Lesson Explainer: Antiderivatives Mathematics • Higher Education

In this explainer, we will learn how to find the antiderivative of a function. The antiderivative of a function ๐‘“(๐‘ฅ) is the function ๐น(๐‘ฅ) where ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ).

An antiderivative, also known as an inverse derivative or primitive, of a function ๐‘“ is another function ๐น whose derivative is equal to the original function ๐‘“.

Definition: An Antiderivative of a Function

For a differentiable function ๐น(๐‘ฅ), if we have ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ), then we say that ๐น(๐‘ฅ) is an antiderivative of ๐‘“(๐‘ฅ).

We note that since ๐‘“โ€ฒ(๐‘ฅ) is the derivative of ๐‘“(๐‘ฅ), ๐‘“(๐‘ฅ) is an antiderivative of ๐‘“โ€ฒ(๐‘ฅ); similarly, ๐‘“โ€ฒ(๐‘ฅ) is an antiderivative of ๐‘“โ€ฒโ€ฒ(๐‘ฅ) and ๐‘“โ€ฒโ€ฒ(๐‘ฅ) is an antiderivative of ๐‘“โ€ฒโ€ฒโ€ฒ(๐‘ฅ). The process of finding an antiderivative of a function is the reverse process of differentiating a function; for example, 2๐‘ฅ is the derivative of ๐‘ฅ๏Šจ, so we can say that ๐‘ฅ๏Šจ is an antiderivative of 2๐‘ฅ.

There are many real-world applications of antiderivatives, for example, when we consider the equations of motion in Newtonian mechanics. The velocity, ๐‘ฃ(๐‘ก), is defined as the rate of change of the displacement, ๐‘ (๐‘ก), with respect to time, ๐‘ก. In other words, velocity is the derivative of displacement: ๐‘ฃ(๐‘ก)=๐‘ โ€ฒ(๐‘ก).

This means the reverse is also true; displacement is an antiderivative of velocity and so the function ๐‘ (๐‘ก) is an antiderivative of the function ๐‘ฃ(๐‘ก). Similarly, acceleration, ๐‘Ž(๐‘ก), is the derivative of velocity, which means that velocity is an antiderivative of acceleration.

In fact, an antiderivative is not unique and there are many functions, which differ up to a constant, that give the same derivative. In order to see this, letโ€™s consider the constant function ๐น(๐‘ฅ)=1. The derivative of this function with respect to ๐‘ฅ is ๐‘“(๐‘ฅ)=๐นโ€ฒ(๐‘ฅ)=(1)=0.๏Ž˜

This is expected since the function ๐น does not vary as ๐‘ฅ varies; thus its derivative is zero. This means that an antiderivative of ๐‘“(๐‘ฅ)=0 is ๐น(๐‘ฅ)=1. This is also true for ๐น(๐‘ฅ)=2, and in fact any real constant ๐น(๐‘ฅ)=C, since the derivative will always be ๐‘“(๐‘ฅ)=0, as Cโ€ฒ=0. This means an antiderivative of ๐‘“(๐‘ฅ)=0 will be some constant ๐น(๐‘ฅ)=C; or we can say that the antiderivative of ๐‘“(๐‘ฅ)=0 is ๐น(๐‘ฅ)=C, for all Cโˆˆโ„.

What about the derivative of ๐น(๐‘ฅ)+C? Since the derivative is linear, that is, we can differentiate linear combinations of functions separately, in particular, (๐‘“+๐‘”)=๐‘“โ€ฒ+๐‘”โ€ฒ๏Ž˜, we have, (๐น(๐‘ฅ)+)=๐นโ€ฒ(๐‘ฅ)+โ€ฒ=๐นโ€ฒ(๐‘ฅ)+0=๐‘“(๐‘ฅ).CC๏Ž˜

The constant C, also known as the constant of antidifferentiation, is very important as it produces a family of antiderivatives, ๐น(๐‘ฅ)+C, parameterized by C. This means that ๐‘“(๐‘ฅ) is the derivative of ๐น(๐‘ฅ)+C, so that ๐น(๐‘ฅ)+C is the antiderivative or the most general antiderivative of ๐‘“(๐‘ฅ), for all Cโˆˆโ„; this is why we always add a +C when finding the (general) antiderivative of any function as without it we only have an antiderivative, which is not unique. In other words, ๐น(๐‘ฅ)+C is the most general function that has a derivative ๐‘“(๐‘ฅ), for all Cโˆˆโ„; for example, while ๐‘ฅ๏Šจ is an antiderivative of 2๐‘ฅ, ๐‘ฅ+๏ŠจC is the general antiderivative of 2๐‘ฅ. This means that ๐‘ฅ+1๏Šจ, ๐‘ฅ+7๏Šจ, ๐‘ฅ+โˆš2๏Šจ, or ๐‘ฅ+๐œ‹๏Šจ, and so on, are all also antiderivatives of 2๐‘ฅ.

By using this, we can also say that the general antiderivative of a function ๐‘“โ€ฒ(๐‘ฅ) is the function ๐‘“(๐‘ฅ)+C, for Cโˆˆโ„, and similarly the general antiderivative of the function ๐‘“โ€ฒโ€ฒ(๐‘ฅ) is the function ๐‘“โ€ฒ(๐‘ฅ)+C and this process continues.

If we are also given ๐น(๐‘ฅ)=๐น๏Šฆ๏Šฆ, also known as a boundary condition, then we can also determine the constant C to give a unique antiderivative that satisfies the boundary condition.

We also note that a constant multiple in front of a function does not affect its antiderivative. This follows from the definition of the derivative applied to ๐‘Ž๐น(๐‘ฅ): (๐‘Ž๐น(๐‘ฅ))=๐‘Ž๐นโ€ฒ(๐‘ฅ)=๐‘Ž(๐น(๐‘ฅ))=๐‘Ž๐‘“(๐‘ฅ).๏Ž˜๏Ž˜

Therefore, ๐‘Ž๐น(๐‘ฅ) is an antiderivative of ๐‘Ž๐‘“(๐‘ฅ), or ๐‘Ž๐น(๐‘ฅ)+C is the general antiderivative of ๐‘Ž๐‘“(๐‘ฅ), for all Cโˆˆโ„. This means we can always bring a constant multiple inside the derivative as ๐‘Ž(๐น(๐‘ฅ))=(๐‘Ž๐น(๐‘ฅ))๏Ž˜๏Ž˜; in other words, if the derivative gets multiplied by a constant, the antiderivative also gets multiplied by the same constant and vice versa.

If we want to find the antiderivative of ๐‘“(๐‘ฅ)=๐‘Ž๐‘ข(๐‘ฅ)+๐‘๐‘ฃ(๐‘ฅ) for some functions ๐‘ข(๐‘ฅ),๐‘ฃ(๐‘ฅ)โˆถโ„โ†’โ„ and constants ๐‘Ž,๐‘โˆˆโ„, we can find the antiderivatives of ๐‘Ž๐‘ข(๐‘ฅ) and ๐‘๐‘ฃ(๐‘ฅ) separately, say ๐‘Ž๐‘ˆ(๐‘ฅ) and ๐‘๐‘‰(๐‘ฅ), and add the results. In other words, the antiderivative of a sum is the sum of antiderivatives; the antiderivative is a linear operation, which follows from the linearity of the derivative.

In order to see this, we let ๐น(๐‘ฅ)=๐‘Ž๐‘ˆ(๐‘ฅ)+๐‘๐‘‰(๐‘ฅ), with ๐‘ข(๐‘ฅ)=๐‘ˆโ€ฒ(๐‘ฅ) and ๐‘ฃ(๐‘ฅ)=๐‘‰โ€ฒ(๐‘ฅ), and using the fact that we can differentiate linear combinations of functions separately, we have ๐‘“(๐‘ฅ)=๐นโ€ฒ(๐‘ฅ)=(๐‘Ž๐‘ˆ(๐‘ฅ)+๐‘๐‘‰(๐‘ฅ))=๐‘Ž๐‘ˆโ€ฒ(๐‘ฅ)+๐‘๐‘‰โ€ฒ(๐‘ฅ)=๐‘Ž๐‘ข(๐‘ฅ)+๐‘๐‘ฃ(๐‘ฅ).๏Ž˜

Thus, an antiderivative of the sum ๐‘“(๐‘ฅ)=๐‘Ž๐‘ข(๐‘ฅ)+๐‘๐‘ฃ(๐‘ฅ) is ๐น(๐‘ฅ)=๐‘Ž๐‘ˆ(๐‘ฅ)+๐‘๐‘‰(๐‘ฅ), the sum of antiderivatives. Therefore, in order to work in reverse and find an antiderivative of a sum of functions, we just find an antiderivative of each part separately and add the results together, not forgetting the +C at the end. Usually, we would obtain multiple constants for each part from the process of antidifferentiation, but we can combine these into one constant. The most general antiderivative of ๐‘“(๐‘ฅ)=๐‘Ž๐‘ข(๐‘ฅ)+๐‘๐‘ฃ(๐‘ฅ) is ๐น(๐‘ฅ)+=๐‘Ž๐‘ˆ(๐‘ฅ)+๐‘๐‘‰(๐‘ฅ)+.CC

We note that while we have been writing the most general antiderivative as ๐น(๐‘ฅ)+C to make the constant appear explicitly, we can usually absorb this in the function ๐น(๐‘ฅ) so that it already contains +C. For example, we can say that the general antiderivative of ๐‘“(๐‘ฅ)=2๐‘ฅ is ๐น(๐‘ฅ)=๐‘ฅ+๏ŠจC, instead of writing the general antiderivative as ๐น(๐‘ฅ)+C with ๐น(๐‘ฅ)=๐‘ฅ๏Šจ.

Now, suppose ๐น(๐‘ฅ)=๐‘ฅ๏Œ for ๐‘โˆˆโ„; the derivative of this function can be found by the power rule for differentiation as (๐‘ฅ)=๐‘๐‘ฅ.๏Œ๏Ž˜๏Œ๏Šฑ๏Šง

It will be useful to rewrite this as 1๐‘(๐‘ฅ)=๏€ฝ๐‘ฅ๐‘๏‰=๐‘ฅ,๏Œ๏Ž˜๏Œ๏Ž˜๏Œ๏Šฑ๏Šง where we have divided it by the constant ๐‘โ‰ 0, as a constant multiple of a function does not affect the derivative or antiderivative. But what if we want to work in reverse? That is, given ๐‘“(๐‘ฅ)=๐‘ฅ๏Œบ, we want to determine the antiderivative. This means we want to find the most general function ๐น(๐‘ฅ) which differentiates to give ๐‘ฅ๏Œบ.

We have already shown that the derivative of ๐‘ฅ๐‘๏Œ is ๐‘ฅ๏Œ๏Šฑ๏Šง, for ๐‘โ‰ 0. If we let ๐‘=๐‘Ž+1, then we have ๏€พ๐‘ฅ๐‘Ž+1๏Š=๐‘ฅ,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏Šง๏Ž˜๏Œบ

Thus, ๐‘ฅ๐‘Ž+1๏Œบ๏Šฐ๏Šง is an antiderivative of ๐‘ฅ๏Œบ, provided ๐‘Žโ‰ โˆ’1. The most general antiderivative of ๐‘“(๐‘ฅ)=๐‘ฅ๏Œบ can be found by adding the constant of antidifferentiation as ๐น(๐‘ฅ)=๐‘ฅ๐‘Ž+1+,๐‘Žโ‰ โˆ’1,๏Œบ๏Šฐ๏ŠงC which we can verify by differentiating this expression with respect to ๐‘ฅ. For the case when ๐‘Ž=โˆ’1, that is, the antiderivative of ๐‘ฅ๏Šฑ๏Šง, we note that (|๐‘ฅ|)=๐‘ฅ.ln๏Ž˜๏Šฑ๏Šง

Thus, the general antiderivative of ๐‘“(๐‘ฅ)=๐‘ฅ๏Œบ when ๐‘Ž=โˆ’1 is ๐น(๐‘ฅ)=|๐‘ฅ|+.lnC

Letโ€™s look at a few examples where we apply these rules, starting with finding an antiderivative of a polynomial function.

Example 1: Finding the General Antiderivative of a Polynomial Function

Find the most general antiderivative ๐น(๐‘ฅ) of the function ๐‘“(๐‘ฅ)=2๐‘ฅโˆ’3๐‘ฅโˆ’๐‘ฅ๏Šญ๏Šซ๏Šจ.

Answer

In this example, we want to find the most general antiderivative of a given polynomial function.

The general antiderivative of a function ๐‘“(๐‘ฅ) is the function ๐น(๐‘ฅ)+C, for all Cโˆˆโ„, where ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ). We can also absorb the constant C into the function ๐น(๐‘ฅ) for simplicity.

We recall that the antiderivative is linear; the antiderivative of a sum is the sum of the antiderivatives. This means we can find the antiderivative of each term separately and combine the constants of antidifferentiation to find the general antiderivative of a function.

By using the power rule for differentiation, we can show that ๏€พ๐‘ฅ๐‘Ž+1๏Š=๐‘ฅ,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏Šง๏Ž˜๏Œบ

Therefore, the general antiderivative of ๐‘ฅ๏Œบ is ๐‘ฅ๐‘Ž+1+,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏ŠงC

The general antiderivative of ๐‘“(๐‘ฅ)=2๐‘ฅโˆ’3๐‘ฅโˆ’๐‘ฅ๏Šญ๏Šซ๏Šจ can be found using these rules as ๐น(๐‘ฅ)=2๏€พ๐‘ฅ7+1๏Šโˆ’3๏€พ๐‘ฅ5+1๏Šโˆ’๏€พ๐‘ฅ2+1๏Š+=28๐‘ฅโˆ’36๐‘ฅโˆ’13๐‘ฅ+=๐‘ฅ4โˆ’๐‘ฅ2โˆ’๐‘ฅ3+.๏Šญ๏Šฐ๏Šง๏Šซ๏Šฐ๏Šง๏Šจ๏Šฐ๏Šง๏Šฎ๏Šฌ๏Šฉ๏Šฎ๏Šฌ๏ŠฉCCC

Now letโ€™s consider an example where we have to determine the constant of antidifferentiation by applying a boundary condition.

Example 2: Finding the Antiderivative of a Polynomial Function

Determine the antiderivative ๐น of the function ๐‘“(๐‘ฅ)=5๐‘ฅ+4๐‘ฅ๏Šช๏Šฉ, where ๐น(1)=โˆ’2.

Answer

In this example, we want to determine the unique antiderivative of a function with a given boundary condition.

The general antiderivative of a function ๐‘“(๐‘ฅ) is the function ๐น(๐‘ฅ)+C, for all Cโˆˆโ„, where ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ). We can also absorb the constant C into the function ๐น(๐‘ฅ) for simplicity.

We recall that the antiderivative is linear; the antiderivative of a sum is the sum of the antiderivatives. This means we can find the antiderivative of each term separately and combine the constants of antidifferentiation to find the general antiderivative of a function.

By using the power rule for differentiation, we can show that ๏€พ๐‘ฅ๐‘Ž+1๏Š=๐‘ฅ,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏Šง๏Ž˜๏Œบ

Therefore, the general antiderivative of ๐‘ฅ๏Œบ is ๐‘ฅ๐‘Ž+1+,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏ŠงC

The general antiderivative of ๐‘“(๐‘ฅ)=5๐‘ฅ+4๐‘ฅ๏Šช๏Šฉ can be found from these rules as ๐น(๐‘ฅ)=5๏€พ๐‘ฅ4+1๏Š+4๏€พ๐‘ฅ3+1๏Š+=๐‘ฅ+๐‘ฅ+.๏Šช๏Šฐ๏Šง๏Šฉ๏Šฐ๏Šง๏Šซ๏ŠชCC

If we are also given ๐น(๐‘ฅ)=๐น๏Šฆ๏Šฆ, also known as a boundary condition, then we can also determine the constant C to give a unique antiderivative that satisfies the boundary condition.

Using the boundary condition ๐น(1)=โˆ’2, we obtain ๐น(1)=(1)+(1)+=โˆ’2,๏Šซ๏ŠชC which after rearranging gives C=โˆ’4.

Therefore, the unique antiderivative which satisfies the boundary condition ๐น(1)=โˆ’2 is given by ๐น(๐‘ฅ)=๐‘ฅ+๐‘ฅโˆ’4.๏Šซ๏Šช

The next example involves taking an antiderivative twice, in order to determine the function from its second derivative.

Example 3: Finding the Expression of a Function given Its Second Derivative Using Integration

If ๐‘“โ€ฒโ€ฒ(๐‘ฅ)=3๐‘ฅ+3๐‘ฅ+5๐‘ฅ+2๏Šซ๏Šฉ, determine ๐‘“(๐‘ฅ).

Answer

In this example, we want to determine the function ๐‘“(๐‘ฅ) from the expression of its second derivative ๐‘“โ€ฒโ€ฒ(๐‘ฅ) using the process of antidifferentiation.

The general antiderivative of a function ๐‘“(๐‘ฅ) is the function ๐น(๐‘ฅ)+C, for all Cโˆˆโ„, where ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ). We can also absorb the constant C into the function ๐น(๐‘ฅ) for simplicity.

Since ๐‘“โ€ฒ(๐‘ฅ) is the derivative of ๐‘“(๐‘ฅ), ๐‘“(๐‘ฅ) is an antiderivative of ๐‘“โ€ฒ(๐‘ฅ) and similarly ๐‘“โ€ฒ(๐‘ฅ) is an antiderivative of ๐‘“โ€ฒโ€ฒ(๐‘ฅ). Therefore, in order to determine ๐‘“(๐‘ฅ) from ๐‘“โ€ฒโ€ฒ(๐‘ฅ), we need to perform the process of antidifferentiation twice.

We recall that the antiderivative is linear; the antiderivative of a sum is the sum of the antiderivatives. This means we can find the antiderivative of each term separately and combine the constants of antidifferentiation to find the general antiderivative of a function.

By using the power rule for differentiation, we can show that ๏€พ๐‘ฅ๐‘Ž+1๏Š=๐‘ฅ,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏Šง๏Ž˜๏Œบ

Therefore, the general antiderivative of ๐‘ฅ๏Œบ is ๐‘ฅ๐‘Ž+1+,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏ŠงC

The general antiderivative of ๐‘“โ€ฒโ€ฒ(๐‘ฅ)=3๐‘ฅ+3๐‘ฅ+5๐‘ฅ+2๏Šซ๏Šฉ can be found from these rules allowing us to determine ๐‘“โ€ฒ(๐‘ฅ) as ๐‘“โ€ฒ(๐‘ฅ)=3๏€พ๐‘ฅ5+1๏Š+3๏€พ๐‘ฅ3+1๏Š+5๏€พ๐‘ฅ1+1๏Š+2๏€พ๐‘ฅ1+0๏Š+=๐‘ฅ2+3๐‘ฅ4+5๐‘ฅ2+2๐‘ฅ+.๏Šซ๏Šฐ๏Šง๏Šฉ๏Šฐ๏Šง๏Šง๏Šฐ๏Šง๏Šง๏Šฐ๏Šฆ๏Šฌ๏Šช๏ŠจCC

Repeating this process to find the general antiderivative of ๐‘“โ€ฒ(๐‘ฅ), we find ๐‘“(๐‘ฅ) as ๐‘“(๐‘ฅ)=12๏€พ๐‘ฅ6+1๏Š+34๏€พ๐‘ฅ4+1๏Š+52๏€พ๐‘ฅ2+1๏Š+2๏€พ๐‘ฅ1+1๏Š+๏€พ๐‘ฅ1+0๏Š+=๐‘ฅ14+3๐‘ฅ20+5๐‘ฅ6+๐‘ฅ+๐‘ฅ+.๏Šฌ๏Šฐ๏Šง๏Šช๏Šฐ๏Šง๏Šจ๏Šฐ๏Šง๏Šง๏Šฐ๏Šง๏Šง๏Šฐ๏Šฆ๏Šญ๏Šซ๏Šฉ๏ŠจCDCD

The next example involves finding the unique expression of a function using its first derivative and a boundary condition.

Example 4: Finding the Expression of a Function given Its First Derivative and the Value of the Function at a Point

Determine the function ๐‘“ if ๐‘“โ€ฒ(๐‘ฅ)=โˆ’3๐‘ฅ+1โˆš๐‘ฅ and ๐‘“(1)=4.

Answer

For this example, we want to determine the unique antiderivative of a function with a given boundary condition.

The general antiderivative of a function ๐‘“(๐‘ฅ) is the function ๐น(๐‘ฅ)+C, for all Cโˆˆโ„, where ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ). We can also absorb the constant C into the function ๐น(๐‘ฅ) for simplicity.

Since ๐‘“โ€ฒ(๐‘ฅ) is the derivative of ๐‘“(๐‘ฅ), ๐‘“(๐‘ฅ) is an antiderivative of ๐‘“โ€ฒ(๐‘ฅ).

We recall that the antiderivative is linear; the antiderivative of a sum is the sum of the antiderivatives. This means we can find the antiderivative of each term separately and combine the constants of antidifferentiation to find the general antiderivative of a function.

By using the power rule for differentiation, we can show that ๏€พ๐‘ฅ๐‘Ž+1๏Š=๐‘ฅ,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏Šง๏Ž˜๏Œบ

Therefore, the general antiderivative of ๐‘ฅ๏Œบ is ๐‘ฅ๐‘Ž+1+,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏ŠงC

We begin by rewriting ๐‘“โ€ฒ(๐‘ฅ) as ๐‘“โ€ฒ(๐‘ฅ)=โˆ’3๐‘ฅ+1๐‘ฅ=โˆ’3๐‘ฅ๐‘ฅ+1๐‘ฅ=โˆ’3๐‘ฅ+๐‘ฅ.๏Ž ๏Žก๏Ž ๏Žก๏Ž ๏Žก๏Ž ๏Žก๏Žช๏Ž ๏Žก

We can now take the antiderivative by applying the power rule for antidifferentiation given above: ๐‘“(๐‘ฅ)=โˆ’3๏‚๐‘ฅ+1๏Ž+๏‚๐‘ฅ+1๏Ž+=โˆ’2๐‘ฅ+2โˆš๐‘ฅ+.๏Ž ๏Žก๏Žช๏Ž ๏Žก๏Žข๏Žก๏Šฐ๏Šง๏Šง๏Šจ๏Šฐ๏Šง๏Šฑ๏Šง๏ŠจCC

If we are given ๐‘“(๐‘ฅ)=๐‘“๏Šฆ๏Šฆ, also known as a boundary condition, then we can also determine the constant C to give a unique antiderivative that satisfies the boundary condition.

Using the boundary condition ๐‘“(1)=4, ๐‘“(1)=โˆ’2(1)+2โˆš1+=4.๏Žข๏ŽกC

Thus, we find C=4.

Therefore, the unique antiderivative of ๐‘“โ€ฒ(๐‘ฅ) satisfying the given boundary condition is given by ๐‘“(๐‘ฅ)=โˆ’2๐‘ฅ+2โˆš๐‘ฅ+4.๏Žข๏Žก

So far, we have seen how to find the antiderivative of a function involving powers of ๐‘ฅ by working backward from a derivative. Similarly, we can apply the same process to some other common functions (e.g., exponential or trigonometric) that we encounter, by first computing the derivative and working backward to find an antiderivative with the inclusion of the constant +C. This allows us to create a table of general antiderivatives.

These can also be verified directly by computing the derivatives of ๐น(๐‘ฅ) to show that they equal the corresponding ๐‘“(๐‘ฅ). For example, if ๐น(๐‘ฅ)+=1๐‘Ž๐‘’+CC๏Œบ๏—, then the derivative can be found as ๐‘“(๐‘ฅ)=(๐น(๐‘ฅ)+)=๏€ผ1๐‘Ž๐‘’+๏ˆ=1๐‘Ž(๐‘’)+0=1๐‘Žร—๐‘Ž๐‘’=๐‘’.CC๏Ž˜๏Œบ๏—๏Ž˜๏Œบ๏—๏Ž˜๏Œบ๏—๏Œบ๏—

For more complicated functions which are given by the sum or product of these standard functions, we can use a few rules to help us determine an antiderivative, such as the linearity of the antiderivative that we established.

In order to see this, suppose we want to find an antiderivative of ๐‘“(๐‘ฅ)=2๐‘’+๐‘ฅ.๏—sin

This is a sum of two different types of functions: an exponential and trigonometric function. We can determine an antiderivative of 2๐‘’๏— and sin๐‘ฅ separately and add the result. The derivative of โˆ’๐‘ฅcos is sin๐‘ฅ; thus the antiderivative of sin๐‘ฅ is โˆ’๐‘ฅcos and similarly the antiderivative of 2๐‘’๏— is itself. In particular, we find ๐น(๐‘ฅ)+=2๐‘’โˆ’๐‘ฅ+,CcosC๏— which we can verify by taking the derivative.

Now, letโ€™s look at a few more examples in order to practice and deepen our understanding. The next example involves trigonometric function and a noninteger power of ๐‘ฅ.

Example 5: Finding the General Antiderivative of a Given Function Involving Trigonometric and Root Functions

Find the most general antiderivative of the function ๐‘“(๐‘ฅ)=4๐‘ฅ+3โˆ’23โˆš๐‘ฅsin.

Answer

In this example, we want to determine the most general antiderivative of a function involving a root and trigonometric function.

The general antiderivative of a function ๐‘“(๐‘ฅ) is the function ๐น(๐‘ฅ)+C, for all Cโˆˆโ„, where ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ). We can also absorb the constant C into the function ๐น(๐‘ฅ) for simplicity.

We recall that the antiderivative is linear; the antiderivative of a sum is the sum of the antiderivatives. This means we can find the antiderivative of each term separately and combine the constants of antidifferentiation to find the general antiderivative of a function.

By using the power rule for differentiation, we can show that ๏€พ๐‘ฅ๐‘Ž+1๏Š=๐‘ฅ,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏Šง๏Ž˜๏Œบ

Therefore, the general antiderivative of ๐‘ฅ๏Œบ is ๐‘ฅ๐‘Ž+1+,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏ŠงC

We also note that (โˆ’๐‘ฅ)=๐‘ฅ.cossin๏Ž˜

Thus, the general antiderivative of sin๐‘ฅ is โˆ’๐‘ฅ+.cosC

We begin by rewriting ๐‘“(๐‘ฅ) as ๐‘“(๐‘ฅ)=4๐‘ฅ+3โˆ’23โˆš๐‘ฅ=4๐‘ฅ+3โˆ’23๐‘ฅ.sinsin๏Žช๏Ž ๏Žก

Now, we can determine the most general antiderivative by using the rules above as ๐น(๐‘ฅ)=โˆ’4๐‘ฅ+3๐‘ฅโˆ’23๏‚๐‘ฅ+1๏Ž+=โˆ’4๐‘ฅ+3๐‘ฅโˆ’43๐‘ฅ+.cosCcosC๏Žช๏Ž ๏Žก๏Ž ๏Žก๏Šฐ๏Šง๏Šฑ๏Šง๏Šจ

Therefore, the most general antiderivative is given by ๐น(๐‘ฅ)=โˆ’4โˆš๐‘ฅ3+3๐‘ฅโˆ’4๐‘ฅ+.cosC

Now, letโ€™s look at an example where we have to determine a function by its third derivative involving a trigonometric and square root function, by finding an antiderivative three times.

Example 6: Finding the Expression of a Function given Its Third Derivative

Determine ๐‘“(๐‘ก) if ๐‘“โ€ฒโ€ฒโ€ฒ(๐‘ก)=โˆ’4โˆš๐‘ก+5๐‘กcos.

Answer

In this example, we want to find the most general expression of a function from its third derivative ๐‘“โ€ฒโ€ฒโ€ฒ(๐‘ก) using the process of antidifferentiation.

The general antiderivative of a function ๐‘“(๐‘ก) is the function ๐น(๐‘ก)+C, for all Cโˆˆโ„, where ๐นโ€ฒ(๐‘ก)=๐‘“(๐‘ก). We can also absorb the constant C into the function ๐น(๐‘ก) for simplicity.

Since ๐‘“โ€ฒ(๐‘ก) is the derivative of ๐‘“(๐‘ก), ๐‘“(๐‘ก) is an antiderivative of ๐‘“โ€ฒ(๐‘ก); similarly, ๐‘“โ€ฒ(๐‘ก) is the antiderivative of ๐‘“โ€ฒโ€ฒ(๐‘ก) and ๐‘“โ€ฒโ€ฒ(๐‘ก) is the antiderivative of ๐‘“โ€ฒโ€ฒโ€ฒ(๐‘ก). Therefore, in order to determine ๐‘“(๐‘ก) from ๐‘“โ€ฒโ€ฒโ€ฒ(๐‘ก), we need to perform the process of antidifferentiation three times.

We recall that the antiderivative is linear; the antiderivative of a sum is the sum of the antiderivatives. This means we can find the antiderivative of each term separately and combine the constants of antidifferentiation to find the general antiderivative of a function.

By using the power rule for differentiation, we can show that ๏€พ๐‘ก๐‘Ž+1๏Š=๐‘ก,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏Šง๏Ž˜๏Œบ Therefore, the general antiderivative of ๐‘ก๏Œบ is ๐‘ก๐‘Ž+1+,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏ŠงC

We also note that (โˆ’๐‘ก)=๐‘ก,(๐‘ก)=๐‘ก.cossinsincos๏Ž˜๏Ž˜

Thus, the general antiderivative of sin๐‘ก is โˆ’๐‘ก+,cosC and the general antiderivative of cos๐‘ก is sinC๐‘ก+.

We can also write the root function as โˆš๐‘ก=๐‘ก๏Ž ๏Žก.

The general antiderivative of ๐‘“โ€ฒโ€ฒโ€ฒ(๐‘ก)=โˆ’4๐‘ก+5๐‘ก๏Ž ๏Žกcos can be found from these rules allowing us to determine ๐‘“โ€ฒโ€ฒ(๐‘ก) as ๐‘“โ€ฒโ€ฒ(๐‘ก)=โˆ’4๏‚๐‘ก+1๏Ž+5๐‘ก+๏‚—=โˆ’83๐‘ก+5๐‘ก+๏‚—.๏Ž ๏Žก๏Žข๏Žก๏Šฐ๏Šง๏Šง๏ŠจsinCsinC

Repeating the process to find the general antiderivative of ๐‘“โ€ฒโ€ฒ(๐‘ก), we find ๐‘“โ€ฒ(๐‘ก) as ๐‘“โ€ฒ(๐‘ก)=โˆ’83๏‚๐‘ก+1๏Žโˆ’5๐‘ก+๏‚—๐‘ก+=โˆ’1615๐‘กโˆ’5๐‘ก+๏‚—๐‘ก+.๏Žข๏Žก๏Žค๏Žก๏Šฐ๏Šง๏Šฉ๏ŠจcosCDcosCD

Once more, to determine the antiderivative of ๐‘“โ€ฒ(๐‘ก), we find ๐‘“(๐‘ก): ๐‘“(๐‘ก)=โˆ’1615๏‚๐‘ก+1๏Žโˆ’5๐‘ก+๏‚—๏€พ๐‘ก1+1๏Š+๐‘ก+=โˆ’32๐‘ก105โˆ’5๐‘ก+๏‚—๐‘ก2+๐‘ก+=โˆ’32๐‘ก105โˆ’5๐‘ก+๐‘ก+๐‘ก+.๏Žค๏Žก๏Žฆ๏Žก๏Žฆ๏Žก๏Šฐ๏Šง๏Šซ๏Šจ๏Šง๏Šฐ๏Šง๏Šจ๏ŠจsinCDEsinCDEsinCDE

We note that we made the replacement CC=๏‚—2, since this is just another constant which we can rewrite for simplicity.

If we have a product of functions ๐น(๐‘ฅ)=๐‘ˆ(๐‘ฅ)๐‘‰(๐‘ฅ)+C, where ๐‘ˆ(๐‘ฅ) and ๐‘‰(๐‘ฅ) are the antiderivatives of ๐‘ข(๐‘ฅ) and ๐‘ฃ(๐‘ฅ) with ๐‘ข(๐‘ฅ)=๐‘ˆโ€ฒ(๐‘ฅ) and ๐‘ฃ(๐‘ฅ)=๐‘‰โ€ฒ(๐‘ฅ), we know from the product rule that ๐‘“(๐‘ฅ)=๐นโ€ฒ(๐‘ฅ)=(๐‘ˆ(๐‘ฅ)๐‘‰(๐‘ฅ)+)=๐‘ˆโ€ฒ(๐‘ฅ)๐‘‰(๐‘ฅ)+๐‘‰โ€ฒ(๐‘ฅ)๐‘ˆ(๐‘ฅ)=๐‘ข(๐‘ฅ)๐‘‰(๐‘ฅ)+๐‘ฃ(๐‘ฅ)๐‘ˆ(๐‘ฅ).C๏Ž˜

Therefore, an antiderivative of the combination ๐‘ข(๐‘ฅ)๐‘‰(๐‘ฅ)+๐‘ฃ(๐‘ฅ)๐‘ˆ(๐‘ฅ) is ๐‘ˆ(๐‘ฅ)๐‘‰(๐‘ฅ) and the most general antiderivative of ๐‘“(๐‘ฅ)=๐‘ข(๐‘ฅ)๐‘‰(๐‘ฅ)+๐‘ฃ(๐‘ฅ)๐‘ˆ(๐‘ฅ) is ๐น(๐‘ฅ)=๐‘ˆ(๐‘ฅ)๐‘‰(๐‘ฅ)+.C

In fact, we can follow this process for any derivative rule to find a corresponding rule for the antiderivative.

Now, suppose we want to find an antiderivative of ๐‘“(๐‘ฅ)=2๐‘’๐‘ฅ+2๐‘’๐‘ฅ.๏—๏—sincos

This is a sum of two functions, each a product of two different types of functions: an exponential and trigonometric function. In order to find the antiderivative, we have to reverse the product rule by identifying ๐‘ข(๐‘ฅ) and ๐‘ฃ(๐‘ฅ) and taking the product to compute an antiderivative. By comparing the first term in ๐‘“(๐‘ฅ) with the first term in the product rule, ๐‘ข(๐‘ฅ)๐‘‰(๐‘ฅ), we can have ๐‘ข(๐‘ฅ)๐‘‰(๐‘ฅ)=2๐‘’๐‘ฅ๏—sin and thus the two choices ๐‘ข(๐‘ฅ)=๐‘’,๐‘‰(๐‘ฅ)=2๐‘ฅ,๏—sin or ๐‘ข(๐‘ฅ)=2๐‘ฅ,๐‘‰(๐‘ฅ)=๐‘’.sin๏—

Only one of these choices will work for our antiderivative. We can find an antiderivative of ๐‘ˆ(๐‘ฅ) to determine ๐‘ข(๐‘ฅ) for both of these choices and then compare the derivative of the product ๐‘ˆ(๐‘ฅ)๐‘‰(๐‘ฅ) to see which would give us the function ๐‘“(๐‘ฅ).

The first choice would give ๐‘ˆ(๐‘ฅ)=๐‘’๏— since the antiderivative of ๐‘’๏— is itself and the product ๐‘ˆ(๐‘ฅ)๐‘‰(๐‘ฅ)=2๐‘’๐‘ฅ๏—sin: ddsinsincos๐‘ฅ(2๐‘’๐‘ฅ)=2๐‘’๐‘ฅ+2๐‘’๐‘ฅ.๏—๏—๏—

Clearly this choice works as the right-hand side is equivalent to ๐‘“(๐‘ฅ). For completeness, letโ€™s also examine the second choice. An antiderivative of ๐‘ข(๐‘ฅ)=2๐‘ฅsin is ๐‘ˆ(๐‘ฅ)=โˆ’2๐‘ฅcos, which would give us the product ๐‘ˆ(๐‘ฅ)๐‘‰(๐‘ฅ)=โˆ’2๐‘ฅ๐‘’cos๏—. By taking the derivative of this product, we have ddcossincos๐‘ฅ(โˆ’2๐‘’๐‘ฅ)=2๐‘’๐‘ฅโˆ’2๐‘’๐‘ฅ.๏—๏—๏—

This does not work since the second term has a different sign compared to the function ๐‘“(๐‘ฅ). Thus, the most general antiderivative of ๐‘“(๐‘ฅ)=2๐‘’๐‘ฅ+2๐‘’๐‘ฅ๏—๏—sincos is ๐น(๐‘ฅ)=2๐‘’๐‘ฅ+.๏—sinC

We note that we could have also done this with the second term in the product rule and ๐‘“(๐‘ฅ) with ๐‘ฃ(๐‘ฅ)๐‘ˆ(๐‘ฅ)=2๐‘’๐‘ฅ๏—cos, but this would give us the same result; thus, it is sufficient to check just one of the terms to find the appropriate product, which is the antiderivative that we require, and then differentiating to check if this gives back the function ๐‘“(๐‘ฅ).

Finally, consider the function ๐‘“(๐‘ฅ)=๐‘ฅ๐‘’+2๐‘ฅ๐‘’.๏Šจ๏—๏—

Again this is the sum of two functions, each a product of a power of ๐‘ฅ and an exponential. Therefore, upon identification of ๐‘ˆ(๐‘ฅ)=๐‘ฅ๏Šจ and ๐‘‰(๐‘ฅ)=๐‘’๏—, we obtain ๐น(๐‘ฅ)=๐‘ฅ๐‘’+,๏Šจ๏—C which we can also verify by taking the derivative ๐นโ€ฒ(๐‘ฅ)=๏€น๐‘ฅ๐‘’+๏…=๏€น๐‘ฅ๐‘’๏…+0=๐‘ฅ๐‘’+2๐‘ฅ๐‘’=๐‘“(๐‘ฅ).๏Šจ๏—๏Ž˜๏Šจ๏—๏Ž˜๏Šจ๏—๏—C

Finally, letโ€™s consider an example where we have to reverse the product rule on a function involving a square root and exponential.

Example 7: Using the Product Rule to Find an Antiderivative

By considering the product rule, find a function ๐‘“ so that ๐‘“โ€ฒ(๐‘ฅ)=๐‘’โˆš๐‘ฅ+2๐‘’โˆš๐‘ฅ๏—๏—.

Answer

In this example, we want to determine the antiderivative of ๐‘“โ€ฒ(๐‘ฅ) by considering the product rule.

The general antiderivative of a function ๐‘“(๐‘ฅ) is the function ๐น(๐‘ฅ)+C, for all Cโˆˆโ„, where ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ). We can also absorb the constant C into the function ๐น(๐‘ฅ) for simplicity.

We want to determine an antiderivative of a function by considering the product rule in reverse. Recall that the product rule, for two differentiable functions ๐‘ข and ๐‘ฃ, is given by (๐‘ข๐‘ฃ)=๐‘ขโ€ฒ๐‘ฃ+๐‘ฃโ€ฒ๐‘ข.๏Ž˜

By using the power rule for differentiation, we can show that ๏€พ๐‘ฅ๐‘Ž+1๏Š=๐‘ฅ,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏Šง๏Ž˜๏Œบ

Therefore, the general antiderivative of ๐‘ฅ๏Œบ is ๐‘ฅ๐‘Ž+1+,๐‘Žโ‰ โˆ’1.๏Œบ๏Šฐ๏ŠงC

We also note the derivative of ๐‘’๏— is itself; thus the general antiderivative of ๐‘’๏— is ๐‘’+๏—C.

If we compare terms, we can identify ๐‘ฃโ€ฒ(๐‘ฅ)=2๐‘’,๐‘ข(๐‘ฅ)=โˆš๐‘ฅ,๏— where an antiderivative of ๐‘ฃโ€ฒ(๐‘ฅ) is ๐‘ฃ(๐‘ฅ)=2๐‘’,๏— or ๐‘ฃโ€ฒ(๐‘ฅ)=โˆš๐‘ฅ,๐‘ข(๐‘ฅ)=2๐‘’,๏— where an antiderivative of ๐‘ฃโ€ฒ(๐‘ฅ) is ๐‘ฃ(๐‘ฅ)=๐‘ฅ+1=23๐‘ฅ.๏Ž ๏Žก๏Žข๏Žก๏Šฐ๏Šง๏Šง๏Šจ

The first choice gives our product as ๐‘ข(๐‘ฅ)๐‘ฃ(๐‘ฅ)=2๐‘’โˆš๐‘ฅ๏—, which we can verify directly by computing the derivative with the product rule with ๐‘ข(๐‘ฅ)=2๐‘’๏— and ๐‘ฃ(๐‘ฅ)=โˆš๐‘ฅ: ๐‘ขโ€ฒ(๐‘ฅ)=๐‘ฅ(๐‘’)=๐‘’,dd๏—๏— and ๐‘ฃโ€ฒ(๐‘ฅ)=๐‘ฅ๏€บโˆš๐‘ฅ๏†=๐‘ฅ๏€ฝ๐‘ฅ๏‰=12๐‘ฅ=12๐‘ฅ=12โˆš๐‘ฅ.dddd๏Ž ๏Žก๏Ž ๏Žก๏Žช๏Ž ๏Žก๏Šฑ๏Šง

Thus, using the product rule, we have dd๐‘ฅ๏€บ2โˆš๐‘ฅ๐‘’๏†=๐‘’โˆš๐‘ฅ+2โˆš๐‘ฅ๐‘’,๏—๏—๏— which is clearly equal to ๐‘“โ€ฒ(๐‘ฅ) and we can write ๐‘“โ€ฒ(๐‘ฅ)=๐‘ฅ๏€บ2โˆš๐‘ฅ๐‘’๏†.dd๏—

Therefore, the general antiderivative of ๐‘“โ€ฒ(๐‘ฅ) is given by ๐‘“(๐‘ฅ)=2โˆš๐‘ฅ๐‘’+.๏—C

In this explainer, we determined antiderivatives of a function by essentially reversing the process of differentiation. There are more direct ways to compute an antiderivative, using something called an integral. In particular, an antiderivative of a function ๐‘“(๐‘ฅ) is equivalent to the indefinite integral of ๐‘“(๐‘ฅ). Thus, if ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ), then ๐น(๐‘ฅ)=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ+,dC where C is also called the constant of integration. To determine indefinite integrals, there are many tools we can use, which makes finding an antiderivative easier. The fundamental theorem of calculus also provides the connection between derivatives and definite integrals, which can be interpreted as the area under the curve of ๐‘“(๐‘ฅ) within an interval.

These are beyond the scope of this explainer and will be covered in another lesson in more detail.

Key Points

  • Antidifferentiation is the reverse process of differentiation.
  • If ๐น(๐‘ฅ) is an antiderivative of ๐‘“(๐‘ฅ), then the general antiderivative is ๐น(๐‘ฅ)+C, for all Cโˆˆโ„. In other words, antiderivatives always include a +C, which give a family of (general) antiderivatives parameterized by the constant C.
  • A unique antiderivative can be determined with the inclusion of a boundary condition.
  • Antiderivatives satisfy certain properties, similar to derivatives. We can summarize these properties in a table, where ๐‘ˆ, ๐‘‰, and ๐น are the antiderivatives of ๐‘ข, ๐‘ฃ, and ๐‘“ respectively.
    Original FunctionGeneral Antiderivative
    Constant multiple rule: ๐‘Ž๐‘“(๐‘ฅ)๐‘Ž๐น(๐‘ฅ)+C
    Sum rule: ๐‘ข(๐‘ฅ)+๐‘ฃ(๐‘ฅ)๐‘ˆ(๐‘ฅ)+๐‘‰(๐‘ฅ)+C
    Product rule: ๐‘ข(๐‘ฅ)๐‘‰(๐‘ฅ)+๐‘ˆ(๐‘ฅ)๐‘ฃ(๐‘ฅ)๐‘ˆ(๐‘ฅ)๐‘‰(๐‘ฅ)+C

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