smc-2564-10-Linear

Calculus, SPEC2-NHT 2019 VCAA 9 MC

With a suitable substitution, `int_1^2sqrt(5x - 1)\ dx` can be expressed as

`5int_1^2sqrtu\ du`
`1/5int_1^2sqrtu\ du`
`5int_4^9sqrtu\ du`
`1/5int_4^9sqrtu\ du`
`5int_4^9sqrt(5u - 1)\ du`

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`D`

Show Worked Solution

`text(Let)\ \ u = 5x – 1`

`(du)/(dx) = 5\ \ =>\ 1/5\ du = dx`

`text(When)\ \ x = 2, u = 9`

`text(When)\ \ x = 1, u = 4`

`int_1^2sqrt(5x – 1)\ dx = 1/5 int_4^9 sqrtu\ du`

`=>\ D`

Calculus, SPEC2 2019 VCAA 8 MC

With a suitable substitution, `int_1^5(2x - 1)sqrt(2x + 1)\ dx` can be expressed as

`1/2 int_1^5 (u^(3/2) + u^(1/2))\ du`
`2 int_3^11 (u^(3/2) + u^(1/2))\ du`
`2 int_1^5 (u^(3/2) - 2u^(1/2))\ du`
`2 int_3^11 (u^(3/2) - 2u^(1/2))\ du`
`1/2 int_3^11 (u^(3/2) - 2u^(1/2))\ du`

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`E`

Show Worked Solution

`text(Let)\ \ u = 2x + 1 \ => \ u – 2 = 2x – 1`

`(du)/(dx) = 2 \ => \ dx = 1/2 du`

`text(When)\ \ x = 5, \ u = 11`

`text(When)\ \ x = 1, \ u = 3`

`:. int_1^5 (2x – 1)sqrt(2x + 1)\ dx`

`= 1/2 int_3^11 (u – 2)u^(1/2)\ du`

`= 1/2 int_3^11 u^(3/2) – 2u^(1/2)\ du`

`=>E`

Calculus, SPEC2 2015 VCAA 10 MC

Using a suitable substitution, the definite integral `int_0^1(x^2sqrt(3x + 1))\ dx` is equivalent to

A. `1/9int_0^1(u^(5/2) - 2u^(3/2) + u^(1/2))\ du`

B. `1/27int_1^4(u^(5/2) - 2u^(3/2) + u^(1/2))\ du`

C. `1/9int_1^4(u^(5/2) - 2u^(3/2) + u^(1/2))\ du`

D. `1/27int_0^1(u^(5/2) - 2u^(3/2) + u^(1/2))\ du`

E. `1/3int_1^4(u^(5/2) - 2u^(3/2) + u^(1/2))\ du`

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`B`

Show Worked Solution

`text(Let)\ \ \ u = 3x + 1`

`x=(u – 1)/3\ \ =>\ \ x^2=(u^2 – 2u + 1)/9`

`(du)/dx = 3\ \ =>\ \ 1/3 du = dx`

`text(When)\ \ x=1,\ u=4`

`text(When)\ \ x=0,\ u=1`

`:. int_0^1(x^2sqrt(3x + 1))\ dx`

`=int_1^4 1/9(u^2 – 2u + 1)*u^(1/2) xx 1/3\ du`

`= 1/27 int_1^4 (u^(5/2) – 2u^(3/2) + u^(1/2))\ du`

`=> B`

Calculus, SPEC2-NHT 2018 VCAA 8 MC

Using a suitable substitution, `int_1^2(3/(2 + (4x + 1)^2))\ dx` can be expressed as

A. `3/4 int_1^2 (1/(2 + u^2))\ du`

B. `3/4 int_5^9 (1/(2 + u^2))\ du`

C. `3 int_5^9 (1/(2 + u^2))\ du`

D. `3 int_1^2 (1/(2 + u^2))\ du`

E. `-12 int_9^5 (1/(2 + u^2))\ du`

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`B`

Show Worked Solution

`u = 4x + 1 qquad qquad qquad u(1) = 5`

`(du)/(dx) = 4 qquad qquad qquad qquad \ \ u(2) = 9`

`3/4 \ du =3\ dx`

`int_1^2(3/(2 + (4x + 1)^2))\ dx`

`=int_5^9 ((3/4)/(2 + u^2))\ du`

`= 3/4 int_5^9 (1/(2 + u^2))\ du`

`=> B`

Calculus, SPEC1-NHT 2018 VCAA 5

Evaluate `int_1^(2 sqrt(3) - 1) (1/(x^2 + 2x + 5))\ dx`. (4 marks)

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`pi/24`

Show Worked Solution

` int_1^(2 sqrt(3) – 1) (1/(x^2 + 2x + 5))\ dx`

`= int_1^(2 sqrt(3) – 1) 1/((x + 1)^2 + 4)\ dx`

`text(Let)\ \ u = x + 1, quad (du)/(dx) = 1\ \ =>\ \ du=dx`

`text(Limits:)\ \ u(1) = 2, quad u(2 sqrt 3 – 1) = 2 sqrt 3`

`:. int_1^(2 sqrt(3) – 1) 1/(x^2 + 2x + 5)\ dx`

`= int_2^(2 sqrt 3) (1/(u^2 + 4))\ du`

`= 1/2 int_2^(2 sqrt 3) 2/(u^2 + 4)\ du`

`= 1/2 [tan^(-1) (u/2)]_2^(2 sqrt 3)`

`= 1/2(tan^(-1) (sqrt 3) – tan^(-1)(1))`

`= 1/2 (pi/3 – pi/4)`

`= 1/2 ((4 pi – 3 pi)/12)`

`= pi/24`

Calculus, SPEC2 2017 VCAA 7 MC

With a suitable substitution `int_1^2 x^2 sqrt(2 - x)\ dx` can be expressed as

`−int_1^2(4u^(1/2) - 4u^(3/2) + u^(5/2)) du`
`int_1^2(4u^(1/2) - 4u^(3/2) + u^(5/2)) du`
`int_0^1(−4u^(1/2) + 4u^(3/2) - u^(5/2)) du`
`−int_1^0(4u^(1/2) - 4u^(3/2) + u^(5/2)) du`
`int_1^0(4u^(1/2) - 4u^(3/2) + u^(5/2)) du`

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`D`

Show Worked Solution

`u`	`= 2 – x`
`x`	`= 2 – u\ \ =>\ \ x^2 = (2 – u)^2`

`(du)/(dx) = −1\ \ =>\ \ du = – dx`

`u(2)`	`= 0`
`u(1)`	`= 2 – 1=1`

`int_1^2 – (2 – u)^2sqrtu\ du`

`= −int_1^0(2 – u)^2 u^(1/2)\ du`

`= −int_1^0(4 – 4u + u^2) u^(1/2)\ du`

`= −int_1^0 (4u^(1/2) – 4u^(3/2) + u^(5/2))\ du`

`=>D`