation
\( \text{Area} = \int_0^2 (x^2 + 1) dx \)
Integrate
\( = \left[ \frac{x^3}{3} + x \right]_0^2 \)
Compute the result
\( = \left( \frac{8}{3} + 2 \right) – \left( 0 + 0 \right) = \frac{8}{3} + 2 = \frac{8}{3} + \frac{6}{3} = \frac{14}{3} \)
Final Answer:
The area is: \( \frac{14}{3} \text{ square units} \)
Using Technology to Understand Area Under a Curve
Using Technology to Understand Area Under a Curve
Dynamic geometry tools (such as GeoGebra) and graphing calculators (such as TI-Nspire, Casio GDC, or Desmos) are valuable in exploring and understanding the concept of area under a curve.
Example of using technology:
In Desmos, you can enter: integrate(f(x), a, b) and visually display the area by shading.
In TI-Nspire / Casio GDC: Menu → Calculus → Integral → select bounds → shaded region appears automatically.
Example:
Use graphing software to find the area under \( y = 2x + 1 \) from \( x = 1 \) to \( x = 4 \).
▶️ Answer/Explanation
- Enter \( f(x) = 2x + 1 \) into the graphing calculator or Desmos.
- Set integral bounds: \( x = 1 \) and \( x = 4 \).
- Use integral tool:
- TI-Nspire: Menu → Calculus → Integral → select bounds → shaded region appears + area displayed.
- Desmos: Type:
integrate(2x + 1, 1, 4).
- The technology displays: \( \int_1^4 (2x + 1) dx = 18 \)
Velocity-Time Graphs (with Technology)
A velocity-time graph represents how velocity \( v(t) \) of an object changes over time.
- The gradient of the graph: gives the acceleration \( a(t) \).
- The area under the graph (between the curve and the time axis): gives the displacement or distance travelled.
Dynamic geometry and graphing software (TI-Nspire, Casio fx-CG50, Desmos, GeoGebra) help:
- Visualize how velocity changes over time
- Automatically compute gradients (acceleration) at points
- Compute areas under curves (displacement)
- Animate motion to link graph to physical movement
Technology Workflow:
- Enter \( v(t) \) into graphing tool.
- Plot the velocity-time graph.
- Use slope/derivative tools to find acceleration at specific times.
- Use integral tools to compute displacement over intervals.
Example:
A particle moves with velocity \( v(t) = t^2 – 2t + 3 \) m/s. Use technology to:
- Find its displacement from \( t = 0 \) to \( t = 3 \) seconds.
- Find its acceleration at \( t = 2 \) seconds.
▶️ Answer/Explanation
- Enter \( v(t) = t^2 – 2t + 3 \) into the GDC or Desmos.
- Plot graph for \( 0 \le t \le 3 \).
- Use integral tool:
TI-Nspire: Menu → Calculus → Integral → lower: 0, upper: 3 → function: \( v(t) \)
Casio: Interactive → Calculate → ∫ → enter 0, 3, \( v(t) \)
Desmos:integrate(t^2 - 2t + 3, 0, 3) - The calculator shows: \( \int_0^3 (t^2 – 2t + 3) dt = 7.5 \text{ m} \)
- Use derivative tool at \( t = 2 \):
TI-Nspire: Menu → Analyze Graph → Derivative → point: 2
Casio: Interactive → Calculate → d/dt → enter 2
Or compute manually: \( a(t) = v'(t) = 2t – 2 \) \( a(2) = 2(2) – 2 = 2 \text{ m/s}^2 \)