Applications of integrals #4 - Questions and Answers

Question 1 Find the indefinite integral. ∫(2csc²x - 9sin x)dx Answer ∫(2csc²x - 9sin x)dx = 2∫csc²xdx - 9∫sin x dx = 2(-cotx) - 9(-cosx) +C = -2cotx + 9cosx + C Question 2 Is the algebraic expression a polynomial? If it is, write the polynomial in standard form: 2x + 3x²-5 Answer Step 1 A polynomial is defined as an expression which is composed of variables, constants and exponents, that are combined using the mathematical operations such as addition, subtraction, multiplication and division. A polynomial function is an expression constructed with one or more terms of variables with constant exponents. If there are real numbers denoted by a, then function with one variable and of degree n can be written as: f(x) = a_n x^n + a₁ x^(n-1) + a₂ x^(n-2) + ... + a_(n-2) x² + a_(n-1) x + a_n Given expression is 2x + 3x²-5. Step 2 This can be in the form of algebraic expression of a polynomial that is: f(x) = 3x² + 2x – 5 Also, it is written in the standard form of polynomial i.e, f(x) = 3x²+2x - 5 It means decreasing power of x is there standard form = 3x² + 2x - 5 Question 3 Factor each expression completely. a. y+2xy b. tan u + 2cos u tan u Answer Step 1 Factorization is the process of taking common from the terms of given expression and the common term is known as factor of that expression. Given expression is: a. y+2xy Here, y is common in both terms so taking y from y and 2xy, we get y+2xy =y(1+2x) Hence, factorization of given expression is y(1+2x) and factors are y and 1+2x. Step 2 And, b.tan u + 2cos u tan u Here, tan u is common in both terms so taking tan u from tanu and 2cos u tan u,we get tan u + 2cos u tan u = tan u(1 + 2cos u) Hence, factorization of given expression is tanu (1 + 2cos u) and factors are tan u and 1 + 2cos u. Hence, factorization of given expression is tan u(1 + 2cos u) and factors are tan u and 1 + 2cos u. Question 4 What is the integral of the constant function f(x, y) = 5 over the rectangle [-2,3] × [2,4] ? Answer Step 1 Integration of a function gives the result as the area covered by the function with an appropriate axis. INtegrals are of two types, definite and indefinite integrals. Definite integrals are defined for a limit of variable from a lower limit to an upper limit. This is represented by the expression:- I = ∫(from a to b) f(x)dx = F(b) - F(a) Here, F(x) is the anti-derivative of the function f(x). Step 2 The value of the definite integral is constructed by using the limits of x from -2 to 2, and the limts of y from 3 to 4 as I = ∫(from 2 to 4) ∫(from -2 to 2) f(x, y)dxdy. This integral is calculated as follows:- I = ∫(from 2 to 4) ∫(from -2 to 2) 5dxdy = 5 ∫(from 2 to 4) [x](from -2 to 2) dy = 5 ∫(from 2 to 4) (2 - (-2)) dy = 20 ∫(from 2 to 4) dy = 20 [y](from 2 to 4) = 20 * (4-2) = 20 * 2 = 40 Question 5 Evaluate this indefinite integral: ∫(3x + 5)dx Answer The value of indefinite integration must be determined in this case. We are aware that indefinite integrals are those integrals without an upper or lower bound. Here we use the formula of integration ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C And we know that ∫ kdx = k(x) + C Consider the indefinite integral ∫ (3x + 5)dx. We can integrate the above expression using the formula ∫ xⁿ dx = xⁿ⁺¹/(n+1) + C. Here we have (3x¹ + 5) thus we have n=1. Integrating the above expression, we get ∫ (3x + 5)dx = 3 ∫ xdx + 5 ∫dx = 3x²/2 + 5x = 3x²/2 + 5x + C hence we have ∫ (3x + 5)dx = 3x²/2 + 5x + C.