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Calculus - Introduction to Integration (Anti-differentiation)

Grade 11IB

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Definition of Anti-differentiation: Integration is the reverse process of differentiation. If we have a derivative function fβ€²(x)f'(x), integration allows us to find the original function f(x)f(x). Visually, while differentiation finds the slope of a tangent line at a specific point, integration reconstructs the curve whose steepness is described by that gradient function.

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The Constant of Integration (CC): When we differentiate a constant, the result is zero. Consequently, when we integrate, we must add an arbitrary constant +C+ C because we cannot know the original constant value from the derivative alone. Visually, this creates a 'family of curves' where each curve has the same shape but is shifted vertically upward or downward on the Cartesian plane.

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The Power Rule for Integration: To integrate a power of xx, you increase the exponent by 11 and then divide the term by this new exponent. This rule is applicable to all real number exponents nn except for n=βˆ’1n = -1. Visually, this rule reverses the 'power down' effect of differentiation to restore the degree of the polynomial.

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Linearity of Integration: The integral of a sum or difference of functions is equal to the sum or difference of their individual integrals. Additionally, constants can be moved outside the integral sign: ∫kβ‹…f(x)dx=k∫f(x)dx\int k \cdot f(x) dx = k \int f(x) dx. This allows complex expressions to be broken down into simpler, manageable parts.

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Integration of 1/x1/x: The power rule fails when n=βˆ’1n = -1 because it would lead to division by zero. Instead, the integral of 1x\frac{1}{x} is the natural logarithm ln⁑∣x∣+C\ln|x| + C. Visually, this relates to the area under the hyperbola y=1xy = \frac{1}{x}, which behaves differently than standard polynomial curves.

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Exponential Integration: The function exe^x is unique because it is its own derivative and, consequently, its own integral (plus the constant CC). Visually, the slope of the function y=exy = e^x at any point is exactly equal to the yy-value at that point, a property preserved through the integration process.

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Boundary Conditions and Particular Solutions: If we are given a specific point (x,y)(x, y) that the original curve passes through, we can solve for the specific value of CC. This turns an indefinite integral (a family of curves) into a particular solution (one specific curve). Visually, this is like picking one single line out of an infinite stack of parallel vertical shifts.

πŸ“Formulae

∫xndx=xn+1n+1+C,Β whereΒ nβ‰ βˆ’1\int x^n dx = \frac{x^{n+1}}{n+1} + C, \text{ where } n \neq -1

∫kdx=kx+C\int k dx = kx + C

∫1xdx=ln⁑∣x∣+C\int \frac{1}{x} dx = \ln|x| + C

∫exdx=ex+C\int e^x dx = e^x + C

∫[f(x)±g(x)]dx=∫f(x)dx±∫g(x)dx\int [f(x) \pm g(x)] dx = \int f(x) dx \pm \int g(x) dx

∫kf(x)dx=k∫f(x)dx\int k f(x) dx = k \int f(x) dx

πŸ’‘Examples

Problem 1:

Find the indefinite integral: ∫(6x2βˆ’4x+3)dx\int (6x^2 - 4x + 3) dx

Solution:

  1. Apply the sum rule to integrate each term separately: ∫6x2dxβˆ’βˆ«4xdx+∫3dx\int 6x^2 dx - \int 4x dx + \int 3 dx
  2. Apply the power rule to 6x26x^2: 6x2+12+1=6x33=2x3\frac{6x^{2+1}}{2+1} = \frac{6x^3}{3} = 2x^3
  3. Apply the power rule to 4x14x^1: 4x1+11+1=4x22=2x2\frac{4x^{1+1}}{1+1} = \frac{4x^2}{2} = 2x^2
  4. Apply the constant rule to 33: 3x3x
  5. Combine the terms and add the constant of integration: 2x3βˆ’2x2+3x+C2x^3 - 2x^2 + 3x + C

Explanation:

This example demonstrates the basic application of the power rule and the linearity of integration across a polynomial expression.

Problem 2:

Find the particular equation of the curve y=f(x)y = f(x) given that fβ€²(x)=3x2βˆ’2f'(x) = 3x^2 - 2 and the curve passes through the point (2,10)(2, 10).

Solution:

  1. Find the general solution by integrating fβ€²(x)f'(x): f(x)=∫(3x2βˆ’2)dxf(x) = \int (3x^2 - 2) dx
  2. Calculate the integral: f(x)=3x33βˆ’2x+C=x3βˆ’2x+Cf(x) = \frac{3x^3}{3} - 2x + C = x^3 - 2x + C
  3. Use the given point (2,10)(2, 10) to find CC. Substitute x=2x = 2 and f(x)=10f(x) = 10: 10=(2)3βˆ’2(2)+C10 = (2)^3 - 2(2) + C
  4. Simplify: 10=8βˆ’4+Cβ‡’10=4+Cβ‡’C=610 = 8 - 4 + C \Rightarrow 10 = 4 + C \Rightarrow C = 6
  5. State the final equation: f(x)=x3βˆ’2x+6f(x) = x^3 - 2x + 6

Explanation:

This example shows how to use a specific coordinate (boundary condition) to solve for the constant CC, identifying a unique function from the family of possible integrals.