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Calculus, 2ADV C4 2025 HSC 25

  1. Show that  \(\dfrac{d}{d x}(\sin x-x\, \cos x)=x\, \sin x\).   (2 marks)
  2. Hence, find the value of  \(\displaystyle\int_0^{2025 \pi} x\, \sin x \, dx\).   (2 marks)

  3. The regions bounded by the \(x\)-axis and the graph of  \(y=x\, \sin x\)  for  \(x \geq 0\)  are shown.
     

  1. Let  \(A_n=\displaystyle \int_{(n-1) \pi}^{n \pi} x\, \sin x \,dx\),  where \(n\) is a positive integer.
  2. It can be shown that  \(\left|A_n\right|=(2 n-1) \pi\).  (Do NOT prove this.)
  3. Find the exact total area of the regions bounded by the curve  \(y=x \sin x\), and the \(x\)-axis between  \(x=0\)  and  \(x=2025 \pi\).   (2 marks)  
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a.   \(\text{See Worked Solutions}\)

b.   \(2025 \pi\)

c.   \(4\,100\,625 \pi \ \text{units}^2\)

Show Worked Solution
a.     \(\dfrac{d}{dx}(\sin x-x\, \cos x)\) \(=\dfrac{d}{dx} \sin x-\dfrac{d}{dx} x\, \cos x\)
    \(=\cos x+x\, \sin x-\cos x\)
    \(=x\, \sin x\)

 

b.     \(\displaystyle\int_0^{2025 \pi} x\, \sin x\) \(=\Big[\sin x-x\, \cos x\Big]_0^{2025 \pi}\)
    \(=\Big[(\sin (2025\pi)-2025 \pi \times \cos (2025 \pi))-0\Big]\)
    \(=0-2025 \pi \times -1\)
    \(=2025 \pi\)

 

c.    \(\text{Area}=\displaystyle \int_0^\pi x\, \sin x \, dx+\left|\int_\pi^{2 \pi} x\, \sin x \, dx\right|+\cdots+\int_{2024 \pi}^{2025 \pi} x\, \sin x \, dx\)

\(\text{Using}\ \ \left|A_n\right|=(2n-1) \pi:\)

\(A_1=\pi, A_2=3 \pi, A_3=5 \pi, \ldots, A_{2025}=4049 \pi\)

\begin{aligned}
\rule{0pt}{2.5ex} \text{Area}& =\underbrace{\pi+3 \pi+5 \pi+\ldots+4049 \pi}_{\text {AP where } a=\pi, \ l=4049 \pi, \ n=2025} \\
\rule{0pt}{4.5ex} & =\frac{2025}{2}(\pi+4049 \pi) \\
\rule{0pt}{3.5ex} & =4\,100\,625 \pi \ \text{units }^2
\end{aligned}

Calculus, 2ADV C4 SM-Bank 8

Evaluate \(\displaystyle \int_0^5\abs{x^2-4 x+3} dx\).   (3 marks)

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\(\dfrac{28}{3}\)

Show Worked Solution

\(\text{Shaded Area}\)

\(=\displaystyle \int_0^5\abs{x^2-4 x+3} dx\)

\(=\left[\dfrac{x^3}{3}-2 x^2+3 x\right]_0^1+\left|\left[\dfrac{x^3}{3}-2 x^2+3 x\right]_1^3\right|+\left[\dfrac{x^3}{3}-2 x^2+3 x\right]_3^5\)

\(=\left(\dfrac{1}{3}-2+3\right)+\left|(9-18+9)-\left(\dfrac{1}{3}-2+3\right)\right|+ …\)

\(\left[\left(\dfrac{125}{3}-50+15\right)-(9-18+9)\right]\)

\(=\dfrac{4}{3}+\left|-\dfrac{4}{3}\right|+\dfrac{20}{3}\)

\(=\dfrac{28}{3}\)