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13.1 Antiderivatives and Indefinite Integrals The Antiderivative The reverse operation of finding a derivative is called the antiderivative. A function F is an antiderivative of a function f if F ’(x) = f (x). 1) Find the antiderivative of f(x) = 5 Theorem 1: Find several functions that have the derivative of 5 If a function has more than one antiderivative, then the antiderivatives differ by at most a constant. Answer: 5x; 5x+ 1; 5x -3; 2) Find the antiderivative of f(x) = x2 Find several functions that have the derivative of x2 1 3 1 3 1 3 x ; or x ; or x e Answer: 3 3 3 • The graphs of antiderivatives are vertical translations of each other. • For example: f(x) = 2x Find several functions that are the antiderivatives for f(x) Answer: x2, x2 + 1, x2 + 3, x2 - 2, x2 + c (c is any real number) Indefinite Integrals Let f (x) be a function. The family of all functions that are antiderivatives of f (x) is called the indefinite integral and has the symbol f ( x) dx The symbol is called an integral sign, and the function f (x) is called the integrand. The symbol dx indicates that antidifferentiation is performed with respect to the variable x. By the previous theorem, if F(x) is any antiderivative of f, then f ( x) dx F ( x) C The arbitrary constant C is called the constant of integration. Indefinite Integral Formulas and Properties n 1 x 1. x n dx C , n 1 (power rule) n 1 2. x x e dx e C 3. 1 x dx ln | x | C 4. 5. k f ( x) dx k f ( x) dx f ( x) g ( x) dx f ( x) dx g ( x) dx It is important to note that property 4 states that a constant factor can be moved across an integral sign. A variable factor cannot be moved across an integral sign. Example 1: A) 2dx 2 x C B) 16e dt 16e C) t 3x dx 4 t C x5 3 5 3 C x C 5 5 Example 1 (continue) D) (2 x 5 3 x 1)dx 2 2 x dx 3 x dx 1dx 5 2 x6 x3 2 3 1x C 6 3 1 6 3 x x xC 3 Example 1 (continue) E) 5 x 4 e dx x 5 x dx 4e dx x 1 x 5 dx 4 e dx x 5 ln x 4e C x Example 2 A) 23 3 2 x dx 4 x 2 3 2 3 2x dx 3x 4dx 2 x dx 3 x 4dx 5 5 x 3 x 3 6 3 3 C x x C 2 3 5 5 3 3 5 6 3 1 x 3 C 5 x Example 2 (continue) B) 4 5 3 5 w dw 4 w dw 3 8 w5 4 8 5 8 5 C x5 C 2 Example 2 (continue) x 8x x 2 4 C) 3 dx x 2 8 x dx x dx 8 xdx 2 2 x x 8 C 3 2 3 x 2 4x C 3 3 Example 2 (continue) D) 1 3 1 2 6 3 8 x x dx 8 x dx 6 x dx 4 x3 8 4 3 1 x2 6 1 2 4 3 C 6 x 12 x C Example 2 (continue) E) ( x 2)( x 3)dx 2 x 3 3 x 2 x 6 dx 2 x dx 3 x dx 2 xdx 6dx 3 2 x 4 x3 x 2 3 2 6 x C 4 3 2 4 x x3 x 2 6 x C 4 Example 3 Find the equation of the curve that passes through (2,6) if its slope is given by dy/dx = 3x2 at any point x. The curve that has the derivative of 3x2 is 3x dx 2 x3 3 C x 3 C 3 Since we know that the curve passes through (2, 6), we can find out C 3 y x C 6 23 C 6 8C C 2 Therefore, the equation is y = x3 - 2 Example 4 Find the revenue function R(x) when the marginal revenue is R’(x) = 400 - .4x and no revenue results at a 0 production level. What is the revenue at a production of 1000 units? The marginal revenue is the derivative of the function so to find the revenue function, we need to find the antiderivative of that function (400 .4 x)dx x2 400 dx .4 xdx 400x .4 C 400x .2 x 2 2 So R(x) = 400x -.2x2, we know need to find R(1000) R(1000) 400(1000) .2(1000) 200,000 2 Therefore, the revenue at a production level of 1000 units is $200,000 Example 5 The current monthly circulation of the magazine is 640,000 copies. Due to the competition from a new magazine, the monthly circulation is expected to decrease at a rate of C’(t)= -6000t1/3 copies per month, t is the # of months. How long will it take the circulation of the magazine to decrease to 460,000 copies per month? We must solve this equation: C(t) = 460,000 with C(0) = 640,000 To find the function C(t), take the antiderivative 4 t3 6000t dt 6000 4 3 43 C (0) 4500 0 C 1 3 43 640,000 4500 0 C 640,000 C 4 3 C (t ) 4500t 640,000 4 C 4500t 3 C 4 3 460,000 4500t 640,000 180,000 4500t 40 t 4 3 t 15.9 t 40 3 4 4 3 3 4 So, it takes about 16 months 4 3