5.4 Integration Techniques

(a) Write `2x-x^2` in the form `a(x-h)^2+k`, where `a,h,k ∈ R`.

(b) Hence, find the value of `∫_(1/2)^(3/2)1/sqrt (2x-x^2 ) dx.`

(a) attempt to complete the square or multiplication and equating coefficients

   `2x-x^2=-(x-1)^2+1`

 `a=-1,h=1,k=1`

(b) use of their identity from part (a)  `(∫_(1/2)^(3/2)1/sqrt(1-(x-1)^2 ) dx)`

`=[arcsin(x-1)]_(1/2)^(3/2)` or `[arcsin(u)]_(-1/2)^(1/2)`

`=arcsin(1/2)-arcsin(-1/2)`

`=π/6-(-π/6)`

`=π/3`

Find, in terms of `k`, the area bounded by the curve `y=sqrtx`, the `x`-axis and the line `x=k`, where `k > 0`.

Area `=∫_0^ksqrtx dx=[x^(3/2)/(3/2)]_0^k=2/3 k^(3/2)`

Answer:  Area `=2/3 k^(3/2)`

Find `∫dx/sqrt(6x-x^2-5)`.

`6x-x^2-5=4-(x-3)^2`

`⇒∫dx/sqrt(6x-x^2-5)=∫dx/sqrt(4-(x-3)^2 )=arcsin (x-3)/2+c`

Given that `∫_(-2)^2f(x)dx=10` and `∫_0^2f(x)dx=12`, find

(a) `∫_(-2)^0( f(x)+2)dx`

(b) `∫_(-2)^0f(x+2)dx`

(a) `∫_(-2)^0f(x)dx=10-12=-2`

`∫_(-2)^(0)2 dx=[2x]_(-2)^0=4`

`∫_(-2)^0( f(x)+2)dx=2`

(b) `∫_(-2)^0f(x+2)dx=∫_0^2f(x)dx`

`=12`

By using the substitution `x=tanu`, find the value of `∫_0^1x^2/(1+x^2 )^3 dx`.

`x=tanu⇒( dx)/( du)=sec^2 u` OR `u=arctanx⇒( du)/( dx)=1/(1+x^2 )`

attempt to write the integral in terms of `u`

`∫_0^(π/4)(tan^2 usec^2 u du)/(1+tan^2 u)^3`

`∫_0^(π/4)(tan^2 usec^2 u du)/(sec^2 u)^3`

`=∫_0^(π/4)sin^2 ucos^2 u du`

`=1/4 ∫_0^(π/4)sin^2 2u du`

`=1/8 ∫_0^(π/4)( 1-cos4u)du`

`=1/8 [u-(sin4u)/4]_0^(π/4) =1/8 [π/4-sinπ/4-0-0]`

`=π/32`

Given that `∫_0^lnke^(2x) dx=12`, find the value of `k`.

`1/2 e^(2x) " seen "`

attempt at using limits in an integrated expression `([1/2 e^(2x) ]_0^lnk=1/2 e^2lnk-1/2 e^0 )`

`=1/2 e^(lnk^2 )-1/2 e^0`

Setting their equation  `=12`

`1/2 k^2-1/2=12`

`k^2=25⇒k=5`

Let `y=arccos(x/2)`.

(a) Find `(dy)/dx`

(b) Find `∫_0^1arccos (x/2)dx.`.

(a) `y=arccos(x/2)⇒dy/( dx)=-1/(2sqrt(1-(x/2)^2 )) (=-1/sqrt(4-x^2 ))`

(b) attempt at integration by parts

`u=arccos(x/2)⇒(du)/( dx)=-1/sqrt(4-x^2 )`

`(dv)/( dx)=1⇒v=x`

`∫_0^1arccos (x/2)dx=[xarccos(x/2)]_0^1+∫_0^1x/sqrt(4-x^2 ) dx`

using integration by substitution or inspection

`[xarccos(x/2)]_0^1+[-(4-x^2 )^(1/2) ]_0^1`

attempt to substitute limits into their integral

`=π/3-sqrt3+2`

Find the indefinite integral `∫(x^2-1)^3 x dx`

`∫(x^2-1)^3 x dx=1/2∫(x^2-1)^3 2x dx=(x^2-1)^4/8+c`

Answer: `(x^2-1)^4/8+c`

The interior of a vase is modelled by rotating the region bounded by the curve `y=1/2 x^2-1`, and the lines `x=0,y=0` and `y=15`, through `2π` radians about the `y`-axis. The values of `x` and `y` are measured in centimetres.

The vase is filled with water to a height of `hcm`.

(a) Find an explicit expression for the volume of water in terms of `h`.

The vase is filled at a rate of `20 "cm" ^3 "s" ^(-1).`.

(b) Find the time taken to completely fill the vase.

(c) Find the rate at which the height is changing when h=10 cm .

(a) attempt to use `V=π∫x^2 dy`

`x^2=2y+2`or any reasonable attempt to find `x` in terms of `y`

`V=π∫_0^h2 y+2 dy`

`∫2y+2 dy=y^2+2y`

`V=π[y^2+2y]_0^h`

`=π(h^2+2h)`

(b) volume of vase `=π(15^2+2×15) (=801.106…)`

(time to fill vase `=(801.106…)/20=) 40.1(40.0553…)` (seconds)

(c)

`(dh)/( dt)=(dV)/( dt)×(dh)/( dV)`

`( dV)/( dh)=π(2h+2)`

`(dh)/( dt)=20×1/(π(2h+2))`

substituting h=10 seen anywhere

`0.289(0.289372…)cms^(-1)`

Find the indefinite integral `∫x^2 e^(-2x) dx`

`∫x^2 e^(-2x) dx=-1/2 e^(-2x) x^2+1/2∫e^(-2x) 2x dx`

`=-1/2 e^(-2x) x^2+∫e^(-2x) x dx`

`=-1/2 e^(-2x) x^2-1/2 e^(-2x) x+1/2∫e^(-2x) dx`

`=-1/2 e^(-2x) (x^2+x+1/2)+c`

Answer: `-1/2 e^(-2x) (x^2+x+1/2)+c`