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Using the substitution `u=1+e^x, \int_0^{\log _e 2} \frac{1}{1+e^x}dx` can be expressed as
`D`
`u=1+e^x\ \ =>\ \ \frac{du}{dx}=e^x\ \ =>\ \ dx=\frac{du}{e^x}=\frac{du}{u-1}`
`text{When}\ \ x=0,\ \ u=2`
`text{When}\ \ x=\log_e 2,\ \ u=1+e^{\log_e 2} = 3`
`\int_0^{\log _e 2} \frac{1}{1+e^x}dx = \int_2^3\left(\frac{1}{u}*\frac{1}{u-1}\right) du= \int_2^{1+e^2}\left(\frac{1}{u-1}-\frac{1}{u}\right) du`
`=>D`
Evaluate `int_(e^3) ^(e^4) (1)/(x log_e (x))\ dx`. (3 marks)
`log_e ((4)/(3))`
`text(Let)\ \ u = log_e x`
`(du)/(dx) = (1)/(x) \ => \ du = (1)/(x) dx`
`text(When) \ \ x = e^4 \ => \ u = 4`
`text(When) \ \ x = e^3 \ => \ u = 3`
| `int_(e^3) ^(e^4) (1)/(x log_e (x))` | `= int_3 ^4 (1)/(u)\ du` |
| `= [ log_e u]_3 ^4` | |
| `= log_e 4 – log_e 3` | |
| `= log_e ((4)/(3))` |
Evaluate `int_0^1 e^x cos (e^x)\ dx.` (2 marks)
`sin(e) – sin (1)`
`text(Let)\ \ u = e^x`
`(du)/(dx) = e^x\ \ =>\ \ du=e^x\ dx`
`text(When)\ \ x=1,\ \ u=e`
`text(When)\ \ x=0,\ \ u=^0=1`
| `:. int_0^1 e^xcos(e^x)\ dx` | `= int_1^e cos(u)\ du` |
| `= [sin(u)]_1^e` | |
| `= sin (e) – sin1` |
The definite integral `int_(e^3)^(e^4)1/(xlog_e(x))\ dx` can be written in the form `int_a^b1/u\ du` where
A. `u = log_e(x), a = log_e(3), b = log_e(4)`
B. `u = log_e(x), a = 3, b = 4`
C. `u = log_e(x), a = e^3, b = e^4`
D. `u = 1/x, a = e^(−3), b = e^(−4)`
E. `u = 1/x, a = e^3, b = e^4`
`B`
`text(Let)\ \ u = ln(x)`
`(du)/(dx) = 1/x\ \ =>\ \ du=1/x dx`
`text(When)\ \ x=e^3\ \ =>\ \ u=lne^3=3`
`text(When)\ \ x=e^4\ \ =>\ \ u=lne^4=4`
`:. int_(e^3)^(e^4)1/(xlog_e(x))\ dx = int_3^4 1/udu`
`:. a = 3, b = 4, u = lnx`
`=> B`