Section 4 8 Antiderivatives A function F x is an antiderivative of the function f x if F x f x For example the function F x x 2 is an antiderivative of f x 2x 2 2 2 x 1 x 2 are also antiderivatives of 2x as is x C for any constant C x 2 C is called the most general antiderivative of 2x Reversing the Power rule For n not equal to 1 d 1 dx n 1 x n 1 1 n 1 1 n x n 1 x n x n so 1 n 1 x n 1 n C is the most general antideriva tive of x most general antiderivative function 1 n 1 x n 1 1 x n 1 C ln x C x e x e x C sin x cos x C cos x sin x C sec 2 tan x C x sec x C sec x tan x There is no product rule or quotient rule for antiderivatives The linear rule and the shift rule work just as they do for derivatives Examples Find the most general antiderivative F of each function f 1 f x x 2 F x 1 3 x C 2 f x 5x 2 F x 3 3 f x x 1 2 2 F x x 3 2 C 4 f x 4 x 1 2 5 3 x C 3 F x 3 5 f x 5x 2 4x 1 2 6 f x 4e x sin x 3 8 x 3 2 C 3 F x 5 3 8 32 x x C 3 3 F x 4e x cos x 3 x C Example Find the most general antiderivative of x 1 x The antiderivative of the product is not the product of the antiderivatives We must multiply it out x 1 x x x x x 3 2 x 1 2 F x 2 5 x 5 2 2 3 x 3 2 C Example Find the most general antiderivative of x x 1 x 1 1 x 1 x 1 x 1 F x 2 x 1 5 2 5 2 x 1 3 2 x 1 x x 1 x 1 3 2 x 1 1 2 C 3 Finding a specific antiderivative that meets a given condition Example 1 Find an antiderivative F x for f x x 2 2x 3 so F 1 0 x There is no quotient rule for antiderivatives The antiderivative is not the quotient of the antiderivatives We must divide f x x 2 3 F x x 1 x 2 2 x 3 ln x C is the most general antiderivative 2 Now plug in x 1 and solve for C F 1 0 5 2 0 C 0 so C 2 5 F x 1 x 2 2 x 3 ln x 2 5 2 Example 2 An object is fired straight up from a height of 6 m with initial velocity 20 m sec Given the acceleration due to gravity is 9 8 m sec 2 find the height at time t seconds a t 9 8 h t 9 8 v t 9 8 t C 1 2 t 20 t C 2 and and v 0 20 h t 6 so so v t 9 8 t 20 h t 4 9 t 2 20 t 6 Example 3 If the acceleration of an object moving in a straight line is given by t a t 2 e 1 and v 0 4 and s 0 10 find s t t v t 2e t C v 0 2 0 C 4 so C 2 t 1 2 t 2t C 2 t 1 2 t 2t 8 2 s t 2e s t 2e t v t 2 e t 2 s 0 2 C 10 so C 8
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