SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Calculus, 2ADV EQ-Bank 22

The diagram shows the graph of  \(y=\log _e(x+1)\)
 

  1. Express \(x\) as a function of \(y\).   (1 mark)

  2. Hence, or otherwise, find the exact area of the shaded region bounded by the curve  \(y=\log _e(x+1)\), the \(x\)-axis, and the line  \(x=3\).   (3 marks)

Show Answers Only

a.    \(x=e^y-1\)

b.    \(A=4 \ln 4-3 \ \text{u}^2\)

Show Worked Solution

a.    \(y=\ln (x+1) \ \Rightarrow \ x+1=e^y \ \Rightarrow \ x=e^y-1\)
 

b.    \(\text{Area of rectangle}=\ln 4 \times 3=3 \ln 4\)

\(\text{Find the area between curve and \(y\)-axis from  \(\ y=0\ \)  to  \(\ y=\ln 4\):}\)

\(A\) \(=\displaystyle \int_0^{\ln 4} e^y-1\, d y\)
  \(=\Big[e^y-y\Big]_0^{\ln 4}\)
  \(=\left(e^{\ln 4}-\ln 4\right)-(1)\)
  \(=4-\ln 4-1\)
  \(=3-\ln 4\)

 

\(\text{Shaded Area}\) \(=3 \ln 4-(3-\ln 4)\)
  \(=4 \ln 4-3 \ \text{u}^2\)

Calculus, 2ADV EQ-Bank 29

The diagram shows the graph of  \(y=\log _2 2 x\)
 

 

Determine the exact value of the shaded area bounded by the \(x\)-axis, the \(y\)-axis, and the curve  \(y=\log _2 2 x\).

Express your answer is the form \(\dfrac{a}{\ln b}\), where \(a\) and \(b\) are integers.   (4 marks)

Show Answers Only

\(A=\dfrac{7}{\ln 4}\ \text{u}^2\)

Show Worked Solution

\(y=\log _2(2 x) \ \Rightarrow \ 2 x=2^y \ \Rightarrow \ x=\dfrac{1}{2} \times 2^y\)

\(A\) \(=\dfrac{1}{2} \displaystyle \int_0^3 2^y\, d y\)
  \(=\dfrac{1}{2}\left[\dfrac{2^y}{\ln 2}\right]_0^3\)
  \(=\dfrac{1}{2}\left[\dfrac{2^3}{\ln 2}-\dfrac{1}{\ln 2}\right]\)
  \(=\dfrac{7}{2 \ln 2}\)
  \(=\dfrac{7}{\ln 4}\ \text{u}^2\)

Calculus, 2ADV C4 EQ-Bank 19

The diagram shows the graph of  \(y=\ln (x+2)\).
  

Find the exact value of the shaded area bounded by the \(y\)-axis, the line  \(y=\ln 6\)  and the curve  \(y=\ln (x+2)\). Express your answer in the form  \(a+b\,\ln c\) where  \(a, b\) and \(c\) are integers.   (4 marks)

Show Answers Only

\(A=(4-2 \ln 3) \ \text{u}^2\)

Show Worked Solution

\(y=\ln (x+2)\ \ \Rightarrow\ \ x+2=e^y\ \ \Rightarrow\ \ x=e^y-2\)

\(y\text{-intercept occurs at}\ (0,\ln 2).\) 

\(A\) \(=\displaystyle \int_{\ln 2}^{\ln 6}\left(e^y-2\right) d y\)
  \(=\Big[e^y-2 y\Big]_{\ln 2}^{\ln 6}\)
  \(=e^{\ln 6}-2 \ln 6-e^{\ln 2}+2 \ln 2\)
  \(=6-2-2\left(\ln \dfrac{6}{2}\right)\)
  \(=(4-2 \ln 3) \ \text{u}^2\)

Calculus, 2ADV C4 2008 HSC 10a


 

In the diagram, the shaded region is bounded by  `y = log_e (x-2)`, the  `x`-axis and the line  `x = 7`.

Find the exact value of the area of the shaded region.   (5 marks)

Show Answers Only

`5 log_e 5-4\ \ \ text(u²)`

Show Worked Solution

`text(Shaded Area)\ text{(} A_1 text{)}` `=\ text(Rectangle)-A_2`
`text(Area of Rectangle)\ ` `= 7 xx log_e 5`

 

`text(Finding the Area of)\ \ A_2`

`y` `= log_e (x-2)` 
`x-2` `= e^y`
`x` `= e^y + 2`
`:. A_2` `= int_0^(log_e 5) x\ dy`
  `= int_0^(log_e 5) e^y + 2\ dy`
  `= [e^y + 2y]_0^(log_e 5)`
  `= [(e^(log_e 5) + 2 log_e 5)-(e^0 + 0)]`
  `= (5 + 2 log_e 5)-1`
  `= 4 + 2 log_e 5`
   
`:.\ A_1` `= 7 log_e 5-(4 + 2 log_e 5)`
  `= 5 log_e 5-4\ \ \ text(u²)`