Integration as reverse differentiation | CAIE 9709 Pure Mathematics 2 Notes

Chapters

Pure Mathematics 2

Mathematics9709

Reverse the derivative, then check it

Integration runs differentiation backwards

integrand

f(x)

⇄

family

F(x) + C

→ integrate

← differentiate to check

C disappears when differentiated.

An antiderivative is correct only if differentiating it returns the integrand.

F'(x)=f(x)

Therefore:

\int f(x)\,dx=F(x)+C

An indefinite integral gives a family of functions, so it ends with $+C$ . Different constants disappear when differentiated.
A definite integral has limits and gives one value; the constant cancels, so do not add $+C$ .
Differentiate your answer. If it does not return the original integrand, repair the coefficient or sign.

Key idea

Integration is not a new list of unrelated tricks. Every rule in this chapter can be rebuilt from a derivative you already know.

Cancel the derivative of the inside

Balance the linear inside

inside

u = ax + b

→

inside derivative

du/dx = a

Reverse the outside, then divide by a

Read the sign: 3 − 2x has a = −2.

The inside derivative creates a; integration cancels it with 1/a.

\int e^{ax+b}\,dx=\frac1a e^{ax+b}+C

Here $a\ne0$ .

Worked example

Keep the inside unchanged

\int 6e^{3x-1}\,dx

The inside is $3x-1$ , whose derivative is $3$ .
Keep $e^{3x-1}$ and divide the outside coefficient by 3.

\int 6e^{3x-1}\,dx=2e^{3x-1}+C

Check: $(2e^{3x-1})'=2(3)e^{3x-1}$ , which returns the integrand.

Common mistake

Read

a

with its sign. For

e^{\,2-x}

a=-1

, so the antiderivative is

-e^{\,2-x}+C

Use limits only after integrating

In $\int_a^b f(x)\,dx$ , $a$ is the lower limit and $b$ is the upper limit. If $F'(x)=f(x)$ , the value is $F(b)-F(a)$ .

Find an antiderivative without adding rounded decimals.
Substitute the upper limit.
Subtract the value at the lower limit.

Your turnIndependent check [4 marks]

Evaluate $\displaystyle\int_0^{\ln2}4e^{2x}\,dx$ exactly.

Show worked answer

\int4e^{2x}\,dx=2e^{2x}

\left[2e^{2x}\right]_0^{\ln2}=2e^{2\ln2}-2

Since $e^{2\ln2}=e^{\ln4}=4$ , the exact value is:

Answer

6