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โ™พ๏ธap calculus ab/bc review

6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation

Verified for the 2025 AP Calculus AB/BC examโ€ขCitation:

6.8 Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation

Just like how there are basic rules for calculating derivatives, there are rules for calculating antiderivatives. Since antiderivatives are the inverses of derivatives, these rules are mostly the reverse of the basic derivative rules. ๐Ÿ’ก


๐Ÿ“ˆ Indefinite Integrals: Notation

Letโ€™s first talk about a family of functions before we dive into reversing the derivative process.

Imagine we have two different antiderivatives, $F(x) = x^2+3$ and $G(x) = x^2-2$.

If we were to take the derivative of both of these functions, we would find that they both have the same derivative, $2x$. If we reverse the derivative process through integration, how do we account for arriving at these two different antiderivatives? Introducing the magical constant $C$! ๐Ÿช„

When we integrate $2x$, the antiderivative is $2x+C$ where $C$ is any constant. This result is often referred to as a family of functions because they vary only in the value of their constants and all share the same derivative.

This type of integral is referred to as an indefinite integral because we canโ€™t be sure which member of the family of antiderivatives is at play. If the bounds of the integral are not specified as they are in a definite integral, always add โ€˜+Cโ€™ to the end of your antiderivative!

Hereโ€™s a general look at the notation!

โˆซf(x)dx=F(x)+C\int f(x)dx=F(x)+C

Where $F'(x)=f(x)$ and $C$ represents the integration constant.


๐Ÿ“ Indefinite Integrals: Basic Rules

Now letโ€™s look at how to reverse the process of some of the derivatives we learned early in our study of calculus.

Reverse Power Rule

First up, we have the reverse power rule. This essentially refers to how to take the indefinite integral of a function, which is the reverse of the power rule used for differentiation. Suppose we have the following function:

f(x)=xn+1n+1+Cf(x) = \frac{x^{n+1}}{n+1}+C

Where $n \neq -1$ since $n=-1$ causes $f(x)$ to be undefined.

What is its derivative?

If we recall the power rule for derivatives, we see that the derivative of $f(x)$ is

fโ€ฒ(x)=xnf'(x) = x^{n}

Now, what is the antiderivative of the derivative of $f(x)$?

Using the fact that antiderivatives and derivatives are inverses, we see that...

โˆซxndx=xn+1n+1+C\int x^{n}dx = \frac{x^{n+1}}{n+1}+C

This is the reverse power rule. Youโ€™re basically adding one to the exponent of each term and dividing by the new exponent!

Reverse Power Rule Example 1

Evaluate the following integral:

โˆซx3dx\int x^3dx

Using the reverse power rule, we see that

โˆซx3dx=x3+13+1+C=x44+C\int x^3dx=\frac{x^{3+1}}{3+1}+C=\frac{x^{4}}{4}+C

Reverse Power Rule Example 2

Give the following a try! A useful tip is to rewrite fractions with negative exponents. You can also apply this logic to radical functions, since they can be rewritten with fractional exponents.

โˆซ(1x2โˆ’7x3+2x2โˆ’x+4)โ€‰dx\int(\frac{1}{x^2}-7x^3+2x^2-x+4)\, dx

When we rewrite the first term, we see thatโ€ฆ

โˆซ(xโˆ’2โˆ’7x3+2x2โˆ’x+4)โ€‰dx\int(x^{-2}-7x^3+2x^2-x+4)\, dx

Once we use the reverse power rule and evaluate this integral term by term, we get:

โˆซ(1x2โˆ’7x3+2x2โˆ’x+4)โ€‰dx=โˆ’1xโˆ’7x44+2x33โˆ’x22+4x+C\int(\frac{1}{x^2}-7x^3+2x^2-x+4)\, dx = -\frac{1}{x} -\frac{7x^4}{4}+\frac{2x^3}{3}-\frac{x^2}{2}+4x+C

Sums and Multiples Rules for Antiderivatives

If you recall, we learned in Unit 2 that there were the sums and multiples rules for derivatives. Similarly, there are the sums and multiples rules for antiderivatives.

The sums rule states that

โˆซ[f(x)+g(x)]dx=โˆซf(x)dx+โˆซg(x)dx\int \left[f(x)+g(x)\right]dx=\int f(x)dx + \int g(x)dx

The multiples rule states that

โˆซcโ‹…f(x)dx=cโˆซf(x)dx\int c \cdot f(x)dx=c \int f(x)dx

Sums and Multiples Examples

Here are examples of these two rules in action, the first covering the sums rule and the second covering the multiples rule.

โˆซ[x4+x2]dx=โˆซx4dx+โˆซx2dx\int \left[x^4+x^2\right]dx=\int x^4dx + \int x^2dx โˆซ5x6dx=5โˆซx6dx\int 5x^{6}dx=5\int x^6dx

Antiderivatives of Trigonometric Functions

When youโ€™re first learning your trig antiderivatives, you may find it useful to think to yourself, โ€œWhat has a derivative ofโ€ฆ?โ€

Antiderivative of $\sin(x)$

If you recall, $\frac{d}{dx}[\cos(x)]=-\sin(x)$. This means that $\frac{d}{dx}[-\cos(x)]=\sin(x)$. Therefore,

โˆซsinโก(x)dx=โˆ’cosโก(x)+C\int \sin(x)dx=-\cos(x)+C

Antiderivative of $\cos(x)$

If you recall, $\frac{d}{dx}[\sin(x)]=\cos(x)$. Therefore,

โˆซcosโก(x)dx=sinโก(x)+C\int \cos(x)dx=\sin(x)+C

Other Antiderivatives of Trig Functions

I would also know the following trig integrals for the AP exam:

โˆซsec2(x)โ€‰dx=tan(x)+C\int sec^2(x) \, dx = tan(x) +C โˆซcsc2(x)โ€‰dx=โˆ’cot(x)+C\int csc^2(x) \, dx = -cot(x) +C โˆซsec(x)tan(x)โ€‰dx=sec(x)+C\int sec(x)tan(x) \, dx = sec(x) +C โˆซcsc(x)cot(x)โ€‰dx=โˆ’csc(x)+C\int csc(x)cot(x) \, dx = -csc(x) +C

Antiderivatives of Inverse Trig Functions

These integrals arenโ€™t nearly as common on the AP test, but below are the forms you may encounter on the AP test.

โˆซ11โˆ’x2โ€‰dx=sinโˆ’1(x)+C\int \frac{1}{\sqrt{1-x^2}} \, dx = sin^{-1}(x) +C โˆซ11+x2โ€‰dx=tanโˆ’1(x)+C\int \frac{1}{{1+x^2}} \, dx = tan^{-1}(x) +C

Antiderivatives of Transcendental Functions ($\frac{1}{x}$, $e^x$)

Finally, we have the integrals for the transcendental functions you are likely to encounter on the AP exams.

Antiderivative of $\frac{1}{x}$

If you recall, $\frac{d}{dx}[\ln(x)]=\frac{1}{x}$. Therefore, it is not a bad guess to say that $\int \frac{1}{x}dx=\ln(x)+C$.

However, because of the domain of $\ln(x)$, which is $(0, \infty)$, if we want to be able to take the antiderivative of $\frac{1}{x}$ for any positive or negative $x$, we need to rewrite this rule as

โˆซ1xdx=lnโกโˆฃxโˆฃ+C\int \frac{1}{x}dx=\ln\mid x \mid + C

Antiderivative of $e^{x}$

If you recall, $\frac{d}{dx}[e^{x}]=e^{x}$. Therefore,

โˆซexdx=ex+C\int e^xdx=e^x+C

๐Ÿ“ Indefinite Integrals Practice Problems

Now that you know all the basic rules for antiderivatives, letโ€™s do some practice problems!

โ“ Indefinite Integrals Problems

Evaluate each of the following integrals.

1.โˆซx7dx=?1. \int x^7dx=? 2.โˆซ[x4+cosโก(x)]dx=?2.\int \left[x^4+\cos(x)\right]dx=? 3.โˆซ[4cosโก(x)+ex]dx=?3.\int \left[4\cos(x)+e^x\right]dx=? 4.โˆซ(3x+x2)โ€‰dx4.\int (\frac{3}{x}+x^2) \, dx

โœ… Indefinite Integrals Question Solutions

Indefinite Integrals Question 1

When we take a look at question 1, we can quickly tell that we have to use the reverse power rule!

โˆซx7dx=x7+17+1+C=18x8+C\int x^7dx=\frac{x^{7+1}}{7+1}+C=\boxed{\frac{1}{8}x^{8}+C}

Indefinite Integrals Question 2

Using the sums rule for antiderivatives, we see that

โˆซ[x4+cosโก(x)]dx=โˆซx4dx+โˆซcosโก(x)dx\int \left[x^4+\cos(x)\right]dx=\int x^4dx+\int \cos(x)dx

This means that we can take the antiderivatives of the two terms separately and then sum their individual antiderivatives together afterward.

Using the reverse power rule, we see that

โˆซx4dx=x4+14+1+C=15x5+C\int x^4dx = \frac{x^{4+1}}{4+1}+C=\frac{1}{5}x^5 + C

And using the antiderivative of $\cos(x)$, we see that

โˆซcosโก(x)dx=sinโก(x)+C\int \cos(x)dx = \sin(x)+C

Combining these two parts, we get

โˆซ[x4+cosโก(x)]dx=15x5+sinโก(x)+C\int \left[x^4+\cos(x)\right]dx=\boxed{\frac{1}{5}x^5 +\sin(x)+ C}

Indefinite Integrals Question 3

Using the sums rule for antiderivatives, we see that

โˆซ[4cosโก(x)+ex]dx=โˆซ4cosโก(x)dx+โˆซexdx\int \left[4\cos(x)+e^x\right]dx=\int 4\cos(x)dx + \int e^xdx

This means that we can take the antiderivatives of the two terms separately and then sum their individual antiderivatives together afterward.

Using the multiples rule and the antiderivative of $\cos(x)$, we see that

โˆซ4cosโก(x)dx=4โˆซcosโก(x)dx=4sinโก(x)+C\int 4\cos(x)dx=4\int \cos(x)dx=4\sin(x)+C

And using the antiderivative of $e^x$, we see that

โˆซexdx=ex+C\int e^xdx=e^x+C

Combining these two parts, we get

โˆซ[4cosโก(x)+ex]dx=4sinโก(x)+ex+C\int \left[4\cos(x)+e^x\right]dx=\boxed{4\sin(x)+e^x+C}

Indefinite Integrals Question 4

Using the sums rule for antiderivatives, we see that

โˆซ(3x+x2)โ€‰dx=โˆซ3xdx+โˆซx2โ€‰dx\int (\frac{3}{x}+x^2) \, dx=\int \frac{3}{x}dx+ \int x^2 \, dx

We can again take the antiderivatives of the two terms separately and then sum their individual antiderivatives together afterward.

Letโ€™s take the integral of the first term, using the following rule $\int \frac{1}{x}dx=\ln\mid x \mid + C$

โˆซ(3x)dx=3ln(โˆฃxโˆฃ)+Cโˆซ (\frac{3}{x}) dx = 3 ln(|x|) + C

And using the reverse power rule, we see that

โˆซx2dx=x33+Cโˆซ x^2 dx = \frac{x^3}{3} + C

Combining these two parts, we get

โˆซ(3x+x2)โ€‰dx=3ln(โˆฃxโˆฃ)+x33+C\int (\frac{3}{x}+x^2) \, dx =\boxed{3ln(|x|) + \frac{x^3}{3}+C}

โœจ Closing

Woah! We've covered the reverse power rule, sums and multiples rules for antiderivatives, antiderivatives of trigonometric functions, inverse trig functions, transcendental functions, and practiced.

My biggest tip? Remember that taking integrals involves the reverse process of differentiation and you must add $+C$ to the end of your answer of an indefinite integral. Good luck! ๐Ÿ€