IB Mathematics AA SL IB Mathematics AA SL Definite integrals Study Notes - New Syllabus
IB Mathematics AA SL Definite integrals Study Notes
LEARNING OBJECTIVE
- Definite integrals
Key Concepts:
- Definite integrals
- Areas of a region
- IBDP Maths AA SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IBDP Maths AA SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AA HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Maths AA HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Maths AA HL- IB Style Practice Questions with Answer-Topic Wise-Paper 3
Definite Integrals
Definite Integrals
A definite integral is the evaluation of the integral of a function over a specific interval \([a, b]\), denoted by:
$ \boxed{\int_a^b f(x)\, dx }$
This represents the signed area under the curve \( f(x) \) from \( x = a \) to \( x = b \). A positive value indicates the area above the x-axis, while a negative value represents area below it.
The Fundamental Theorem of Calculus
If \( f(x) \) is continuous on \([a, b]\), and \( F(x) \) is any antiderivative of \( f(x) \), then:
$ \boxed{\int_a^b f(x)\, dx = F(b) – F(a) }$
This theorem allows us to evaluate definite integrals analytically if we know the antiderivative \( F(x) \).
Special Case: Derivative of a Function
If \( f(x) = g'(x) \), then the definite integral simplifies as:
$ \boxed{\int_a^b g'(x) \, dx = g(b) – g(a) }$
Analytical Approach
To evaluate \( \int_a^b f(x)\, dx \) analytically:
- Find the antiderivative \( F(x) \) of \( f(x) \)
- Apply \( F(b) – F(a) \)
Example (Analytical Approach)
Evaluate \( \int_1^3 2x \, dx \)
▶️ Answer/Explanation
Antiderivative of \( 2x \) is \( x^2 \)
Now evaluate from 1 to 3:
Using Technology
When the antiderivative cannot be found easily (e.g., \( e^{-x^2} \), \( \sin(x^2) \)), or the function is given as a data set or graph, technology is used to approximate the integral value.
- Graphing Calculators (GDC): Use the
fnIntcommand or integral function. - Desmos or CAS tools: Visually estimate or calculate the exact value.
Example (Technology Required)
Evaluate \( \int_0^1 e^{-x^2} \, dx \)
▶️ Answer/Explanation
The function \( e^{-x^2} \) has no elementary antiderivative.
Use your GDC:
- TI-nSpire:
∫(e^(-x^2), x, 0, 1) - Result ≈ 0.7468
Final Answer: Approximately 0.7468
Key Notes:
- If the function is not integrable algebraically, always mention: “Use technology to evaluate.”
- Always include the constant of integration \( C \) for indefinite integrals, but not for definite integrals.
- Pay attention to area below the x-axis; the definite integral will be negative in that region.
Area Under a Curve
Area Under a Curve – Signed and Total Area
When calculating the area between a curve \( y = f(x) \) and the x-axis, the definite integral gives the signed area:
$ \boxed{\int_a^b f(x)\, dx} $
This value is positive where the curve lies above the x-axis (i.e. \( f(x) > 0 \)) and negative where it lies below the x-axis (i.e. \( f(x) < 0 \)).
When Total (Geometric) Area is Required
If the curve crosses the x-axis within the interval \([a, b]\), the integral must be split at the points where \( f(x) = 0 \), and the absolute value of each part must be taken to ensure area is positive throughout:
$ \boxed{\text{Total Area} = \int_a^c f(x)\, dx – \int_c^b f(x)\, dx \quad \text{(if } f(x) < 0 \text{ on } [c, b])} $ or $\boxed{ \text{Total Area} = \int_a^c |f(x)|\, dx + \int_c^b |f(x)|\, dx }$
To evaluate this without technology, follow these steps:
- Identify the values of \( x \) where \( f(x) = 0 \) (i.e., x-intercepts)
- Determine sign of \( f(x) \) on each subinterval
- Compute definite integrals over each subinterval
- Take absolute values and sum them to get total area
Example
Find the total area enclosed by the curve \( f(x) = x^2 – 4 \) and the x-axis over the interval \( [-3, 3] \).
▶️ Answer/Explanation
Step 1: Find where \( f(x) = 0 \)
$ x^2 – 4 = 0 \Rightarrow x = \pm2 $ So split the interval at \( x = -2 \) and \( x = 2 \).
From \( x = -3 \) to \( x = -2 \):
\( f(x) < 0 \), so area is \( -\int_{-3}^{-2} (x^2 – 4)\, dx \)
From \( x = -2 \) to \( x = 2 \):
\( f(x) \geq 0 \), so area is \( \int_{-2}^{2} (x^2 – 4)\, dx \)
From \( x = 2 \) to \( x = 3 \):
\( f(x) < 0 \), so area is \( -\int_{2}^{3} (x^2 – 4)\, dx \)
Let’s compute all three: $ \int_{-3}^{-2} (x^2 – 4)\, dx = \left[\frac{x^3}{3} – 4x\right]_{-3}^{-2} = \left( \frac{-8}{3} + 8 \right) – \left( \frac{-27}{3} + 12 \right) = \frac{16}{3} – \frac{9}{3} = \frac{7}{3} $
Important Notes:
- Signed area is found directly via definite integral.
- Total (geometric) area requires splitting at x-intercepts and taking the absolute value.
- When not using technology, break into regions where \( f(x) \) is entirely positive or negative.
Area Between Two Curves
Area Between Two Curves
To find the area enclosed between two curves, say \( y = f(x) \) and \( y = g(x) \), over the interval \([a, b]\), the formula is:
$ \boxed{\text{Area} = \int_a^b \left[ f(x) – g(x) \right] \, dx }$
This assumes that \( f(x) \geq g(x) \) on the interval \([a, b]\), so that the result is positive and represents the actual area.
Steps (Analytical Approach):
- Sketch or analyze the curves to determine which one is on top (i.e., has greater y-values) over the interval.
- Find the points of intersection, if not given.
- Set up the definite integral \( \int_a^b (f(x) – g(x))\, dx \).
- Evaluate the integral using algebraic methods or technology.
Example
Find the area between the curves \( y = x^2 \) and \( y = x + 2 \).
▶️ Answer/Explanation
Find points of intersection by solving: $ x^2 = x + 2 \Rightarrow x^2 – x – 2 = 0 \Rightarrow (x – 2)(x + 1) = 0 $ So \( x = -1 \) and \( x = 2 \)
Determine which function is on top between \( x = -1 \) and \( x = 2 \).
Try \( x = 0 \): \( x + 2 = 2 \), \( x^2 = 0 \) → \( x + 2 > x^2 \)
So the area is: $ \int_{-1}^{2} \left[ (x + 2) – x^2 \right] \, dx $ $ = \int_{-1}^{2} (-x^2 + x + 2)\, dx $
$ = \left[ -\frac{x^3}{3} + \frac{x^2}{2} + 2x \right]_{-1}^{2} $ Compute values:
- At \( x = 2 \): \( -\frac{8}{3} + 2 + 4 = -\frac{8}{3} + 6 = \frac{10}{3} \)
- At \( x = -1 \): \( \frac{1}{3} + \frac{1}{2} – 2 = \frac{5}{6} – 2 = -\frac{7}{6} \)
So, $ \text{Area} = \frac{10}{3} – (-\frac{7}{6}) = \frac{10}{3} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2} $
Technology and Visualization:
- Graphing calculators or tools like Desmos, GeoGebra, or TI-Nspire help visualize the region.
- Use intersection functions to quickly find points of intersection.
- Use the definite integral tool to compute area directly by selecting the region between two curves.
- This approach is useful when curves are complicated or intersect multiple times.
Note: If the functions are given in terms of \( y \) (i.e., \( x = f(y) \)), you can integrate with respect to \( y \) instead:
$ \boxed{\text{Area} = \int_c^d \left[ f(y) – g(y) \right] \, dy }$
Historical Notes on Volume Calculation
- Liu Hui (3rd century CE): Developed an accurate method for calculating the volume of a cylinder using the method of exhaustion, an early precursor to integral calculus.
- Ibn Al-Haytham (965–1040 CE): The first mathematician to calculate the integral of a function to find the volume of a paraboloid, pioneering concepts in integral calculus.
Concept: Volumes of curved solids (like paraboloids) can be found by integrating the cross-sectional area along an axis.
Example:
Find the volume of the solid formed by rotating the parabola \( y = x^2 \) about the x-axis from \( x=0 \) to \( x=2 \).
▶️ Answer/Explanation
$ A(x) = \pi r^2 = \pi (x^2)^2 = \pi x^4 $
Volume is the integral of the area from \( 0 \) to \( 2 \):
$V = \int_0^2 \pi x^4 \, dx = \pi \int_0^2 x^4 \, dx = \pi \left[ \frac{x^5}{5} \right]_0^2 = \pi \times \frac{32}{5} = \frac{32\pi}{5} \approx 20.106 $
Final Answer: \( \displaystyle \frac{32\pi}{5} \approx 20.106 \) units³
