Find the value of \(n\), given
\(\displaystyle \int_0^n-\dfrac{1}{\sqrt{1-x^2}}=-\dfrac{\pi}{6}\). (2 marks)
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Find the value of \(n\), given
\(\displaystyle \int_0^n-\dfrac{1}{\sqrt{1-x^2}}=-\dfrac{\pi}{6}\). (2 marks)
\(n=\dfrac{1}{2}\)
| \(\displaystyle\int_0^n-\dfrac{1}{\sqrt{1-x^2}}\) | \(=-\dfrac{\pi}{6}\) |
| \(\left[\cos ^{-1} x\right]_0^n\) | \(=-\dfrac{\pi}{6}\) |
| \(\cos ^{-1}(n)-\cos ^{-1} 0\) | \(=-\dfrac{\pi}{6}\) |
| \(\cos ^{-1}(n)-\dfrac{\pi}{2}\) | \(=-\dfrac{\pi}{6}\) |
| \(\cos ^{-1}(n)\) | \(=\dfrac{\pi}{2}-\dfrac{\pi}{6}\) |
| \(n\) | \(=\cos \left(\dfrac{\pi}{3}\right)\) |
| \(=\dfrac{1}{2}\) |
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a. \(\sin ^{-1}\left(\frac{x-2}{2}\right)+c\)
b. \(0<x<4\)
| a. | \(\displaystyle\int \frac{1}{\sqrt{4 x-x^2}}\, d x\) | \(=\displaystyle\int \frac{1}{\sqrt{4-4+4 x-x^2}}\, d x\) |
| \(=\displaystyle\int \frac{1}{\sqrt{4-(x-2)^2}}\, d x\) | ||
| \(=\sin ^{-1}\left(\dfrac{x-2}{2}\right)+c\) |
b. \(4 x-x^2>0\)
\(x(4-x)>0\)
\(0<x<4\)
Find \(\displaystyle \int \frac{1}{\sqrt{15-2 x-x^2}}\, d x\). (3 marks)
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\(\displaystyle\sin ^{-1}\left(\frac{x+1}{4}\right)+c\)
| \(\displaystyle\int \frac{1}{\sqrt{15-2 x-x^2}}\, d x\) | \(=\displaystyle\int \frac{1}{\sqrt{16-1-2 x-x^2}}\, d x\) |
| \(=\displaystyle\int \frac{1}{\sqrt{16-(x+1)^2}}\, d x\) | |
| \(=\displaystyle\sin ^{-1}\left(\frac{x+1}{4}\right)+c\) |
\(\displaystyle \int \dfrac{1}{\sqrt{5-4 x-x^2}}\, d x=\)
\(\Rightarrow A\)
| \(\displaystyle \int \frac{1}{\sqrt{5-4 x-x^2}}\, d x\) | \(=\displaystyle\int \frac{1}{\sqrt{9-4-4 x-x^2}}\,dx\) |
| \(=\displaystyle \int \frac{1}{\sqrt{9-(x+2)^2}}\,dx\) | |
| \(=\sin ^{-1}\left(\dfrac{x+2}{3}\right)+c\) |
\(\Rightarrow A\)
Find \( {\displaystyle \int} \dfrac{1}{\sqrt{4-9x^2}}\ dx\) (2 marks)
\(\dfrac{1}{3} \sin^{-1} \Big{(}\dfrac{3x}{2} \Big{)} +c \)
| \({\displaystyle \int} \dfrac{1}{\sqrt{4-9x^2}}\ dx\) | \(=\dfrac{1}{3} {\displaystyle \int} \dfrac{3}{\sqrt{2^2-(3x)^2}}\ dx\) | |
| \(=\dfrac{1}{3} \sin^{-1} \Big{(}\dfrac{3x}{2} \Big{)} +c \) |
Evaluate `int_0^sqrt3 1/(sqrt(4 - x^2))\ dx`. (2 marks)
`pi/3`
| `int_0^sqrt3 1/(sqrt(4 – x^2))\ dx` | `= [sin^(-1)\ x/2]_0^sqrt3` | |
| `= sin^(-1)\ sqrt3/2 – sin^(-1) 0` | ||
| `= pi/3` |
Find `int_0^1(dx)/(sqrt(4 − x^2))`. (2 marks)
`pi/6`
| `int_0^1(dx)/(sqrt(4 − x^2))` | `= [sin^(−1)\ x/2]_0^1` |
| `= sin^(−1)(1/2) − sin^(−1)(0)` | |
| `= pi/6` |
What is the value of `k` such that `int_0^k 1/sqrt(4 − x^2) \ dx= pi/3 ?`
`B`
| `int_0^1 1/sqrt(4 − x^2)dx` | `= pi/3` |
| `[sin^(−1)\ x/2]_0^k` | `= pi/3` |
| `sin^(−1)\ k/2 − sin^(−1)\ 0` | `= pi/3` |
| `sin^(−1)\ k/2` | `= pi/3` |
| `k/2` | `= sin\ pi/3` |
| `= sqrt3/2` | |
| `:.k` | `= sqrt3` |
`⇒ B`
Evaluate `int_-1^1 1/sqrt(4 - x^2)\ dx`. (2 marks)
`pi/3`
`int_-1^1 1/sqrt(4 – x^2)\ dx`
`= [sin^(-1) (x/2)]_(-1)^1`
`= sin^(-1) (1/2) – sin^(-1) (-1/2)`
`= pi/6 – (- pi/6)`
`= pi/3`
Find `int 1/sqrt (49 - 4x^2)\ dx`. (2 marks)
`1/2 sin^(-1) ((2x)/7) + c`
`int 1/sqrt(49 – 4x^2)\ dx`
`= int 1/(2 sqrt(49/4 – x^2))\ dx`
`= 1/2 int 1/sqrt((7/2)^2 – x^2)\ dx`
`= 1/2 sin^(-1) ((2x)/7) + c`
Use the table of standard integrals to find `int 1/sqrt(4 - x^2)\ dx`. (1 mark)
`sin^(-1)\ x/2 + c`
`int 1/sqrt(4\ – x^2)\ dx`
`= sin^(-1)\ x/2 + c`