5.7 Antiderivatives We know how to solve the derivative problem: given a function, find its derivative. But many problems in mathematics and its applications require us to sole the inverse of the derivative problem: given a function f , find a function F whose derivative is f . If such a function F exists, it is called an antiderivative of f . Definition. A function F is called an antiderivative of f on an interval I if F '( x) f ( x) for all x in I . For example, x 2 is an antiderivative of 2x . We discover that x 2 1 is an antiderivative of 2x , so the antiderivative of a function is not unique. In fact, x 2 c ( c is a constant) is an antiderivative of 2x too. In general, we have the following. Theorem. If F is an antiderivative of f on an interval I , then the most general antiderivative of f on I is F ( x) c where c is a constant. Properties: Given F ( x ) is an antiderivative of f ( x ) , and G( x ) is an antiderivative of g ( x ) , then F ( x) G( x) is an antiderivative of f ( x) g ( x) , and cF ( x ) is an antiderivative of cf ( x ) . Table of the antiderivatives. function x n (n 1) 1 x ex sin x cos x sec2 x csc2 x sec x tan x csc x cot x 1 1 x2 1 1 x2 antiderivative x n 1 c n 1 ln x c ex c cos x c sin x c tan x c cot x c sec x c csc x c sin 1 x c ( cos1 x c ) tan 1 x c ( cot 1 x c) Example 1. Find all functions f such that f '( x) sin x 2 x 2 x 4 x 3 4(1 x 2 ) 1 1 3 ' Solution. Notice that ( cos x ) ' sin x, x 3 x 2 , (tan 1 x ) ' 1 , 1 x2 ' x x x 4 3 13/4 x 7/4 1 1 74 x x 7/4 7 1 4 ' 4 11 or x 4 x 7/4 11 2 3 4 114 4 Thus, we have f ( x ) cos x x x c 3 11 1 x2 Example 2. Find the general antiderivative r(t ) 2 i 3t j Solution. The general antiderivative is 3 3 2t i t 2 j c 2t i t 2 j c1 i c2 j 2 2 Example 3. Find f such that f ''( x ) x 2 , f (1) 1 , f '(1) 0 Solution. f '( x ) f ( x) 1 3 x c 3 1 4 x cx c1 12 Since f (1) 1 and f '(1) 0 , we have 1 1 f '(1) 0 (1)3 c 0 c 3 3 1 1 5 f (1) 1 (1)4 c(1) c1 1 c c1 1 c1 12 12 4 Thus, f ( x ) 1 4 1 5 x x 12 3 4 Example 3. Given that the graph of f passes through the point (1, 2) and that the slope of its tangent line at ( x, f ( x )) is x 2 1. Find f ( x ) and f (0) Solution. f '( x ) x 2 1 f ( x) 1 3 x xc 3 The graph passes through (1, 2) which means f (1) 2 1 3 (1) 1 c 2 3 8 c 3 1 8 f ( x) x3 x 3 3 8 f (0) 3 Example 4. A particle moves in a straight line and has acceleration given by a(t ) 4t 1 . Its initial velocity is v(0 2 cm/s and its initial displacement is s(0 5. Find its position function s(t ). Solution. v '(t ) a(t ) v '(t ) 4t 1 v ( t ) 2t 2 t c v (0) 2 2(0) 2 0 c 2 c2 v ( t ) 2t 2 t 2 s '(t ) v (t ) s '(t ) 2t 2 t 2 2 3 1 2 t t 2t c1 3 2 s(0) 5 s(t ) c1 5 s(t ) 2 3 1 2 t t 2t 5 3 2