IB Applications and Interpretation SL · Differential and integral calculus · Section 6 of 7

Antiderivatives of powers and indefinite integrals

3 worked examples · By Álvaro López

Key idea

Integrating undoes differentiation: given $f$ , we look for an antiderivative $F$ whose derivative is $f$ . In the exam it appears mostly with powers and polynomials, and almost always comes with a condition that allows you to work out the constant in the answer.

Antiderivatives and the constant of integration

If $F$ is an antiderivative of $f$ , so is $F (x) + C$ for any number $C$ , because the derivative of a constant is zero. That is why a function has infinitely many antiderivatives, which differ only by a constant.

To check a result, differentiate your answer: you should get the integrand. It is the quickest check and the one that avoids almost all mistakes.

The power rule

Integrating a power is the reverse of differentiating it. When differentiating, the exponent goes down by one and multiplies the power; when integrating, the exponent goes up by one and the power is divided by the new exponent.

Constants, sums and multiples

With the power rule and three more properties, any polynomial can be integrated term by term.

Finding an antiderivative with a condition

A general antiderivative has a constant $C$ with no value. To determine a particular curve, the question gives a point it passes through, and that point fixes $C$ .

With a parameter in the derivative, each condition gives one equation. Two conditions allow you to find the constant of integration and the parameter at the same time.

Worked examples

Example 1 Basic

An antiderivative through a point

The function $f (x) = 6 x^{2} - 4 x + 1$ is the derivative of a function $F$ . Find the antiderivative $F$ that satisfies $F (0) = 5$ .

Solution

We integrate term by term: $F (x) = \int (6 x^{2} - 4 x + 1) d x = \frac{6 x ^{3}}{3} - \frac{4 x ^{2}}{2} + x + C = 2 x^{3} - 2 x^{2} + x + C$ The condition $F (0) = 5$ gives $F (0) = C$ , so $C = 5$ .

Check: $F (0) = 0 - 0 + 0 + 5 = 5$ , and the derivative of $2 x^{3} - 2 x^{2} + x + 5$ is $6 x^{2} - 4 x + 1$ , which is $f$ .

Answer: $F (x) = 2 x^{3} - 2 x^{2} + x + 5$ .

Example 2 Medium

Water flowing into a tank

A water tank receives water at a rate of $v (t) = 3 t^{2} - 12 t + 20$ litres per minute, for $0 \leq t \leq 10$ , where $t$ is the time in minutes. When measuring starts ( $t = 0$ ), the tank holds $150$ litres.

a) Find the function $V (t)$ that gives the amount of water in the tank at time $t$ .

b) Calculate the amount of water in the tank after $4$ minutes.

Solution

a) The amount of water is the antiderivative of the inflow rate: $V (t) = \int (3 t^{2} - 12 t + 20) d t = t^{3} - 6 t^{2} + 20 t + C$ Since $V (0) = 150$ and $V (0) = C$ , we have $C = 150$ . Therefore $V (t) = t^{3} - 6 t^{2} + 20 t + 150$

b) Substitute $t = 4$ : $V (4) = 4^{3} - 6 \cdot 4^{2} + 20 \cdot 4 + 150 = 64 - 96 + 80 + 150 = 198$ After $4$ minutes there are $198$ litres in the tank.

Answer: $V (t) = t^{3} - 6 t^{2} + 20 t + 150$ , and after $4$ minutes there are $198$ litres.

Example 3 Hard

A parameter in the derivative

The derivative of a function $f$ is $f^{'} (x) = 3 x^{2} + 2 a x$ , where $a$ is a real number. It is known that $f (0) = 1$ and $f (1) = 4$ .

a) Find the value of $a$ .

b) Write the complete function $f (x)$ .

Solution

a) Integrate the derivative: $f (x) = \int (3 x^{2} + 2 a x) d x = x^{3} + a x^{2} + C$ With $f (0) = 1$ we get $C = 1$ , so $f (x) = x^{3} + a x^{2} + 1$ . The second condition gives $f (1) = 1 + a + 1 = 4 \Rightarrow a = 2$

b) With $a = 2$ , the function is $f (x) = x^{3} + 2 x^{2} + 1$ .

Check: $f (0) = 1$ , $f (1) = 1 + 2 + 1 = 4$ , and $f^{'} (x) = 3 x^{2} + 4 x = 3 x^{2} + 2 \cdot 2 \cdot x$ , which matches the given derivative.

In the exam: each condition gives one equation. With a parameter and a constant, you need two conditions, and the exam awards marks for both the value of $a$ and the final function.

Common mistakes

Summary

An antiderivative of $f$ is a function whose derivative is $f$ . All antiderivatives differ by a constant $C$ .
Power rule: $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ for $n \neq = - 1$ .
Constant factors come outside, and the integral of a sum is the sum of the integrals.
To find $f$ with a condition, integrate, substitute the point and calculate $C$ .
Always check your result by differentiating it.

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