Antiderivative

# Antiderivative - f x is f x = e x-cos x 3 sin x x 2 C Since...

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ANTIDERIVATIVES Recall that an antiderivative of f is a function whose derivative is f . For example, if F ( x ) = 1 3 x 3 , then F 0 ( x ) = x 2 ), thus F ( x ) is an antiderivative of x 2 . We should notice, however, that the function G ( x ) = 1 3 x 3 + 1 also satisfies G 0 ( x ) = x 2 and hence an antiderivative of x 2 . In fact, we have a following theorem. Theorem 1. 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 an arbitrary constant. Example 1. Find the most general antiderivative of x - 2 . Solution Using the power rule, we get d dx ( - x - 1 ) = ( - 1)( - x - 1 - 1 ) = x - 2 . Thus, the general antiderivative of x - 2 is - x - 1 + C . Since x - 2 is discontinuous at x = 0, the domain of the function x - 2 is a union of two disjoint intervals, ( -∞ , 0) and (0 , ). Thus, we have to use the theorem above separately on those intervals, and can get the most general antiderivative is F ( x ) = ( - x - 1 + C 1 if x > 0 - x - 1 + C 2 if x < 0 Example 2. Find f if f 0 ( x ) = e x + sin x + 3 cos x + 2 x and f (0) = 2. Solution
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Unformatted text preview: f ( x ) is f ( x ) = e x-cos x +3 sin x + x 2 + C . Since f (0) = 1-1 + 0 + C = C = 2, we have C = 2, so the particular solution is f ( x ) = e x-cos x + 3 sin x + x 2 + 2 . In physics, antidiﬀerentiation is useful to recover the position function from a given velocity function or acceleration function. Example 3. A particle moves in a straight line with constant acceleration -32 ft/s 2 . Its initial velocity is 60 ft/s and its initial displacement is 100 ft. Find its position function s ( t ). Solution Since s 00 ( t ) =-32, s ( t ) =-32 t + C 1 and s ( t ) =-16 t 2 + C 1 t + C 2 . We know that v (0) = s (0) = 60, and it gives C 1 = 60. We also have a (0) = s 00 (0) = 100, hence C 2 = 100. Therefore, s ( t ) =-16 t 2 + 60 t + 100. 1...
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