Year 12 Integration For A Level Maths
Year 12 Integration For A Level Maths – Introduction
In this article we will continue to look at Year 12 Integration for A Level Maths. The focus is on working with indefinite integrals and also solving first order differential equations.
It is important to remember the rules of indices when performing integral calculations as well as simplifying single algebraic fractions into more than one term so that the expression is in a form which can be used for year 12 integration for A Level Maths.
There are 14 questions in total which are then followed by a detailed solution.
It is important to be able to practise these types of questions and to be fully confident when doing the question. Always remember the constant “c” when doing an indefinite integral.
Year 12 Integration For A Level Maths - Questions 1 - 5
Q1.
Solution
\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x}=-x^3+\frac{4 x-5}{2 x^3} & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=-x^3+2 x^{-2}-\frac{5}{2} x^{-3} \\ & \Rightarrow y=-\frac{1}{4} x^4+2\left(\frac{x^{-1}}{-1}\right)-\frac{5}{2}\left(\frac{x^{-2}}{-2}\right)+c \\ & \Rightarrow y=-\frac{1}{4} x^4-2 x^{-1}+\frac{5}{4} x^{-2}+c . \end{aligned}Passes (1, 7)
\begin{aligned} &7=-\frac{1}{4}-2+\frac{5}{4}+c \Rightarrow c=8\\ &y=-\frac{1}{4} x^4-2 x^{-1}+\frac{5}{4} x^{-2}+8 . \end{aligned}Q2.
Solution
\begin{aligned} \int\left(10 x^4-4 x-\frac{3}{\sqrt{x}}\right) \mathrm{d} x & =\int\left(10 x^4-4 x-3 x^{-\frac{1}{2}}\right) \mathrm{d} x \\ & =10 \times \frac{1}{5} x^5-4 \times \frac{1}{2} x^2-3 \times 2 x^{\frac{1}{2}}+c \\ & =2 x^5-2 x^2-6 x^{\frac{1}{2}}+c . \end{aligned}Q3.
Solution
\mathrm{f}^{\prime}(x)=\frac{\left(3-x^2\right)^2}{x^2}=\frac{9-6 x^2+x^4}{x^2}=\underline{\underline{9 x^{-2}-6+x^2}} .\begin{aligned} \mathrm{f}^{\prime}(x)=\frac{\left(3-x^2\right)^2}{x^2} & \Rightarrow \mathrm{f}^{\prime}(x)=9 x^{-2}-6+x^2 \\ & \Rightarrow \mathrm{f}(x)=9 \times\left(\frac{x^{-1}}{-1}\right)-6 x+\frac{1}{3} x^3+c \\ & \Rightarrow \mathrm{f}(x)=-9 x^{-1}-6 x+\frac{1}{3} x^3+c . \end{aligned}Passes (-3, 10)
\begin{aligned} &10=3+18-9+c \Rightarrow c=-2\\ &\mathrm{f}(x)=-9 x^{-1}-6 x+\frac{1}{3} x^3-2 . \end{aligned}Q4.
Solution
\begin{aligned} \int\left(3 x^2-\frac{4}{x^2}\right) \mathrm{d} x & =\int\left(3 x^2-4 x^{-2}\right) \mathrm{d} x \\ & =3 \times \frac{1}{3} x^3-4\left(\frac{x^{-1}}{-1}\right)+c \\ & =x^3+4 x^{-1}+c . \end{aligned}Q5.
Solution
\begin{aligned} \mathrm{f}^{\prime}(x)=\frac{x+9}{\sqrt{x}} & \Rightarrow \mathrm{f}^{\prime}(x)=\frac{x+9}{x^{\frac{1}{2}}} \\ & \Rightarrow \mathrm{f}^{\prime}(x)=x^{\frac{1}{2}}+9 x^{-\frac{1}{2}} \\ & \Rightarrow \mathrm{f}(x)=\frac{2}{3} x^{\frac{3}{2}}+9 \times 2 x^{\frac{1}{2}}+c \\ & \Rightarrow \mathrm{f}(x)=\frac{2}{3} x^{\frac{3}{2}}+18 x^{\frac{1}{2}}+c . \end{aligned}Passes (9, 0)
\begin{aligned} &0=18+54+c \Rightarrow c=-72\\ &\mathrm{f}(x)=\frac{2}{3} x^{\frac{3}{2}}+18 x^{\frac{1}{2}}-72 \end{aligned}Year 12 Integration For A Level Maths - Questions 6 - 10
Q6.
Solution
\begin{aligned} \left.\int\left(8 x^3+4\right)\right) \mathrm{d} x & =8 \times \frac{1}{4} x^4+4 x+c \\ & =2 x^4+4 x+c . \end{aligned}Q7.
Solution
\begin{aligned} \mathrm{f}^{\prime}(x)=\frac{3}{8} x^2-10 x^{-\frac{1}{2}}+1 & \Rightarrow \mathrm{f}(x)=\frac{3}{8} \times \frac{1}{3} x^3-10 \times 2 x^{\frac{1}{2}}+x+c \\ & \Rightarrow \mathrm{f}(x)=\frac{1}{8} x^3-20 x^{\frac{1}{2}}+x+c . \end{aligned}Passes (4, 25)
\begin{aligned} &25=8-40+4+c \Rightarrow c=53\\ &f(x)=\frac{1}{8} x^3-20 x^{\frac{1}{2}}+x+53 \end{aligned}Q8.
Solution
\begin{aligned} \int\left(2 x^5+\frac{6}{\sqrt{x}}\right) \mathrm{d} x & =\int\left(2 x^5+6 x^{-\frac{1}{2}}\right) \mathrm{d} x \\ & =2 \times \frac{1}{6} x^6+6 \times 2 x^{\frac{1}{2}}+c \\ & =\frac{1}{3} x^6+12 x^{\frac{1}{2}}+c . \end{aligned}Q9.
Solution
\begin{aligned} \frac{\mathrm{d} y}{\mathrm{~d} x}=6 x^{-\frac{1}{2}}+x \sqrt{x} & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=6 x^{-\frac{1}{2}}+x^{\frac{3}{2}} \\ & \Rightarrow y=6 \times 2 x^{\frac{1}{2}}+\frac{2}{5} x^{\frac{5}{2}}+c \\ & \Rightarrow y=12 x^{\frac{1}{2}}+\frac{2}{5} x^{\frac{5}{2}}+c . \end{aligned}Passes (4, 37)
\begin{aligned} &37=24+12 \frac{4}{5}+c \Rightarrow c=\frac{1}{5}\\ &y=12 x^{\frac{1}{2}}+\frac{2}{5} x^{\frac{5}{2}}+\frac{1}{5} . \end{aligned}Q10.
\begin{aligned} \int\left(4 x^3-\frac{5}{x^2}\right) \mathrm{d} x & =\int\left(4 x^3-5 x^{-2}\right) \mathrm{d} x \\ & =4 \times \frac{1}{4} x^4-5\left(\frac{x^{-1}}{-1}\right)+c \\ & =x^4+5 x^{-1}+c . \end{aligned}Year 12 Integration For A Level Maths – Questions 11 – 14
Solution
\begin{aligned} \int\left(2 x^4-\frac{4}{\sqrt{x}}+3\right) \mathrm{d} x & =\int\left(2 x^4-4 x^{-\frac{1}{2}}+3\right) \mathrm{d} x \\ & =2 \times \frac{1}{5} x^5-4 \times 2 x^{\frac{1}{2}}+3 x+c \\ & =\frac{2}{5} x^5-8 x^{\frac{1}{2}}+3 x+c . \end{aligned}Q13.
Solution
\begin{aligned} \int\left(2 x^5-\frac{1}{4 x^3}-5\right) \mathrm{d} x & =\int\left(2 x^5-\frac{1}{4} x^{-3}-5\right) \mathrm{d} x \\ & =2 \times \frac{1}{6} x^4-\frac{1}{4} \times\left(-\frac{1}{2} x^{-2}\right)-5 x+c \\ & =\frac{1}{3} x^4+\frac{1}{8} x^{-2}-5 x+c . \end{aligned}Q14
Solution
\begin{aligned} \mathrm{f}^{\prime}(x)=30+\frac{6-5 x^2}{\sqrt{x}} & \Rightarrow \mathrm{f}^{\prime}(x)=30+6 x^{-\frac{1}{2}}-5 x^{\frac{3}{2}} \\ & \Rightarrow \mathrm{f}(x)=30 x+6 \times 2 x^{\frac{1}{2}}-5 \times \frac{2}{5} x^{\frac{5}{2}}+c \\ & \Rightarrow \mathrm{f}(x)=30 x+12 x^{\frac{1}{2}}-2 x^{\frac{5}{2}}+c . \end{aligned}Passes (4, -8)
\begin{aligned} &-8=120+24-64+c \Rightarrow c=-88\\ &\mathrm{f}(x)=30 x+12 x^{\frac{1}{2}}-2 x^{\frac{5}{2}}-88 . \end{aligned}This focus on this article has been to provide additional questions on differential equations and indefinite integration. It is important that you are comfortable with all the questions shown here as the next few articles will look at definite integrals as well as determining areas under a curve and also moving on to year 13 A Level Maths.
Year 12 Integration for A Level Maths creates an important foundation for year 13 and so it is important that you have a strong grasp of the concepts in this article.