Year 11: Introduction to integration

Integration is the reverse process of differentiation. If differentiation tells you how fast something is changing, integration helps you rebuild the original function or calculate accumulated change. In Year 11, the focus is usually on integrating polynomials, understanding the constant of integration, and using definite integrals to measure signed area.

To integrate a polynomial term, increase the power by one and divide by the new power. Then remember to add the constant of integration

+ C

for indefinite integrals.

What to master

Reversing the power rule for polynomial functions.
Understanding why indefinite integrals include $+ C$ .
Evaluating definite integrals by substitution into an antiderivative.
Interpreting definite integrals as signed area or accumulated change.

Indefinite integration

If $d / d x (x^{n + 1}) = (n + 1) x^{n}$ , then the reverse rule is: $\int x^{n} d x = x^{n + 1} / (n + 1) + C$ , for $n \neq - 1$ .

Example 1: Find

\int (3 x^{2} - 4 x + 5) d x

Integrate each term separately.
$\int 3 x^{2} d x = x^{3}$ .
$\int - 4 x d x = - 2 x^{2}$ .
$\int 5 d x = 5 x$ .
Add the constant: $x^{3} - 2 x^{2} + 5 x + C$ .

Answer:

\int (3 x^{2} - 4 x + 5) d x = x^{3} - 2 x^{2} + 5 x + C

The $+ C$ matters because many different functions have the same derivative. For example, the derivative of $x^{2}$ , $x^{2} + 3$ , and $x^{2} - 10$ is the same.

Checking by differentiating

A fast self-check is to differentiate your answer. If you return to the original integrand, your antiderivative is correct.

Example 2: Check

\int 6 x d x = 3 x^{2} + C

Differentiate $3 x^{2} + C$ to get $6 x$ . The result matches, so the integration is correct.

Definite integration

A definite integral has limits, such as $\int_{a}^{b} f (x) d x$ . First find an antiderivative, then substitute the upper bound and subtract the lower bound.

Example 3: Evaluate

\int_{0}^{2} (3 x^{2} + 1) d x

An antiderivative is $x^{3} + x$ .
Substitute the upper bound: $2^{3} + 2 = 10$ .
Substitute the lower bound: $0^{3} + 0 = 0$ .
Subtract: $10 - 0 = 10$ .

Answer: the definite integral is

10

Area interpretation

In many Year 11 questions, a definite integral represents area between a curve and the x-axis. If the graph lies above the x-axis, the value is positive. If part of it lies below, the integral gives signed area, so you need to think carefully about interpretation.

Example 4: Find the area under

y = 2 x + 1

from

x = 0

x = 3

Integrate: $\int (2 x + 1) d x = x^{2} + x$ .
Evaluate from $0$ to $3$ : $(3^{2} + 3) - (0 + 0)$ .
$= 12$ .

Since the graph stays above the x-axis on this interval, the integral is also the area.

Applications of integration

Once the technique is secure, integration can be used in modelling questions. You may be given a rate function and asked for total change, or asked to find the area associated with a polynomial graph. Those questions are usually easier if you first draw a quick sketch and note where the graph is above or below the axis.

Common mistakes

Forgetting the $+ C$ in an indefinite integral.
Increasing the power but forgetting to divide by the new power.
Using the antiderivative correctly but not subtracting lower from upper in a definite integral.
Assuming a definite integral always equals geometric area, even when part of the graph is below the axis.

Study routine

The best revision pattern is to integrate, then differentiate your result immediately. That check catches most power-rule errors. For definite integrals, write the antiderivative first, place square brackets around it, and substitute the upper and lower bounds in a clean separate line.