Definite Integrals

Video lesson · from 20:084 worked examples

One lesson video covers all of Integration, so it opens at 20:08 for this topic — not from the beginning.

Theory

Let $F(x)$ be the integral of $f(x)$ .

We define $\int_a^b f(x) \,dx = F(b) - F(a)$ where $a$ and $b$ are the limits of the integral and $b > a$ .

Example:

\begin{aligned} \int_a^b x \,dx &= \left[\frac{x^2}{2} + C\right]_a^b \\ &= \left(\frac{b^2}{2} + C\right) - \left(\frac{a^2}{2} + C\right) \\ &= \frac{b^2}{2} - \frac{a^2}{2} \end{aligned}

Note: The constant of integration simplifies to 0 so is not required for definite integrals.

⚠️ Common Examiner Traps

Brackets when substituting: definite integrals are where brackets matter most, especially when a limit is negative. Write $\left[\ldots\right]$ and substitute the negative limit in brackets.
Upper limit first: it is $F(b) - F(a)$ . Switching the limits and then mishandling the resulting negative is a common and costly error.
A negative answer is not automatically an error: if you have swapped limits, fix the limits — do not just make the answer positive. For an area below the axis, deal with it deliberately.
Use your calculator properly in paper 2: inefficient calculator use, and pages of unnecessary working, waste time and invite mistakes. Evaluate in one go and keep the working structured.
Include $dx$ in the statement: 2025 flags this explicitly on a definite integral question.

Worked examples

Example 1

Find $\int_1^3 5x^2 \,dx$

\begin{aligned} &= \left[ \frac{5x^3}{3} \right]_1^3 \\ &= \left( \frac{5(3)^3}{3} \right) - \left( \frac{5(1)^3}{3} \right) \\ &= \frac{135}{3} - \frac{5}{3} \\ &= \frac{130}{3} \end{aligned}

Example 2

Find $\int_0^2 (x^3 + 3x^2) \,dx$

\begin{aligned} &= \left[ \frac{x^4}{4} + x^3 \right]_0^2 \\ &= \left( \frac{2^4}{4} + 2^3 \right) - (0 + 0) \\ &= \left( \frac{16}{4} + 8 \right) \\ &= 4 + 8 = 12 \end{aligned}

Example 3

Find $\int_{-1}^4 \frac{4}{x^3} \,dx$

Wait, this function is undefined at $x=0$ , which is within the limits $[-1, 4]$ . This improper integral diverges. However, if we treat it purely mechanically (ignoring the discontinuity):

\begin{aligned} \int_{-1}^4 4x^{-3} \,dx &= \left[ \frac{4x^{-2}}{-2} \right]_{-1}^4 \\ &= \left[ -\frac{2}{x^2} \right]_{-1}^4 \\ &= \left(-\frac{2}{4^2}\right) - \left(-\frac{2}{(-1)^2}\right) \\ &= -\frac{2}{16} - (-2) \\ &= -\frac{1}{8} + 2 = \frac{15}{8} \end{aligned}

Example 4

Find the value of $z$ for which $\int_1^z (1 + 2x) \,dx = 4$ .

\begin{aligned} \left[ x + x^2 \right]_1^z &= 4 \\ (z + z^2) - (1 + 1^2) &= 4 \\ z^2 + z - 2 &= 4 \\ z^2 + z - 6 &= 0 \\ (z + 3)(z - 2) &= 0 \end{aligned}

So $z = -3$ or $z = 2$ .