Resources for Applications
-
Questions
17
With Worked SolutionClick Here -
Video Tutorials
4
Click Here -
HSC Questions
5
With Worked SolutionClick Here
Applications Theory
![Applications - Integration methods can be applied to finding particular solutions to differential equations, determine areas of regions bounded by curves and volumes of solids when parts of curves are rotated about \(x\) or \(y\) axes.\\ \begin{multicols}{2} \textbf{Example 1}\\ %35340 Find the particular solution of \(\dfrac{{dy}}{{dx}} = {e^{ - x}}\cos x\) given that \(y=\dfrac{1}{2}\) when \(x=0\) \\ \textbf{Example 1 solution}\\ $\begin{aligned} &\frac{d y}{d x}=e^{-x} \cos x \\&\therefore y=\int e^{-x} \cos x d x \\&\int u d v=u v-\int v d u \\&u=e^{-x} \quad d v=\cos x \\&d u=-e^{-x} \quad v=\sin x \\&y=e^{-x} \sin x+\int e^{-x} \sin x d x \\ \text{Let }I&=\int e^{-x} \sin x d x \\&u=e^{-x} \quad\quad\quad d v=\sin x \\&d u=-e^{-x} \quad\quad v =-\cos x \\&I=-e^{-x} \cos x-\int e^{-x} \cos x d x \end{aligned}$\\ $\begin{aligned}I&=-e^{-x} \cos x-y \\y&=e^{-x} \sin x+I \\&=e^{-x} \sin x-e^{-x} \cos x-y \\2 y&=e^{-x}(\sin x-\cos x) \\y&=\frac{e^{-x}}{2}(\sin x-\cos x)+c \\&\text { when } x=0, y=\frac{1}{2} \end{aligned}$\\ $\begin{aligned} \qquad\frac{1}{2} &=\frac{1}{2}(0-1)+c \rightarrow \therefore c=1 \\\therefore y &=\frac{e^{-x}}{2}(\sin x-\cos )+1\end{aligned}$\\ \columnbreak \textbf{Example 2}\\ %35341 Find the area of the region bounded by the curve \(y=\sin ^{-1}x\), the \(x\)-axis and the lines \(x=\dfrac{-1}{2}\) and \(x=\dfrac{1}{2}\) \\ \textbf{Example 2 solution}\\ $\begin{aligned} &y=\sin ^{-1} x \text { is an odd function } \\&\therefore \quad \text { Area }=2 \int_{0}^{\frac{1}{2}} \sin ^{-1} x d x \\&\int u d v=u v-\int v d u \\&\quad u=\sin ^{-1} x\qquad d v=1 \\&\quad du=\frac{1}{\sqrt{1-x^{2}}} \quad v=x \end{aligned}$\\ $\begin{aligned}\text { Area }=& 2 \times \frac{\pi}{12}-2 \int_{0}^{\frac{1}{2}} x\left(1-x^{2}\right)^{-\frac{1}{2}} d x \\=& \frac{\pi}{6}+\int_{0}^{\frac{1}{2}}-2 x\left(1-x^{2}\right)^{-\frac{1}{2}} d x \\=& \frac{\pi}{6}+2\left[\sqrt{1-x^{2}}\right]_{0}^{\frac{1}{2}} \\=& \frac{\pi}{6}+\sqrt{3}-2\end{aligned}$\\ \end{multicols}](/media/jkxhdnsq/applications.png)
