08 Antiderivatives and Indefinite Integrals

Understanding Antiderivatives

An antiderivative reverses the process of differentiation. A function $F$ is an Antiderivative of $f$ on an interval when

For example, because

one antiderivative of $3x^2$ is $x^3$. However, it is not the only one. Adding any constant does not change the derivative:

Thus, all antiderivatives of $3x^2$ can be written as $x^3+C$. More generally, any two antiderivatives of the same function differ by a constant on an interval.

For $f(x)=6x-4$, integrate each term:

Differentiating gives

which verifies the result.

Takeaway: Differentiation removes constants, so reversing differentiation requires including an arbitrary constant.

Reading Integral Notation

An Indefinite integral is notation for the complete family of antiderivatives:

where $F'(x)=f(x)$. The parts of the notation have specific roles:

• $\int$ is the integral sign.

• $f(x)$ is the Integrand.

• $dx$ identifies $x$ as the variable of integration.

• $C$ is the arbitrary constant.

For example,

This expression represents infinitely many functions, including $x^4$, $x^4+5$, and $x^4-12$. All have derivative $4x^3$.

Do not confuse an indefinite integral with a Definite integral. An expression such as

has bounds and produces a number when evaluated, whereas an indefinite integral produces a function family.

Takeaway: The absence of bounds signals a family of antiderivatives and therefore requires $+C$.

Core Integration Rules

Integration rules are obtained by reversing familiar differentiation rules. The constant rule is

where $k$ is constant. The Power rule is

The exponent increases by one, and the result is divided by the new exponent. The exceptional case $n=-1$ must be handled separately:

The absolute value is required because the derivative of $\ln|x|$ is $1/x$ for every $x\ne0$.

Constants can be factored out, and sums or differences can be integrated term by term:

Useful formulas include

• $\int e^x\,dx=e^x+C$

• $\int\cos x\,dx=\sin x+C$

• $\int\sin x\,dx=-\cos x+C$

• $\int\sec^2x\,dx=\tan x+C$

• $\int\csc^2x\,dx=-\cot x+C$

• $\int\sec x\tan x\,dx=\sec x+C$

• $\int\csc x\cot x\,dx=-\csc x+C$

• $\int\frac{1}{1+x^2}\,dx=\arctan x+C$

• $\int\frac{1}{\sqrt{1-x^2}}\,dx=\arcsin x+C$

For a polynomial, integrate each term:

Differentiate the result to check every coefficient and exponent.

Takeaway: Apply linearity first, then use the appropriate basic formula, paying special attention to the $1/x$ exception.

Solving Initial-Value Problems

An Initial-value problem supplies both a differential equation and a value of the unknown function. A typical form is

First find the general solution by integrating:

Then substitute the given input and output to determine $C$.

For

integration gives

Applying the condition produces

so the particular solution is

Verify both parts: its derivative is $4x^3-2x$, and its value at $x=1$ is $5$.

Takeaway: Integration gives a family of solutions; the initial condition selects one member of that family.

Motion from Velocity and Acceleration

In one-dimensional motion, position, velocity, and acceleration are linked by differentiation:

Integration reverses these relationships. If velocity is known, integrate once to find position. If acceleration is known, integrate once to find velocity and a second time to find position. Each integration introduces a constant, which must be determined from an initial position or velocity.

Suppose

Since $s'(t)=v(t)$,

Using $s(0)=3$ gives $C=3$, so

If instead

first integrate acceleration:

The velocity condition gives $C_1=7$, so $v(t)=4t+7$. Integrating again,

The position condition gives $C_2=2$, so

A final check is

A negative velocity means motion in the negative direction relative to the chosen positive direction.

Takeaway: Keep the order clear: acceleration integrates to velocity, and velocity integrates to position.

Substitution and the Reverse Chain Rule

Substitution, also called u-substitution, reverses the Chain rule. The chain rule has the form

Therefore, when an integrand contains an inner expression and its derivative, set

The integral then becomes an integral in $u$. Use this sequence:

1. Identify the inner expression $g(x)$.

2. Set $u=g(x)$.

3. Compute $du=g'(x)\,dx$.

4. Rewrite the entire integral using $u$ and $du$.

5. Integrate with respect to $u$.

6. Substitute back for $u$.

7. Differentiate the result to verify it.

For example,

has inner expression $u=3x^2+4$, with $du=6x\,dx$. Thus,

Substituting back gives

When the derivative appears only up to a constant factor, adjust for that factor. For

let $u=x^2+1$. Since $du=2x\,dx$, we have $x\,dx=du/2$. Therefore,

Takeaway: Successful substitution replaces the inner expression and its differential completely, then returns to the original variable at the end.

Checking Work and Avoiding Errors

Several mistakes recur in antiderivative problems:

• Omitting the constant: Every indefinite integral needs $+C$.

• Applying the power rule to $1/x$: Because the exponent is $-1$, use $\ln|x|+C$ instead.

• Ignoring a missing factor: If $du=2x\,dx$, then $x\,dx=du/2$.

• Leaving the answer in terms of $u$: For an indefinite integral, substitute back into the original variable.

• Confusing motion quantities: Integrate acceleration to obtain velocity before integrating velocity to obtain position.

• Skipping verification: Differentiate the final expression whenever possible.

A useful general check is to compare the derivative of the proposed answer with the original integrand:

For an initial-value problem, also substitute the specified input into the final function and confirm the required value.

Takeaway: Constants, exceptional formulas, variable changes, physical quantities, and differentiation checks are the main points requiring deliberate attention.

• Constant of integration