The shaded region is enclosed by the curve `y = x^3-7x` and the line `y = 2x`, as shown in the diagram. The line `y = 2x` meets the curve `y = x^3-7x` at `O(0, 0)` and `A(3, 6)`. Do NOT prove this.
- Use integration to find the area of the shaded region. (2 marks)
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- Use the Trapezoidal rule and four function values to approximate the area of the shaded region. (2 marks)
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The point `P` is chosen on the curve `y = x^3 − 7x` so that the tangent at `P` is parallel to the line `y = 2x` and the `x`-coordinate of `P` is positive
- Show that the coordinates of `P` are `(sqrt 3, -4 sqrt 3)`. (2 marks)
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- Using the perpendicular distance formula `|ax_1 + by_1 + c|/sqrt(a^2 + b^2)`, find the area of `Delta OAP`. (2 marks)
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| i. | `text(Area)` | `= int_0^3 2x-(x^3-7x)\ dx` |
| `= int_0^3 9x-x^3\ dx` | ||
| `= [9/2 x^2-1/4 x^4]_0^3` | ||
| `= [(9/2 xx 3^2-1/4 xx 3^4)-0]` | ||
| `= 81/2-81/4` | ||
| `= 81/4\ text(units)^2` |
ii. `f(x) = 9x-x^3`
`text(Area)~~ 1/2[0 + 2(8 + 10) + 0]~~ 1/2(36)~~ 18\ text(u)^2`
iii. `y = x^3-7x`
`(dy)/(dx) = 3x^2-7`
`text(Find)\ x\ text(such that)\ \ (dy)/(dx) = 2:`
| `3x^2-7` | `= 2` |
| `3x^2` | `= 9` |
| `x^2` | `= 3` |
| `x` | `= sqrt 3 qquad (x > 0)` |
`y= (sqrt 3)^3-7 sqrt 3= 3 sqrt 3-7 sqrt 3= -4 sqrt 3`
`:. P\ \ text(has coordinates)\ (sqrt 3, -4 sqrt 3)`
| iv. |  |
`text(dist)\ OA= sqrt((3-0)^2 + (6-0)^2)= sqrt 45= 3 sqrt 5`
`text(Find)\ _|_\ text(distance of)\ P\ text(from)\ OA:`
`P(sqrt 3, -4 sqrt 3),\ \ 2x-y=0`
`_|_\ text(dist)= |(2 sqrt 3 + 4 sqrt 3)/sqrt (3 + 2)|= (6 sqrt 3)/sqrt 5`
`:.\ text(Area)= 1/2 xx 3 sqrt 5 xx (6 sqrt 3)/sqrt 5= 9 sqrt 3\ text(units)^2`