Four rectangles of equal width are drawn and used to approximate the area under the parabola `y = x^2` from `x = 0` to `x = 1`.
The heights of the rectangles are the values of the graph of `y = x^2` at the right endpoint of each rectangle, as shown in the graph below.
- State the width of each of the rectangles shown above. (1 mark)
- Find the total area of the four rectangles shown above. (1 mark)
- Find the area between the graph of `y = x^2`, the `x`-axis and the line `x=1`. (2 marks)
- The graph of `f` is shown below.
Approximate `int_(-2)^2 f(x)\ dx` using four rectangles of equal width and the right endpoint of each rectangle. (1 mark)
Parts of the graphs of `y = x^2` and `y = sqrtx` are shown below.
- Find the area of the shaded region. (1 mark)
- The graph of `y=x^2` is transformed to the graph of `y = ax^2`, where `a ∈ (0, 2]`.
- Find the values of `a` such that the area defined by region(s) bounded by the graphs of `y = ax^2` and `y = sqrtx` and the lines `x = 0` and `x = a` is equal to `1/3`. Give your answer correct to two decimal places. (4 marks)
a. `0.25`
| b. | `text{Area}` | `= 1/4 (1/16 + 1/4 + 9/16 + 1)` |
| `= 15/32\ text(u)^2` |
| c. | `text{Area}` | `= int_0^1 x^2 dx` |
| `= 1/3\ text(u)^2` |
d.
| `int_-2^2 f(x)\ dx` | `≈ 6 + 2 – 4 – 6` |
| `≈ -2` |
| e. | `text{Area}` | `= int_0^1 (sqrtx – x^2) dx` |
| `= 1/3\ text(u)^2` |
f. `text{Case 1:} \ a ≤ 1`
`int_0^a (sqrtx – ax^2) dx = 1/3`
`a = 0.77 \ \ text{or} \ \ a = 1.00`
`text{Case 2:} \ a > 1`
`sqrtx = ax^2 \ => \ x = a^{- 3/2}`
`text{Solve for}\ a\ text{(by CAS)}:`
`int_0^{a^{- 3/2}} (sqrtx – ax^2) dx + int_{a^{- 3/2}}^a (sqrtx – ax^2) dx = 1/3`
`:. a = 1.13`
