Use the substitution `u = 2x - 1` to evaluate `int_1^2 x/((2x - 1)^2)\ dx`. (3 marks)
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Use the substitution `u = 2x - 1` to evaluate `int_1^2 x/((2x - 1)^2)\ dx`. (3 marks)
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`1/4(ln 3 + 2/3)`
`u = 2x − 1`
`⇒ 2x` | `= u + 1` |
`x` | `= 1/2(u + 1)` |
`(du)/(dx)` | `= 2` |
`dx` | `= (du)/2` |
`text(When)` | `\ \ x = 2,\ ` | `u = 3` |
`\ \ x = 1,\ ` | `u = 1` |
`:. int_1^2 x/((2x − 1)^2) \ dx`
`= int_1^3 1/2(u + 1) · 1/(u^2) · (du)/2`
`= 1/4int_1^3 ((u + 1)/(u^2)) du`
`= 1/4 int_1^3 1/u + u^(−2) du`
`= 1/4 [ln u − u^(−1)]_1^3`
`= 1/4 [(ln 3 − 1/3) − (ln 1 − 1)]`
`= 1/4 (ln 3 − 1/3 + 1)`
`= 1/4(ln 3 + 2/3)`
Use the substitution `u = log_e x` to evaluate `int_e^(e^2) 1/(x (log_e x)^2)\ dx`. (3 marks)
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`1/2`
`u` | `= log_e x` |
`(du)/(dx)` | `= 1/x` |
`:. du` | `= 1/x\ dx` |
`text(When)\ \ \ ` | `x = e^2,\ \ ` | `u = log_e e^2 = 2` |
`x = e,` | `u = 1` |
`:. int_e^(e^2) 1/(x (log_e x)^2)\ dx`
`= int_1^2 1/(u^2)\ du`
`= int_1^2 u^(-2)\ du`
`= [-1/u]_1^2`
`= [(-1/2) – (-1)]`
`= 1/2`
Using the substitution `u = x^3 + 1`, or otherwise, evaluate `int_0^2 x^2 e^(x^3 + 1)\ dx`. (3 marks)
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`1/3 (e^9\ – e)`
`u` | `= x^3 + 1` |
`(du)/(dx)` | `= 3x^2` |
`du` | `= 3x^2\ dx` |
`text(If)\ \ \ ` | `x` | `= 2,\ ` | `u` | `= 9` |
`x` | `= 0,\ ` | `u` | `= 1` |
`:.\ int_0^2 x^2 e^(x^3 + 1)\ dx`
`=1/3 int_0^2 e^(x^3 + 1) * 3x^2\ dx`
`= 1/3 int_1^9 e^u\ du`
`= 1/3 [e^u]_1^9`
`= 1/3 (e^9\ – e)`
Use the substitution `u = e^(3x)` to evaluate `int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`. (3 marks)
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`1/3 (tan^(-1)e\ – pi/4)`
`text(Let)\ \ u = e^(3x)`
`(du)/(dx)` | `= 3e^(3x)` |
`:.dx` | `= (du)/(3e^(3x))` |
`text(When)` | `\ x = 1/3,` | `\ u = e^(3 xx 1/3) = e` |
`\ x = 0,` | `\ u = e^0 = 1` |
`:.int_0^(1/3) (e^(3x))/(e^(6x) + 1)\ dx`
`=int_1^e (e^(3x))/(u^2 + 1) xx (du)/(3e^(3x))`
`= 1/3 int_1^e 1/(u^2 + 1)\ du`
`= 1/3 [tan^(-1)u]_1^e`
`= 1/3 [tan^(-1) e\ – tan^(-1) 1]`
`= 1/3 (tan^(-1)e\ – pi/4)`
Using the substitution `u = sqrtx`, evaluate `int_1^4 (e^(sqrtx))/(sqrtx)\ dx`. (3 marks)
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`2e (e – 1)`
`u` | `= sqrtx = x^(1/2)` |
`(du)/(dx)` | `= 1/2 x^(-1/2) = 1/(2 sqrtx)` |
`du` | `= (dx)/(2sqrtx)` |
`:.2du` | `= (dx)/(sqrtx)` |
`text(When)\ \ \ ` | `x=4,\ \ ` | `u = 2` |
`x = 1,` | `\ x = 1` |
`:. int_1^4 (e^(sqrtx))/(sqrtx)\ dx` |
`= int_1^2 e^u xx 2\ du` |
`= 2 [e^u]_1^2` |
`= 2 [e^2 – e^1]` |
`= 2e (e – 1)` |