Integration A Level Maths
Integration A Level Maths – Introduction
Integration is a fundamental concept in A Level Maths that involves finding the integral (or antiderivative) of a function. It plays a crucial role in various mathematical applications, such as calculating areas, finding volumes, and solving differential equations.
Integration A Level Maths – Definite Integration:
Definite integration is used to find the exact value of the area under a curve between two given points. By evaluating the definite integral, we can determine the precise numerical value of the area enclosed. It is denoted by the integral symbol with limits of integration, such as ∫[a, b] f(x) dx.
Integration A Level Maths – Indefinite Integration:
Indefinite integration, also known as antiderivatives, involves finding a general function whose derivative is equal to the given function. It does not provide a specific numerical value but instead gives a family of functions. The symbol used for indefinite integration is ∫f(x) dx, without any limits of integration.
Integration A Level Maths - Questions 1 - 5
Q1.
\int\left(1+3 \sqrt{x}-\frac{1}{x^2}\right) \mathrm{d} xSolution
\begin{aligned} \int\left(1+3 \sqrt{x}-\frac{1}{x^2}\right) \mathrm{d} x & =\int\left(1+3 x^{\frac{1}{2}}-x^{-2}\right) \mathrm{d} x \\ & =x+3 \times \frac{2}{3} x^{\frac{3}{2}}-\frac{x^{-1}}{-1}+c \\ & =x+2 x^{\frac{3}{2}}+x^{-1}+c . \end{aligned}
Q2.
Solution
\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x}=(3 x-1)^2 & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=9 x^2-6 x+1 \\ & \Rightarrow y=9 \times \frac{1}{3} x^3-6 \times \frac{1}{2} x^2+x+c \\ & \Rightarrow y=3 x^3-3 x^2+x+c . \end{aligned}Passes (1, 4)
\begin{aligned} &4=3-3+1+c \Rightarrow c=3\\ &y=3 x^3-3 x^2+x+3 . \end{aligned}
Q3.
\int\left(6 x-\frac{4}{x^2}\right) \mathrm{d} xSolution
\begin{aligned} \int\left(6 x-\frac{4}{x^2}\right) \mathrm{d} x & =\int\left(6 x-4 x^{-2}\right) \mathrm{d} x \\ & =6 \times \frac{1}{2} x^2-\frac{4 x^{-1}}{-1}+c \\ & =3 x^2+4 x^{-1}+c . \end{aligned}Q4.
Solution
\begin{aligned} \int y \mathrm{~d} x & =\int\left(9 x^{-\frac{1}{2}}-6+x^{\frac{1}{2}}\right) \mathrm{d} x \\ & =9 \times 2 x^{\frac{1}{2}}-6 x+\frac{2}{3} x^{\frac{3}{2}}+c \\ & =18 x^{\frac{1}{2}}-6 x+\frac{2}{3} x^{\frac{3}{2}}+c . \end{aligned}y=\frac{2}{3} \text { at } x=1 \text {, }\frac{2}{3}=18-6+\frac{2}{3}+c \Rightarrow c=-12 \text {. }
y=18 x^{\frac{1}{2}}-6 x+\frac{2}{3} x^{\frac{3}{2}}-12 .
Integration A Level Maths - Questions 6 - 10
Q6.
Passes (1, 6)
\begin{aligned} &6=3+2+4+c \Rightarrow c=-3\\ &\mathrm{f}(x)=3 x+2 x^{\frac{5}{2}}+4 x^{\frac{1}{2}}-3 . \end{aligned}Q7.
\int\left(6 x^2+2+x^{-\frac{1}{2}}\right) \mathrm{d} xSolution
\begin{aligned} \int\left(6 x^2+2+x^{-\frac{1}{2}}\right) \mathrm{d} x & =6 \times \frac{1}{3} x^3+2 x+2 x^{\frac{1}{2}}+c \\ & =2 x^3+2 x+2 x^{\frac{1}{2}}+c . \end{aligned}Q8.
Solution
\begin{aligned} \mathrm{f}^{\prime}(x)=2 x+\frac{3}{x^2} & \Rightarrow \mathrm{f}^{\prime}(x)=2 x+3 x^{-2} \\ & \Rightarrow \mathrm{f}(x)=2 \times \frac{1}{2} x^2+3 \times\left(\frac{x^{-1}}{-1}\right)+c \\ & \Rightarrow \mathrm{f}(x)=x^2-3 x^{-1}+c . \end{aligned}Passes (3, 7.5)
\begin{aligned} &7 \frac{1}{2}=9-1+c \Rightarrow c=-\frac{1}{2}\\ &\mathrm{f}(x)=x^2-3 x^{-1}-\frac{1}{2} . \end{aligned}Q9.
Solution
\begin{aligned} \int(4+3 \sqrt{x})^2 \mathrm{~d} x & =\int\left(16+24 x^{\frac{1}{2}}+9 x\right) \mathrm{d} x \\ & =16 x+24 \times \frac{2}{3} x^{\frac{3}{2}}+9 \times \frac{1}{2} x^2+c \\ & =16 x+16 x^{\frac{3}{2}}+\frac{9}{2} x^2+c . \end{aligned}Q10.
Solution
\begin{aligned} \mathrm{f}^{\prime}(x)=3 x^2-6-\frac{8}{x^2} & \Rightarrow \mathrm{f}^{\prime}(x)=3 x^2-6-8 x^{-2} \\ & \Rightarrow \mathrm{f}(x)=3 \times \frac{1}{3} x^3-6 x-8 \times\left(\frac{x^{-1}}{-1}\right)+c \\ & \Rightarrow \mathrm{f}(x)=x^3-6 x+8 x^{-1}+c . \end{aligned}Pass (2, 1)
\begin{aligned} &1=8-12+4+c \Rightarrow c=1\\ &f(x)=x^3-6 x+8 x^{-1}+1 . \end{aligned}
Integration A Level Maths – Questions 11 – 15
Solution
\begin{aligned} \mathrm{f}^{\prime}(x)=4 x-6 \sqrt{x}+\frac{8}{x^2} & \Rightarrow \mathrm{f}^{\prime}(x)=4 x-6 x^{\frac{1}{2}}+8 x^{-2} \\ & \Rightarrow \mathrm{f}(x)=4 \times \frac{1}{2} x^2-6 \times \frac{2}{3} x^{\frac{3}{2}}+8 \times\left(\frac{x^{-1}}{-1}\right)+c \\ & \Rightarrow \mathrm{f}(x)=2 x^2-4 x^{\frac{3}{2}}-8 x^{-1}+c \end{aligned}Passes (4, 1)
\begin{aligned} &1=32-32-2+c \Rightarrow c=3\\ &f(x)=2 x^2-4 x^{\frac{3}{2}}-8 x^{-1}+3 . \end{aligned}Q15.
\int\left(2+5 x^2\right) \mathrm{d} x .Solution
\begin{aligned} \int\left(2+5 x^2\right) \mathrm{d} x & =2 x+5 \times \frac{1}{3} x^3+c \\ & =2 x+\frac{5}{3} x^3+c . \end{aligned}Conclusion:
In conclusion, integration A Level Maths encompasses both definite and indefinite integration. Definite integration helps us calculate the precise area under a curve between two points, while indefinite integration allows us to find a general function with a given derivative. Understanding the difference between these two concepts is essential for solving various mathematical problems and utilising integration effectively in real-world applications.