Unformatted text preview: Antiderivatives (Section 4.8) Definition: A function F is called an antiderivative of f on an interval I if F '( x ) = f ( x ) for all x in I. Note: An antiderivative is always a function! Example 1: f ( x ) = 3x 2 What are some possible antiderivatives? F(x) = F(x) = F(x) = Notice that you can have any _________________ _____________ that you want. Because of this, for this example we write: where ____ is a __________________ F(x) = We call this the _____________________ antiderivative. Example 2: Find the general antiderivatives of: a) f ( x ) = 2 x 1
= t −4 4
t b) f (t ) = c) f ( x ) = x n where n ≠ −1 1
= x −1 x d) f ( x ) = e) f (t ) = sin t f) f ( x ) = 1
1 − x2 TEXT: Pg 297 has a table of antiderivatives. You must MEMORIZE the antiderivatives in this table. Three properties of antiderivatives: Function General Antiderivative 1. Constant Multiple Rule kf ( x ) − f (x) 2. Negative Rule f ( x ) ± g( x ) 3. Sum/Difference Rule **Practically, these properties mean we can: Indefinite Integrals Definition: If f is a function with antiderivative F , the indefinite integral of f is ∫ f ( x ) dx = F ( x ) + C where C is a constant. NOTE: This is just a convenient notation asking you to find an antiderivative! Verifying formulas We can verify indefinite integral formulas by differentiating the right hand side. Example 3: Verify the following formulas: ( 3x + 5 )−1
+C a) ∫ ( 3x + 5 )−2 dx = −
3 This requires you to evaluate: ⎞
d ⎛ ( 3x + 5 )−1 −
+ C⎟ dx ⎜
3
⎝
⎠ b) ∫ tan θ sec 2 θ dθ = 1 tan 2 θ + C 2 More Antiderivative Practice: Evaluate: 1) ∫ (5 − 6 x ) dx = 2) ∫ ( x −4 + 3) dx = 3) ∫ sec x tan x dx = ⎛2
⎞
4) ∫ ⎜
− y −1/ 4 ⎟ dy = ⎝ 1 + y2
⎠ Initial Value Problems: Suppose we are given f '( x ) and want to find f . (This is asking for a specific function, not a general antiderivative.) What additional information do we need? Example 4: If f '( x ) = x 2 + 3x −2 and f (1) = 1 , find f . Step 1: Find the general antiderivative of f ' . Step 2: Use the initial value to find C. Step 3: Combine the results of steps 1 and 2 to get a single function. d2 f
df
Example 5: If 2 = sin x + cos x and f (0 ) = 3 and
= 4 , find f . dx
dx x = 0 From Physics: If an object has position s(t) at time t, then its velocity at time t is dv d 2 s
ds
= and its acceleration at time t is a(t ) =
. Acceleration due to v(t ) =
dt dt 2
dt
gravity is −9.8 m / s 2 or −32 ft / s 2 . Example 6: A stone is dropped off a cliff and hits the ground with a speed of 120 ft/s. What is the height of the cliff? ...
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This note was uploaded on 10/02/2011 for the course MATH 2214 at Virginia Tech.
 '06
 EDeSturler
 Antiderivatives, Derivative

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