Edexcel IAL - Pure Maths 3- 5.1 Integration of functions and their sums and differences- Study notes - New syllabus
Edexcel IAL – Pure Maths 3- 5.1 Integration of functions and their sums and differences -Study notes- New syllabus
Edexcel IAL – Pure Maths 3- 5.1 Integration of functions and their sums and differences -Study notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- 5.1 Integration of functions and their sums and differences
Integration of Exponential, Logarithmic and Trigonometric Functions
Integration is the inverse process of differentiation.
If:
\( \dfrac{dy}{dx} = f(x) \)
then:
\( y = \displaystyle \int f(x)\,dx \)
Standard Integration Formulae
| Function | Integral |
| \( e^{kx} \) | \( \displaystyle \int e^{kx}\,dx = \dfrac{1}{k}e^{kx} + C \) |
| \( \dfrac{1}{x} \) | \( \displaystyle \int \dfrac{1}{x}\,dx = \ln|x| + C \) |
| \( \sin kx \) | \( \displaystyle \int \sin kx\,dx = -\dfrac{1}{k}\cos kx + C \) |
| \( \cos kx \) | \( \displaystyle \int \cos kx\,dx = \dfrac{1}{k}\sin kx + C \) |
Constant Multiple Rule
If \( a \) is a constant, then:
\( \displaystyle \int a f(x)\,dx = a \int f(x)\,dx \)
Integration of Sums and Differences
Integration is performed term by term.
\( \displaystyle \int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx \)
Common Standard Integrals
- \( \displaystyle \int \sin 3x\,dx = -\dfrac{1}{3}\cos 3x + C \)
- \( \displaystyle \int e^{5x}\,dx = \dfrac{1}{5}e^{5x} + C \)
- \( \displaystyle \int \dfrac{1}{2x}\,dx = \dfrac{1}{2}\ln|x| + C \)
Example
Find:
\( \displaystyle \int \sin 3x\,dx \)
▶️ Answer / Explanation
\( \displaystyle \int \sin 3x\,dx = -\dfrac{1}{3}\cos 3x + C \)
Example
Find:
\( \displaystyle \int \left(e^{5x} + \cos 2x\right)\,dx \)
▶️ Answer / Explanation
\( \displaystyle \int e^{5x}\,dx = \dfrac{1}{5}e^{5x} \)
\( \displaystyle \int \cos 2x\,dx = \dfrac{1}{2}\sin 2x \)
Answer:
\( \dfrac{1}{5}e^{5x} + \dfrac{1}{2}\sin 2x + C \)
Example
Find:
\( \displaystyle \int \left(\dfrac{1}{2x} – 3\sin 4x + e^{2x}\right)\,dx \)
▶️ Answer / Explanation
\( \displaystyle \int \dfrac{1}{2x}\,dx = \dfrac{1}{2}\ln|x| \)
\( \displaystyle \int (-3\sin 4x)\,dx = \dfrac{3}{4}\cos 4x \)
\( \displaystyle \int e^{2x}\,dx = \dfrac{1}{2}e^{2x} \)
Answer:
\( \dfrac{1}{2}\ln|x| + \dfrac{3}{4}\cos 4x + \dfrac{1}{2}e^{2x} + C \)