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Antiderivatives
Definition: Let f be a function. Suppose F is a function such that F ′(x) = f (x), then F is said to be an antiderivative of f.
As its name suggests, if F is an antiderivative of f , then the derivative of F is f.
For example, F (x) = x^2 is an antiderivative of f (x) = 2x because F ′(x) = 2x = f (x).
sin x is an antiderivative of cos x.
Notice we say that sin x is an antiderivative of cos x because the antiderivative of a function f is not unique. Consider sin x + 4. The derivative of sin x + 4 is also cos x, so sin x + 4 is also an antiderivative of cos x. In general, since the derivative of a constant C is 0, if F (x) is an antiderivative of f , then F (x) + C is also going to be an antiderivative of f for any constant C.
You should understand what the above statement says. The statment says that if a function f has an antiderivative F , then it has infinitely many antiderivatives of the form F + C where C is a constant. Each function F + C, where C has a particular value, is one single function which is an antiderivative of F.
For example, F (x) = x^3 is an antiderivative of f (x) = 3x^2. x^3 + 1, x^3 + 3, x^3 − 7, x^3 + π and all other functions of the form x^3 + C where C is a constant are antiderivatives of 3x^2.
Sometimes, when it is understood that the constant C is there when we talk about the antiderivatives of a function, we will omit the C in writing or speaking of the antiderivatives of a function. For example, we might say that the antiderivative of 2x is x^2 , understanding that it is actually x^2 + C, but we omit the C for convenience.
We use the integral symbol
∫ to represent the antiderivatives of a function. That is, if f is
a function of x, then we use the symbol
∫ f (x) dx to mean the (collection of) antiderivatives of f. The symbol dx must be present to indicate that we are looking for the antiderivative of f as a function of x. We will talk more about the dx symbol later. At the meantime, just understand the notation that, if F (x) =
∫ f (x) dx, then it means F (x) is an (collection of) antiderivatives of f.
Eg.
∫ 5 x^4 dx = x^5 + C ∫ sec^2 x dx = tan x + C ∫ x^6 dx =
x^7 + C
How do we find the antiderivative of a function f? In trying to find the derivative of a function, we have rules like the sum rule, power rule, chain rule...etc that help us. Do we have similar rules? It turns out that there are also rules for finding antiderivatives. However, the process of finding antiderivatives is generally much more difficult than the process of finding derivatives. There are sophisticated methods developed to find antiderivatives of various kinds of functions. We will study some of these later.
We know that the derivative of the sum or difference of two functions is the sum or difference of their derivatives. This suggests that the antiderivative of the sum or difference of two functions is the sum or difference of their antiderivatives:
Differentiation: [f (x) ± g(x)]′^ = f ′(x) ± g′(x)
Antidifferenciation: (^) ∫ f (x) ± g(x) dx =
∫ f (x) dx ±
∫ g(x) dx
The above statement says that the antiderivative of the sum/difference is the sum/difference of the antiderivatives.
Example: The antiderivative of 2x is x^2 , and the antiderivative of cos x is sin x, therefore, the antiderivative of 2x + cos x is x^2 + sin x + C. In integral notation, we have:
∫ 2 x + cos x dx =
∫ 2 x dx +
∫ cos x dx = x^2 + C 1 + sin x + C 2 = x^2 + sin x + C
The C 1 comes from the antiderivative of 2x, and the C 2 come from the antiderivative of cos x. Why are we able to combine the C 1 and C 2 into one single constant C? Remember that the C represents only a constant, so is C 1 and C 2. The sum of two constants C 1 and C 2 is just another constant, C, and that’s why we can represent the sum of the two constants with a single term, C. In fact, sometimes we write the above as ∫ 2 x + cos x dx =
∫ 2 x dx +
∫ cos x dx = x^2 + sin x + C
The derivative of a constant times a function is the constant times the derivative of the function. We should expect the same for antiderivatives also.
Differentiation: [cf (x)]′^ = cf ′(x)
Antidifferentiation: (^) ∫ c f (x) dx = c
∫ f (x) dx
the answer and see if it gives us the original function. For example, in the above example,
we can easily verify that we are correct because the derivative of 2
x + C is indeed
x
It is important to note that, just like the derivative of a product (or quotient) is not the product (or quotient) of the derivatives, the antiderivative of a product (or quotient) is also not the product (or quotient) or the antiderivatives. That is,
∫ f (x)g(x) dx 6 =
∫ f (x) dx ·
∫ g(x) dx
and
∫ (^) f (x) g(x)
dx 6 =
∫ f (x) dx ∫ g(x) dx
If we work the differentiation formulas for the trigonometric functions backward, we come up with the following formulas for the antiderivatives:
∫ sin x dx = − cos x + C ∫ cos x dx = sin x + C ∫ sec^2 x dx = tan x + C ∫ csc^2 x dx = − cot x + C ∫ sec x tan x dx = sec x + C ∫ csc x cot x dx = − csc x + C ∫ ex^ dx = ex^ + C ∫ (^1) x
dx = ln |x| + C
Notice the absolute value on x in the formula for the antiderivative of
x
. We know that
the derivative of ln x is
x
. We want the antiderivative of
x
to be defined for both positive
and negative x, but since ln x only defined for x > 0, we put in the absolute value so that the equation is valid for any positive or negative x.
With these rules and formulas we can find antiderivatives of many functions:
E.g. ∫ (^3) x (^4) − x (^2) + 3 x^2
dx =?
This expression might look complicated. However, if we notice that we can actually break the fraction into three terms, we have a easier time with it:
∫ (^3) x (^4) − x (^2) + 3 x^2
dx =
∫ (^3) x 4 x^2
x^2 x^2
x^2
dx =
∫ 3 x^2 − 1 +
x^2
dx
∫ 3 x^2 dx −
∫ dx +
x^2
dx
= 3
∫ x^2 dx −
∫ dx +
∫ 3 x−^2 dx
= 3
x
2+ 2 + 1
(^) − x + 3
x
−2+ −2 + 1
(^) + C
x
3 3
(^) − x + 3
x
− 1 − 1
(^) + C
= x^3 − x − 3 x−^1 + C = x^3 − x −
x
+ C
You should differentiate the answer to convince yourself that this is indeed the correct
antiderivative of 3 x^4 − x^2 + 3 x^2
E.g. Find the antiderivative of
f (x) = 4x^5 − sin x + 3
Notice that the antiderivative of 4x^5 , using the power rule for antiderivatives, is 4
x^6 6
2 x^6 3
The antiderivative of sin x is − cos x, and lastly the antiderivative of 3 is 3x, so we have: ∫ 4 x^5 − sin x + 3 dx = 2 x^6 3
Sometimes, the antiderivative of a function is not immediately obvious. We need to do some algebra to change the expression.
E.g. Find the antiderivatives of
f (x) =
sin 2x cos x
We do not have formulas that can find this antiderivative. However, if we use the double angle formula to change sin 2x to 2 sin x cos x, then it is not hard to find the antiderivative: ∫ (^) sin 2x cos x
dx =
∫ (^) 2 sin x cos x cos x
dx =
∫ 2 sin x dx = −2 cos x + C
To find the distance function s(t), we take the antiderivative of v(t):
s(t) =
∫ v(t) dt =
∫ − 9. 8 t + 50 dt = − 4. 9 t^2 + 50t + s 0
Since s(0) = 200, we see that s 0 = 200, so we have
s(t) = − 4. 9 t^2 + 50t + 200
To find out the heighest point above ground of the ball, the ball reaches its highest point when its upward velocity is 0. (The local, and also global maximum, of s(t)). Setting v(t) = 0 and solving for t we get:
− 9. 8 t + 50 = 0 ⇒ t =
The ball will reach its highest point about 5 seconds after it is being thrown. To find out how high it reaches, we plug this value of t into the displacement function:
s(5) = − 4 .9(25) + 50(5) + 200 ≈ 325
The ball reaches a highest point of about 325 meters above ground. (About 125 meters above the building)
The ball reaches ground when s(t) = 0. Setting s(t) = 0 and solving for t we get:
− 4. 9 t^2 + 50t + 200 = 0 ⇒ t =
√ 502 + 4(4.9)(200) 2(− 4 .9)
After about 13 seconds, the ball will hit the ground.
