Indefinite Integrals, types of integration

Indefinite Integrals, types of integration

• 1.
  Indefinite Integrals - JAIVEERSINGH 12 ‘A’ UID:
• 2.
  It gives meimmense pleasure to present ‘INDEFINITE INTEGERALS ’. It would not have been possible without the kind support of Mr. Ashish Agrawal and teacher in charge, Mr. Rahul Dhakad, under whose supervision the project was brought to present state. I would also like to express my gratitude towards my parents for their kind co-operation and encouragement which helped me in the completion of this project. I am also thankful to ISC for giving me such an amazing opportunity for making this project and giving suitable instructions and guidelines for the project. Last but not the least, I thank my friends who shared necessary information and useful weblinks for preparing my project. Thanks again to all. Acknowledgement Acknowledgement
• 3.
  Introduction Integration is theinverse process of differentiation. The primary problem of Differential Calculus is: Given a function, to find its differential coefficient. But the primary problem of Integral Calculus is its inverse, i.e., ‘Given the differential coefficient of a function, to find the function itself’. Let f(x) be a given function of x. if we x an find a function F(x) such that d/dx [F(x)]=f(x) then F(x) is called an integral of f(x). We write. The sign is called the integral sign. The symbol dx indicates that the integration is to be performed with respect to the variable x. The process of determining an integral of a function is called integration and the function to br integrated is called the Integrand.
• 4.
  Notation Meaning ∫ f(x)dx Integral of f with respect to x f(x) in ∫ f(x) dx Integrand x in ∫ f(x) dx Variable of integration dx in ∫ f(x) dx Differentiation goes in the x direction C Constant of Integration The symbol for integration is S-shaped. Let us get familiar with some of the associated notations.
• 5.
  Types of Integrations; •IndefiniteIntegrals: It is an integral of a function when there is no limit for integration. It contains an arbitrary constant. •Definite Integrals: An integral of a function with limits of integration. There are two values as the limits for the interval of integration. One is the lower limit and the other is the upper limit. It does not contain any constant of integration.
• 6.
  Derivatives Integrals (AntiDerivatives) d⁄dx (x n + 1 ⁄ n + 1) = xn ∫ xn dx = x n + 1 ⁄ (n + 1) + C, n ≠ −1 d ⁄ dx (x) = 1 ∫dx = x + C d⁄dx (sin x) = cos x ∫ cos x dx = sin x + C d ⁄ dx (− cos x) = sin x ∫sinx dx = − cos x + C d⁄dx (tan x) = sec 2 x ∫ sec 2 x dx = tan x + C d ⁄ dx (− cot x) = cosec 2 x ∫ cosec 2 x dx = − cot x+ C d ⁄ dx (sec x) = sec x tan x ∫(sec x + tan x)dx = sec x + C d⁄dx (− cosec x) = cosec x cot x ∫(cosec x cot x)dx = − cosec x + C
• 7.
  d ⁄ dx(sin−1 x) = 1⁄ √(1 − x2) ∫ dx ⁄ √(1 − x2) = sin−1 x+ C d⁄dx (− cos−1 x) = 1⁄ √(1 − x2) ∫dx ⁄ √(1 − x2) = − cos−1 x + C d ⁄ dx (tan−1 x) = 1⁄ (1 + x2) ∫ dx ⁄ (1 + x2) =tan−1 x + C d⁄dx (− cot−1 x) = 1⁄ (1 + x2) ∫dx ⁄ (1 + x2) = − cot−1 x + C d ⁄ dx (sec−1 x) = 1⁄ x√(x2 − 1) ∫ dx ⁄ x√(x2 − 1) = sec−1 x + C d⁄dx (− cosec−1 x) = 1⁄ x√(x2 − 1) ∫dx ⁄ x√(x2 − 1) = − cosec−1 x + C d ⁄ dx (ex) = ex ∫ ex dx = ex + C d⁄dx (log |x|) = 1⁄x ∫1⁄x dx = log |x| + C d ⁄ dx (ax⁄ log a) = ax ∫ ax dx = ax⁄ log a + C
• 8.
  Properties of IndefiniteIntegrals Theorem 1 The process of differentiation and integration are inverses of each other. Proof: Let F be an anti-derivative of f, i.e., d⁄dx F(x) = f(x) Then ∫ f(x) dx = F(x) + C Therefore, d⁄dx ∫ f(x) dx = d⁄dx [F(x) + C] = d ⁄ dx F(x) = f(x). Similarly, f (x) = d⁄dx f(x) and hence ∫ f (x) dx = f(x) + C. C is the constant of integration. ′ ′
• 9.
  Theorem 2 The integrationof the sum of two integrands is the sum of integrations of two integrands. ∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx Proof: Using theorem 1, we have d⁄dx [∫ [f(x) + g(x)] dx] = f(x) + g(x) … (1) Also, d⁄dx [∫ f(x) dx + ∫ g(x) dx] = d⁄dx ∫ f(x) dx + d⁄dx ∫ g(x) dx = f(x) + g(x) … (2) From (1) and (2), we have, ∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx
• 10.
  Example 1: Solve ∫(x5 – 1) ⁄ x2 dx. Solution: ∫(x5 – 1) ⁄ x2 dx = ∫x3 dx – ∫x–2 dx = x 3+1 ⁄ (3 + 1) + C1 – (x–2+1 ⁄ (–2 + 1)) + C2 or, ∫ (x5 – 1) ⁄ x2 dx = x4 ⁄ 4 + 1⁄x + C1 + C2 = x4 ⁄ 4 + 1⁄x + C where, C = C1 + C2
• 11.
  What is theintegration used for? Ans: From a very common and basic point of view, integration is used for measuring things. It might be a length, area, or volume. It can also be probabilities in the context of unplanned variables.
• 12.
  What are theadvantages of integration? Ans. Horizontal integration can benefit the companies and normally takes place when they are challenging in the same industry or sector.