smc-1203-40-Log (Definite)

Calculus, 2ADV C4 SM-Bank 1

If `m = int_1^3 (2)/(x)\ dx`, express `e^m` in its simplest form. (2 marks)

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`9`

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`int_1^3 (2)/(x)\ dx`	`= [2 log_e x]_1^3`
`m`	`= 2 log_e 3 – 2 log_e 1`
	`= 2 log_e 3`
`:. e^m`	`=e^(2 log_e 3)`
	`= e^(log_e 9)`
	`= 9`

Calculus, 2ADV C4 2006 HSC 2bii

Evaluate `int_0^3 (8x)/(1 + x^2) \ dx`. (3 marks)

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`4 log_e 10`

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`int_0^3 (8x)/(1 + x^2) \ dx`

`= 4 int_0^3 (2x)/(1 + x^2) \ dx`

`= 4 [log_e (1 + x^2)]_0^3`

`= 4 [log_e (1 + 9) – log_e (1 + 0)]`

`= 4 [log_e 10 – log_e 1]`

`= 4 log_e 10`

Calculus, 2ADV C4 2011 HSC 4b

Evaluate `int_e^(e^3) 5/x\ dx` (2 marks)

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`10`

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`int_e^(e^3) 5/x\ dx`

`=5int_e^(e^3) 1/x\ dx`

MARKER’S COMMENT: Most common error was `ln(5x)`. Minimize errors by getting the integral in the form of `(f prime(x))/f(x)` before integrating.

`=5[lnx]_e^(e^3)`

`=5(lne^3-lne)`

`=5(3-1)`

`=10`

Calculus, 2ADV C4 2012 HSC 9 MC

What is the value of `int_1^4 1/(3x)\ dx`?

`1/3ln3`
`1/3ln4`
`ln9`
`ln12`

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`B`

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`int_1^4 1/(3x)dx`

TIP: Note that `ln(1)=0`, as `e^0=1`

`=1/3[lnx]_1^4`

`=1/3[ln4-ln1]`

`=1/3ln4`

`=>B`

Calculus, 2ADV C4 2013 HSC 11f

Evaluate `int_0^1x^2/(x^3+1)\ dx` (3 marks)

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`1/3ln2`

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`int_0^1x^2/(x^3+1)dx`

`=1/3int_0^1(3x^2)/(x^3+1)dx`

`=1/3[ln(x^3+1)]_0^1`

TIP: Note that `ln(1)=0`, because `e^0=1`

`=1/3(ln2-ln1)`

`=1/3ln2`