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Evaluate `int_(-1)^0 (1 + x)/sqrt(1 - x)\ dx`. (3 marks)
`(8 sqrt 2)/3 – 10/3`
| `text(Let)\ \ u` | `= 1 – x \ => \ x = 1 – u` |
| `(du)/(dx)` | `= -1 \ => \ dx = -du` |
| `text(When)\ \ x` | `= 0,\ u = 1` |
| `x` | `= -1,\ u = 2` |
| `int_(-1)^0 (1 + x)/sqrt(1 – x)\ dx` | `= -int_2^1 (2 – u)/sqrt u\ du` |
| `= int_1^2 2u^(-1/2) – u^(1/2)\ du` | |
| `= [4u^(1/2) – 2/3u^(3/2)]_1^2` | |
| `= 4 sqrt 2 – (4 sqrt 2)/3 – (4 – 2/3)` | |
| `= (8 sqrt 2)/3 – 10/3` |
Using the substitution `u = sqrt(x + 1)` then `int_0^2(dx)/((x + 2) sqrt (x + 1))` can be expressed as
A. `int_1^sqrt 3 1/(sqrt u (u^2 + 1))\ du`
B. `int_0^2 2/(u^2 + 1)\ du`
C. `int_1^3 1/(sqrt u (u + 1))\ du`
D. `1/4 int_0^2 1/(u^2(u^2 + 1))\ du`
E. `2 int_1^sqrt 3 1/(u^2 + 1)\ du`
`E`
| `u` | `=sqrt(x + 1)` |
| `(du)/(dx)` | `= 1/(2 sqrt (x + 1))\ \ =>\ \ dx=2sqrt(x + 1)\ du` |
| `u^2` | `= x + 1` |
| `u^2 + 1` | `= x + 2` |
| `u(0)` | `= sqrt (0 + 1) = 1` |
| `u(2)` | `= sqrt (2 + 1) = sqrt 3` |
`:. int_0^2 (dx)/((x + 2) sqrt(x + 1))`
`=int_1^sqrt3 (2sqrt(x + 1))/((u^2 + 1) xx sqrt(x + 1))\ du`
`= int_1^sqrt 3 2/(u^2 + 1) du`
`=> E`