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Calculus: A Comprehensive Exploration
Introduction
Welcome to the fascinating world of calculus! This branch of mathematics provides the tools to understand and model change. It has wide-ranging applications in various fields like physics, engineering, economics, and computer science. This course will cover the fundamental concepts of calculus, including limits, derivatives, and integrals.
Chapter 1: Limits and Continuity
1.1 Limits
● Definition : The limit of a function f(x) as x approaches a value c is the value that f(x) gets closer and closer to as x gets closer and closer to c. ● Notation : lim (x→c) f(x) = L ● Epsilon-Delta Definition (More Rigorous): For every ε > 0, there exists a δ > 0 such that if 0 < |x - c| < δ, then |f(x) - L| < ε ● Techniques for finding limits : ○ Direct substitution ○ Factoring ○ Rationalizing ○ L'Hôpital's Rule (for indeterminate forms) ○ Squeeze Theorem ● One-Sided Limits : ○ Limit from the left: lim (x→c-) f(x) ○ Limit from the right: lim (x→c+) f(x) ● Infinite Limits and Limits at Infinity : ○ Vertical asymptotes ○ Horizontal asymptotes Example 1 : Find the limit of the function f(x) = (x² - 1) / (x - 1) as x approaches 1. Solution : lim (x→1) (x² - 1) / (x - 1) = lim (x→1) (x + 1)(x - 1) / (x - 1) = lim (x→1) (x + 1) = 1 + 1 = 2 Example 2 : Find the limit of the function f(x) = sin(x)/x as x approaches 0. Solution: Using L'Hôpital's Rule: lim (x→0) sin(x)/x = lim (x→0) cos(x)/1 = cos(0)/1 = 1 Example 3 : Find the limit of the function f(x) = (√(x+1) - 1)/x as x approaches 0.
Solution: Rationalizing the numerator: lim (x→0) (√(x+1) - 1)/x * (√(x+1) + 1)/(√(x+1) + 1) = lim (x→0) (x+1 - 1)/(x(√(x+1) + 1)) = lim (x→0) x/(x(√(x+1) + 1)) = lim (x→0) 1/(√(x+1) + 1) = 1/(√(0+1) + 1) = 1/ Example 4 : Find the limit of the function f(x) = x^2 * sin(1/x) as x approaches 0. Solution: Using the Squeeze Theorem: Since -1 ≤ sin(1/x) ≤ 1, we have -x^2 ≤ x^2 * sin(1/x) ≤ x^ lim (x→0) -x^2 = 0 and lim (x→0) x^2 = 0 Therefore, by the Squeeze Theorem, lim (x→0) x^2 * sin(1/x) = 0
1.2 Continuity
● Definition : A function f(x) is continuous at x = c if:
- f(c) is defined
- lim (x→c) f(x) exists
- lim (x→c) f(x) = f(c) ● Types of discontinuity : ○ Removable discontinuity (can be "removed" by redefining the function at that point) ○ Jump discontinuity (the function "jumps" from one value to another) ○ Infinite discontinuity (the function approaches infinity) ○ Essential Discontinuity ● Continuity on an Interval: A function is continuous on an open interval (a,b) if it is continuous at each point in the interval. A function is continuous on a closed interval [a,b] if it is continuous on (a,b) and continuous from the right at a and continuous from the left at b. ● The Intermediate Value Theorem: If f is continuous on the closed interval [a, b] and k is any number between f(a) and f(b), then there exists a number c in [a, b] such that f(c) = k. Example 2 : Determine the continuity of the function f(x) = { x², x ≤ 1; 2x - 1, x > 1 } at x = 1. Solution :
- f(1) = 1² = 1
- lim (x→1-) f(x) = lim (x→1-) x² = 1 lim (x→1+) f(x) = lim (x→1+) 2x - 1 = 1 lim (x→1) f(x) = 1
- lim (x→1) f(x) = f(1) = 1 Therefore, the function is continuous at x = 1. Example 3 : Determine the continuity of the function f(x) = 1/x at x = 0. Solution: f(0) is not defined.
= lim (h→0) 1 / [√(x + h) + √x] = 1 / 2√x Example 5 : Find the derivative of f(x) = 1/x using the definition of the derivative. Solution : f'(x) = lim (h→0) [1/(x+h) - 1/x] / h = lim (h→0) [x - (x+h)] / hx(x+h) = lim (h→0) -h / hx(x+h) = lim (h→0) -1 / x(x+h) = -1/x^
2.2 Differentiation Rules
● Power Rule: d/dx (x ) = nxⁿ ⁿ⁻¹ ● Product Rule: d/dx [f(x)g(x)] = f'(x)g(x) + f(x)g'(x) ● Quotient Rule: d/dx [f(x)/g(x)] = [g(x)f'(x) - f(x)g'(x)] / [g(x)]² ● Chain Rule: d/dx [f(g(x))] = f'(g(x))g'(x) ● Derivatives of Trigonometric Functions: ○ d/dx sin(x) = cos(x) ○ d/dx cos(x) = -sin(x) ○ d/dx tan(x) = sec²(x) ○ d/dx csc(x) = -csc(x)cot(x) ○ d/dx sec(x) = sec(x)tan(x) ○ d/dx cot(x) = -csc²(x) ● Derivatives of Exponential and Logarithmic Functions: ○ d/dx e^x = e^x ○ d/dx a^x = a^x ln(a) ○ d/dx ln(x) = 1/x ○ d/dx log (x) = 1/(x ln(a))ₐ ● Implicit Differentiation: Used when the function is not explicitly defined in terms of x. ● Higher Order Derivatives: The second derivative, third derivative, and so on, obtained by repeatedly differentiating the function. Example 4 : Find the derivative of y = x³sin(x). Solution: Using the Product Rule: dy/dx = (d/dx x³)sin(x) + x³(d/dx sin(x)) = 3x²sin(x) + x³cos(x) Example 5 : Find the derivative of y = sin(x²).
Solution: Using the Chain Rule: dy/dx = cos(x²) * (d/dx x²) = cos(x²) * 2x = 2xcos(x²) Example 6 : Find the derivative of y = (x^2 + 1) / (x - 1). Solution: Using the Quotient Rule: dy/dx = [(x-1)(d/dx (x^2 + 1)) - (x^2 + 1)(d/dx (x-1))] / (x-1)^ = [(x-1)(2x) - (x^2 + 1)(1)] / (x-1)^ = (2x^2 - 2x - x^2 - 1) / (x-1)^ = (x^2 - 2x - 1) / (x-1)^ Example 7 : Find the derivative of x^2 + y^2 = 25 using implicit differentiation. Solution: Differentiating both sides with respect to x: d/dx (x^2 + y^2) = d/dx (25) 2x + 2y(dy/dx) = 0 2y(dy/dx) = -2x dy/dx = -x/y
2.3 Applications of Derivatives
● Related rates : Problems involving finding the rate at which a quantity changes by relating it to other quantities whose rates of change are known. ● Optimization : Finding the maximum or minimum values of a function. ○ Critical points ○ First derivative test ○ Second derivative test ○ Absolute maximum and minimum ● Curve sketching : Using derivatives to analyze the behavior of a function and sketch its graph. ○ Increasing and decreasing intervals ○ Concavity and inflection points ○ স্থা Zনী\য় চরম মZনী এবংc চরম মZনী (Local maxima and minima) ○ Asymptotes ● Linear Approximation and Differentials: Using the tangent line to approximate the function near a point. ● Newton's Method: An iterative method for finding the roots of a function.
1 < t < 4 The acceleration is 0 when 12t - 30 = 0, which gives t = 30/12 = 5/2. Chapter 3: Integrals
3.1 Definition of the Integral
● The integral of a function f(x) is the reverse process of differentiation. ● Indefinite integral : ∫ f(x) dx = F(x) + C, where F'(x) = f(x) ○ C is the constant of integration ● Definite integral : ∫ <0,b</0,b> f(x) dx represents the net signed area under the curveₐ of f(x) from x = a to x = b. ○ a and b are the limits of integration ● Riemann Sums: Approximating the definite integral using sums of areas of rectangles. ○ Left Riemann Sum ○ Right Riemann Sum ○ Midpoint Rule ● The Fundamental Theorem of Calculus : ○ Part 1: If f is continuous on [a, b], then the function g defined by g(x) = ∫ <0,x</0,x> f(t) dt is continuous on [a, b] and differentiable on (a, b), and g'(x) =ₐ f(x). ○ Part 2: If f is continuous on [a, b], then ∫ <0,b</0,b> f(x) dx = F(b) - F(a), where Fₐ is any antiderivative of f. Example 6 : Find the indefinite integral of f(x) = 2x. Solution : ∫ 2x dx = x² + C Example 7 : Evaluate the definite integral ∫₀<0,2</0,2> x² dx. Solution : ∫₀<0,2</0,2> x² dx = [x³/3]₀<0,2</0,2> = (2³/3) - (0³/3) = 8/ Example 8 : Evaluate the definite integral ∫₁<0,3</0,3> (x^2 + 2x - 1) dx Solution : ∫₁<0,3</0,3> (x^2 + 2x - 1) dx = [x^3/3 + x^2 - x]_1^ = (3^3/3 + 3^2 - 3) - (1^3/3 + 1^2 - 1) = (9 + 9 - 3) - (1/3 + 1 - 1) = 15 - 1/ = 44/ Example 9 : Evaluate the indefinite integral ∫ (1/x + e^x + sin(x)) dx Solution :
∫ (1/x + e^x + sin(x)) dx = ∫ 1/x dx + ∫ e^x dx + ∫ sin(x) dx = ln|x| + e^x - cos(x) + C
3.2 Techniques of Integration
● Substitution: ∫ f(g(x))g'(x) dx = ∫ f(u) du ● Integration by parts: ∫ u dv = uv - ∫ v du ● Trigonometric integrals ● Trigonometric Substitution ● Partial fractions ● Improper Integrals: Integrals where the interval of integration is unbounded or the integrand has a discontinuity within the interval of integration. Example 8 : Evaluate the integral ∫ x sin(x) dx. Solution: Using integration by parts: Let u = x, dv = sin(x) dx du = dx, v = -cos(x) ∫ x sin(x) dx = -xcos(x) - ∫ -cos(x) dx = -xcos(x) + ∫ cos(x) dx = -xcos(x) + sin(x) + C Example 9 : Evaluate the integral ∫ 1/(x² - 1) dx Solution: Using partial fractions: 1/(x² - 1) = A/(x - 1) + B/(x + 1) 1 = A(x + 1) + B(x - 1) A = 1/2, B = -1/ ∫ 1/(x² - 1) dx = 1/2 ∫ [1/(x - 1) - 1/(x + 1)] dx = 1/2 [ln|x - 1| - ln|x + 1|] + C = 1/2 ln|(x - 1)/(x + 1)| + C Example 10 : Evaluate the integral ∫ x√(x^2 + 1) dx Solution: Using substitution: Let u = x^2 + 1, du = 2x dx ∫ x√(x^2 + 1) dx = 1/2 ∫ √u du = 1/2 * (2/3)u^(3/2) + C = (1/3)(x^2 + 1)^(3/2) + C Example 11 : Evaluate the improper integral ∫₀<0,∞</0,∞> e^(-x) dx Solution : ∫₀<0,∞</0,∞> e^(-x) dx = lim (b→∞) ∫₀<0,b</0,b> e^(-x) dx = lim (b→∞) [-e^(-x)]_0^b
The arc length L is given by: L = ∫₀<0,4</0,4> √(1 + (dy/dx)²) dx dy/dx = (3/2)x^(1/2) (dy/dx)² = (9/4)x L = ∫₀<0,4</0,4> √(1 + (9/4)x) dx Using substitution, let u = 1 + (9/4)x, du = (9/4) dx L = (4/9) ∫₁<0,10</0,10> √u du = (4/9) * (2/3) [u^(3/2)]_1^ = (8/27) (10^(3/2) - 1) = (8/27) (10√10 - 1)
