Unit V: Fundamental Theorem of Integral Calculus and Applications

Fundamental Theorem of Integral Calculus

The Fundamental Theorem of Integral Calculus establishes the connection between the two central operations of calculus: differentiation and integration.

Statement of the Theorem

If a function f(x) is continuous on the closed interval [a, b], and F(x) is an antiderivative of f(x) on [a, b], then: ∫_a^b f(x) dx = F(b) - F(a), where F'(x) = f(x).

Key Definitions and Concepts

• Continuous Function: A function whose graph has no breaks, jumps, or vertical asymptotes on the given interval [a, b].
• Antiderivative: A function F(x) whose derivative is the given function f(x).
• Limits of Integration: The lower limit a and the upper limit b that define the interval of integration.

Exam-Oriented Notes

Remember that the statement requires only the formal definition and understanding without proof. Ensure you correctly identify the antiderivative before substituting the upper and lower limits.

Common Mistakes

Students often forget to subtract the value of the antiderivative at the lower limit from the upper limit, writing only F(b) instead of F(b) - F(a).

Definite Integral and its Properties

A definite integral represents the exact accumulation of quantities, often interpreted as the signed area under a curve between two fixed points.

Properties of Definite Integrals (Without Proof)

• Property 1 (Limits Reversal): ∫_a^b f(x) dx = - ∫_b^a f(x) dx. Reversing the limits of integration changes the sign of the integral.
• Property 2 (Interval Splitting): ∫_a^b f(x) dx = ∫_a^c f(x) dx + ∫_c^b f(x) dx, where a < c < b.
• Property 3 (Additive Property for Sub-intervals): Allows breaking a complex domain into simpler, manageable sub-intervals.
• Property 4 (Reflection Property): ∫_a^b f(x) dx = ∫_a^b f(a + b - x) dx. This is one of the most frequently used properties in solving definite integral problems.
• Property 5 (Zero to a Property): ∫₀^a f(x) dx = ∫₀^a f(a - x) dx. A special case of Property 4 where the lower limit is zero.
• Property 6 (Even and Odd Functions):
  • If f(x) is an even function [f(-x) = f(x)], then ∫_-a^a f(x) dx = 2 ∫₀^a f(x) dx.
  • If f(x) is an odd function [f(-x) = -f(x)], then ∫_-a^a f(x) dx = 0.
• Property 7 (Periodic Functions): ∫₀^nT f(x) dx = n ∫₀^T f(x) dx, where T is the period of the function [f(x + T) = f(x)].

Problems on Definite Integrals Using Properties

Using properties of definite integrals simplifies complex integrations that are otherwise difficult or impossible to evaluate directly.

Step-by-Step Problem Solving Approach

1. Label the given integral as equation (1): I = ∫ f(x) dx.
2. Apply a suitable property (most commonly the reflection property ∫_a^b f(x) dx = ∫_a^b f(a + b - x) dx).
3. Label the resulting integral as equation (2).
4. Add equation (1) and equation (2) together [2I = ∫ (f(x) + f(a + b - x)) dx].
5. Simplify the integrand and solve for I.

Practical Example

Evaluate I = ∫₀^π/2 [ sin(x) / (sin(x) + cos(x)) ] dx.

• Step 1: Write as I = ∫₀^π/2 [ sin(x) / (sin(x) + cos(x)) ] dx --- (1)
• Step 2: Use property ∫₀^a f(x) dx = ∫₀^a f(a - x) dx. Here a = π/2. So, replace x with (π/2 - x).
• Step 3: Since sin(π/2 - x) = cos(x) and cos(π/2 - x) = sin(x), we get I = ∫₀^π/2 [ cos(x) / (cos(x) + sin(x)) ] dx --- (2)
• Step 4: Add (1) and (2): 2I = ∫₀^π/2 [ (sin(x) + cos(x)) / (sin(x) + cos(x)) ] dx = ∫₀^π/2 1 dx
• Step 5: Evaluate: [ x ]₀^π/2 = π/2 - 0 = π/2. Therefore, I = π/4.

Important Observations

Look for symmetry in limits and denominators. Integrals with limits from 0 to π/2 or -a to a are prime candidates for properties 5 and 6 respectively.

Application of Definite Integrals to Find the Area Between Simple Curves

Definite integrals can geometrically determine the area bounded by curves and coordinate axes.

Core Concepts

• Area Under a Curve (with respect to x-axis): The area bounded by the curve y = f(x), the x-axis, and the ordinates x = a and x = b is given by Area = ∫_a^b y dx = ∫_a^b f(x) dx.
• Area with respect to y-axis: The area bounded by the curve x = g(y), the y-axis, and the abscissas y = c and y = d is given by Area = ∫_c^d x dy = ∫_c^d g(y) dy.
• Area Between Two Curves: If f(x) and g(x) are two continuous curves such that f(x) ≥ g(x) on [a, b], the area between them is Area = ∫_a^b [f(x) - g(x)] dx.

Step-by-Step Procedure to Find Area

1. Sketch the Curves: Draw a rough, accurate sketch of the given equations to visualize the region.
2. Identify Points of Intersection: Solve the equations simultaneously to find the limits of integration (x-values or y-values).
3. Determine Upper and Lower Functions: Identify which curve lies above the other in the given interval.
4. Set Up the Integral: Write the definite integral using the correct limits and integrand difference.
5. Evaluate: Compute the definite integral to find the total square units of area.

Area Formulas Summary Table

Scenario	Formula	Visual Description
Area under y = f(x) from x = a to x = b	∫ab f(x) dx	Area between curve and x-axis
Area under x = g(y) from y = c to y = d	∫cd g(y) dy	Area between curve and y-axis
Area between two curves y = f(x) and y = g(x)	∫ab [f(x) - g(x)] dx	Area enclosed between upper curve f(x) and lower curve g(x)

Real-World Applications

• Calculating total physical distance traveled from a velocity-time graph.
• Determining total revenue or cost functions in economics from marginal rates.
• Finding physical areas of irregular 2D shapes in engineering design.

Common Mistakes

Failing to find the correct intersection points leads to wrong integration limits. Always subtract the lower curve equation from the upper curve equation to avoid negative area results.