






Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
This document offers a thorough introduction to fundamental calculus concepts, including limits, continuity, derivatives, and integrals. it explains key definitions, theorems (like the intermediate value theorem and fundamental theorem of calculus), and various techniques for evaluating limits and integrals, supported by numerous examples. The clear explanations and worked-out examples make it an excellent resource for students learning calculus.
Typology: Study notes
1 / 10
This page cannot be seen from the preview
Don't miss anything!
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.
● 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
● Definition : A function f(x) is continuous at x = c if:
= 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^
● 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
● 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
● 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
● 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)