4.7 Antiderivatives
Many of the applications of calculus depend on our ability to deduce facts about a function f from information concerning its derivatives. For example, if we know the velocity of an object and want to know its position at some time, or we know the rate at which a population is increasing or decreasing and want to know what the population was at a particular time, then finding the function from its derivative would be our goal. Finding a function by knowing its derivative is what we call finding the antiderivative.
Consider the function f(x) = 2x. What function could have been differentiated to get this?
Examples: Find the most general antiderivative of each of the following functions.
f(x) = 4x3 f(x) = x2 + 5
f(x) = cos x f(x) = ex f(x) = xn, n≠ ­1
f(x) = 1/x Note that antiderivatives can be checked by taking the derivative!
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Note that going from
particular antiderivative to
general antiderivative
requires the addition of a
constant c.
Example: Find all functions f such that f '(x) = 5 sin x + 6x3 ­ 7ex + 8
Example: Find all functions g such that g'(x) = Example: Find all functions g such that g'(x) = In applications of calculus it is very common to have to find a function f, given information about its derivatives. An equation that involves the derivatives of a function is called a differential equation. These will be studied in more detail in Calculus II, but we can solve some elementary differential equations. We start by finding the general antiderivative of a function, and then solve for the constant by using an additional condition.
Example: Find f if f '(x) = 3x4 + ex ­ 10 and f(0) = 9
Example: Find f if f ''(x) = 20x3 ­ 24x + 16, f(0) = 5, and f(1) = 3.
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Example: Find f if f '(x) = 5x4 ­ 3x2 + 4, f(­1) = 2
Example: Find f if f ''(t) = 2et + 3sin t, f(0) = 0, and f '(0) = 0.
Example: Find f if f ''(x) = 8x3 + 5, f(1) = 0, and f '(1) = 8.
Rectilinear Motion (motion in a straight line): As mentioned before, antidifferentiation is particularly useful in dealing with problems involving motion. Recall that if an object has position function s = f(t), then its velocity function is v = f '(t) and its acceleration function is a = v'(t) = f ''(t).
Example: A particle moves in a straight line and has acceleration given by a = 4t + 10. Its initial velocity is v(0) = ­3 ft/sec and its initial displacement (position) is s(0) = 5 feet.
Find its position function s(t).
Example: A particle is moving with the given data. Find the position of the particle (at time t.)
v(t) = 10 sin t + 3 cos t, s(0) = 0
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