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 Antider

Home , Derivative, Power rule

Antiderivatives Deﬁnition: Let f be a function. Suppose F is a function such that F 0(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) = x2 is an antiderivative of f(x) = 2x because F 0(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) = x3 is an antiderivative of f(x) = 3x2. x3 + 1, x3 + 3, x3 − 7, x3 + π and all other functions of the form x3 + C where C is a constant are antiderivatives of 3x2. 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 x2, understanding that it is actually x2 + C, but we omit the C for convenience. Z We use the integral symbol to represent the antiderivatives of a function. That is, if f is Z 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 Z understand the notation that, if F (x) = f(x) dx, then it means F (x) is an (collection of) antiderivatives of f. Z Eg. 5x4 dx = x5 + C Z sec2 x dx = tan x + C

Z 1 x6 dx = x7 + C 7 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)]0 = f 0(x) ± g0(x)

Antidiﬀerenciation: Z Z Z 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 x2, and the antiderivative of cos x is sin x, therefore, the antiderivative of 2x + cos x is x2 + sin x + C. In integral notation, we have:

Z Z Z 2 2 2x + cos x dx = 2x dx + cos x dx = x + C1 + sin x + C2 = x + sin x + C

The C1 comes from the antiderivative of 2x, and the C2 come from the antiderivative of cos x. Why are we able to combine the C1 and C2 into one single constant C? Remember that the C represents only a constant, so is C1 and C2. The sum of two constants C1 and C2 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 Z Z Z 2x + cos x dx = 2x dx + cos x dx = x2 + 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)]0 = cf 0(x)

Antidiﬀerentiation: Z Z c f(x) dx = c f(x) dx The antiderivative of a constant times a function is the constant times the antiderivative of the function. For example, since the antiderivative of sec2 x is tan x, so the antiderivative of 5 sec2 x is 5 times tan x, which is 5 tan x.

Z Z 5 sec2 x dx = 5 sec2 x dx = 5 tan x + C

What is the antiderivative of the constant function k? If we ask ourself, the derivative of what function (or what kind of functions) is a constant? That is, what kind of function has a constant slope? The answer, obviously, is a line. So the antiderivative of a constant is a line with the constant as its slope. i.e. Z k dx = kx + C

If the constant k is 0, then substitue k in the above formula we see that Z 0 dx = 0x + C = C

That is, the antiderivative of 0 is a (any) constant. If we take k = 1 in the above formula, we have: Z 1 dx = 1x + C = x + C

That is, the antiderivative of 1 is the identity function x (plus a constant). just have the integral sign and the dx symbol without anything else it is understood that we are taking the antiderivative of 1. d The power rule for diﬀerentiation says that xn = nxn−1. If we reverse the process of the dx power rule, we get the power rule for antiderivatives: If n 6= −1, then Z xn+1 xn dx = + C n + 1 The added condition n 6= −1 is necessary because if n = −1, then the denominator of the antiderivative will be 0, which is not valid. Example: Z x5+1 x6 x5 dx = + C = + C 5 + 1 6

− 1 +1 1/2 Z 1 Z −1/2 x 2 x √ √ dx = x dx = 1 + C = + C = 2 x + C x −2 + 1 1/2 Even though it is more diﬃcult to ﬁnd the antiderivative of a function, we can more easily verify our result after we found the antiderivative. We just need to take the derivative of the answer and see if it gives us the original function. For example, in the above example, √ 1 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,

Z Z Z f(x)g(x) dx 6= f(x) dx · g(x) dx and

Z f(x) R f(x) dx dx 6= g(x) R g(x) dx If we work the diﬀerentiation formulas for the trigonometric functions backward, we come up with the following formulas for the antiderivatives:

Z sin x dx = − cos x + C Z cos x dx = sin x + C Z sec2 x dx = tan x + C Z csc2 x dx = − cot x + C Z sec x tan x dx = sec x + C Z csc x cot x dx = − csc x + C Z ex dx = ex + C

Z 1 dx = ln |x| + C x 1 Notice the absolute value on x in the formula for the antiderivative of . We know that x 1 1 the derivative of ln x is . We want the antiderivative of to be defined for both positive x x 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. Z 3x4 − x2 + 3 dx =? x2 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:

Z 3x4 − x2 + 3 Z 3x4 x2 3 Z 3 dx = − + dx = 3x2 − 1 + dx x2 x2 x2 x2 x2

Z Z Z 3 = 3x2 dx − dx + dx x2 Z Z Z = 3 x2 dx − dx + 3x−2 dx  x2+1   x−2+1  = 3   − x + 3   + C 2 + 1 −2 + 1 x3  x−1  = 3   − x + 3   + C 3 −1 = x3 − x − 3x−1 + C 3 = x3 − x − + C x You should diﬀerentiate the answer to convince yourself that this is indeed the correct 3x4 − x2 + 3 antiderivative of . x2 E.g. Find the antiderivative of f(x) = 4x5 − sin x + 3

x6 2x6 Notice that the antiderivative of 4x5, using the power rule for antiderivatives, is 4 = . 6 3 The antiderivative of sin x is − cos x, and lastly the antiderivative of 3 is 3x, so we have: Z 2x6 4x5 − sin x + 3 dx = + cos x + 3x + C 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 sin 2x f(x) = 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: Z sin 2x Z 2 sin x cos x Z dx = dx = 2 sin x dx = −2 cos x + C cos x cos x Sometimes, when the antiderivative of a function is being seek, additional information will be given about the antiderivative, and this will allow us to find the value of the constant C in the antiderivative, thus giving us just one instead of many antiderivatives. E.g. Find a function f such that f 0(x) = x3 − 2 and f(1) = 5. The first condition tells us that we need to find the antiderivative of x3 − 2. The second condition allows us to specify one particular function f that satisfies the condition.

Z x4 x3 − 2 dx = − 2x + C 4 x4 Therefore, f(x) = − 2x + C. But we know that f(1) = 5, so we have: 4 1 27 f(1) = − 2(1) + C = 5 ⇒ C = 4 4 So the particular function we are looking for is: x4 27 f(x) = − 2x + 4 4 E.g. A ball is being thrown upward from the top of the building with an initial velocity of 50 m/s. The building is 200 meters tall. The downward acceleration of the ball due to gravity is −9.8 m/s2. What is the highest point (above ground) the ball will reach? How long does it take the ball to reach the ground? We need to know the relationship between acceleration, velocity, and distance travelled. Remember that acceleration is the derivative of velocity with respect to time, and velocity is the derivative of the distance travelled with respect to time. Let s(t) represent the displacement (location) of the ball as a function of time t. Let v(t) represent the velocity of the ball as a function of time. Let a(t) represent the acceleration of the ball as a function of time. Given the information of the problem we have: a(t) = −9.8 m/s2 v(0) = 50 m/s At time t = 0, the initial velocity of the ball is 50 m/s. s(0) = 200 meters. At time t = 0, the intial height of the ball is 200 meters above ground. To ﬁnd the velocity function, we take the antiderivative of a:

Z Z v(t) = a(t) dt = −9.8 dt = −9.8t + v0

Notice that we used the symbol v0 instead of C to represent the constant in the antiderivative. The reason is that the constant in this case here actually represents the initial velocity of the ball, therefore we use the symbol v0 instead of the generic C to represent the constant.

Since v(0) = 50, we know that v0 = 50 and v(t) = −9.8t + 50. To ﬁnd the distance function s(t), we take the antiderivative of v(t):

Z Z 2 s(t) = v(t) dt = −9.8t + 50 dt = −4.9t + 50t + s0

Since s(0) = 200, we see that s0 = 200, so we have s(t) = −4.9t2 + 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: 50 −9.8t + 50 = 0 ⇒ t = ≈ 5 9.8 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: q −50 ± 502 + 4(4.9)(200) −4.9t2 + 50t + 200 = 0 ⇒ t = ≈ 13.3 2(−4.9)

After about 13 seconds, the ball will hit the ground.