Further calculus

Expand `32/((x+2)^3)` in ascending power of `x`, up to and including the term in`x^3`. 

`32(x+2)^-3`

`=32(2^-3)(1+x/2)^-3` 

`=4{1+(-3)(x/2)+frac{(-3)(-4)}{2!}((x^2)/2)+frac{(-3)(-4)(-5)}{3!}(x/2)^3+…}` 

`=4-6x+6x^2-5x^3+…` 

Expand `1/sqrt(4-2x)` in ascending power of `x`, up to and including the term in `x^2`. 

`(4-2x)^(-1/2)`

`=4^(-1/2)(1-x/2)^(-1/2)`

`=1/2{1+(-1/2)(-x/2)+frac{(-1/2)(-3/2)}{2!}(-x/2)^2+…}` 

`=1/2+1/8x+3/64x^2+…`

Evaluate `int_0^(1/2pi)x^2sin x dx`. 

Using integration by parts:

`int_0^(1/2pi)x^2sin x dx` 

`=[x^2(-cos x )]_0^(pi/2)-int_0^(pi/2)-2xcos x dx`

`=0-0+2int_0^(pi/2)xcos x dx`

`=2[xsin x]_ 0^(pi/2)-2int_0^(pi/2)sin x dx` 

`=2(pi/2-0)-2[-cos x]_0^(pi/2)` 

`=π+2(0-1)`

`=π-2` 

Expand `sqrt(frac{1+2x}{1-x})` in ascending power of`x`, up to and including the term in `x^3`. 

`sqrt(frac{1+2x}{1-x})=(1+2x)^(1/2)(1-x)^(-1/2)`

`(1+2x)^(1/2)`

`=1+(1/2)(2x)+frac{(1/2)(-1/2)}{2!}(2x)^2+frac{(1/2)(-1/2)(-3/2)}{3!}(2x)^3+…` 

`=1+x-1/2x^2+1/2x^3-…` 

`(1-x)^(-1/2)`

`=1+(-1/2)(-x)+frac{(-1/2)(-3/2)}{2!}(-x)^2+frac{(-1/2)(-3/2)(-5/2)}{3!}(-x)^3+…` 

`=1+1/2x+3/8x^2+5/16x^3+…` 

`(1+2x)^(1/2)(1-x)^(-1/2)=(1+x-1/2x^2+1/2x^3-…)(1+1/2x+3/8x^2+5/16x^3+… )`

`=1+(1/2+1)x+(3/8-1/2+1/2)x^2+(5/16+1/2+3/8-1/4)x^3+…`

`=1+3/2x+3/8x^2+15/16x^3+…` 

Show that `int_0^1(x+2)e^(-2x)dx=5/4-7/(4e^2)`. 

Using integration by parts:

`int_0^1(x+2)e^(-2x)dx`

`=[-1/2(x+2)e^(-2x)]_0^1-int_0^1-1/2e^(-2x)dx` 

`=-3/2e^-2+1+1/2int_0^1e^(-2x)dx` 

`=1-3/(2e^2)+1/2[-1/2e^(-2x)]_0^1` 

`=1-3/(2e^2)-1/4(e^-2-1)`  

`=5/4-7/(4e^2)` 

(a) Expand `1/sqrt(1-4x)` in ascending power of `x`, up to and including the term in `x^2`, simplifying the coefficients. 

(b) Hence find the coefficient of `x^2` in the expansion of`( 1+2x)/sqrt(4-16x)`. 

(a)

`(1-4x)^(-1/2)`

`=1+(-1/2)(-4x)+frac{(-1/2)(-3/2)}{2!}(-4x)^2+…` 

`=1+2x+6x^2+…` 

(b)

`frac{1+2x}sqrt{4-16x }`

`=frac{1+2x}{sqrt(4)sqrt(1-4x) }`

`=1/2(1+2x)(1-4x)^(-1/2)`

`=1/2(1+2x)(1+2x+6x^2+… )` 

Coefficient of `x^2`

`=1/2(6+4)=5` 

When `(1+ax)^-2`, where `a` is a positive constant, is expanded in ascending powers of `x`, the coefficients of`x` and `x^3` are equal.

(a) Find the exact value of `a`. 

(b) When `a` has this value, obtain the expansion up to and including the term in `x^2`, simplifying the coefficients. 

(a)

`(1+ax)^-2`

`=1+(-2)(ax)+frac{(-2)(-3)}{2!}(ax)^2+frac{(-2)(-3)(-4)}{3!}(ax^3)+…` 

`=1-2ax+3a^2x^2-4a^3x^3+…` 

Using the fact that the coefficient of `x` and `x^3` are equal:

`-2a=-4a^3`

`2a(2a^2-1)=0` 

`a>0`

`2a^2-1=0`

`a^2=1/2`

`a=sqrt2/2` 

(b)

`(1+ax)^-2=1-2ax+3a^2x^2-4a^3x^3+…`

`(1+(sqrt(2x))/2)^-2` 

`=1-2(sqrt(2)/2)x+3(sqrt(2)/2)^2x^2-…` 

`=1-sqrt2x+3/2x^2-…` 

The diagram shows the curve `y=8sinfrac{1}{2}x -tanfrac{1}{2}x` for `0≤x≤π`. The x-coordinate of the maximum point is and the shaded region is enclosed by the curve and the lines`x=α` and `y=0`.

(a) Show that `α=2/3 pi`. 

(b) Find the exact value of the area of the shaded region. 

(a)

`y=8sinfrac{1}{2}x -tanfrac{1}{2}x`

`(dy)/dx=4cosfrac{1}{2}x -1/2 1/2x` 

At the maximum point,`x=alpha` and `(dy)/dx=0`: 

`4cosfrac{1}{2}alpha -1/2 1/2 alpha =0` 

`8cos frac{1}{2} alpha =1/2 alpha`

`8cosfrac{1}{2} alpha =1/(1/2 alpha)`

`1/2 alpha =1/8`

`cos frac{1}{2} alpha =1/2`

`1/2α=pi/3`

`α=(2π)/3` 

(b)

`int_0^alpha(8 sinfrac{1}(2}x -tanfrac{1}{2}x) dx` 

`=[-16cosfrac{1}{2}x +2ln| cosfrac{1}{2}x|]_ 0^alpha` 

`=-16cosfrac{alpha}{2} +2ln| cos frac{1}{2}| +16-2ln 1` 

`=-16cos frac{pi}{3} +2ln cos frac{pi}{3}  +16`

`=-8+2ln frac{1}{2} +16`

`=8+2ln frac{1}{2}` 

Show that `int_0^5 frac{5-3x}{(x+1)(3x+1)}dx=4ln frac{4}{3}` . 

`frac{5-3x}{(x+1)(3x+1)}=A/(x+1)+B/(3x+1)`

`5-3x=A(3x+1)+B(x+1)`

Letting `x=-1` and `x=-1/3`:

`A=-4` and `B=9` 

`frac{5-3x}{(x+1)(3x+1)}=-4/(x+1)+9/(3x+1 )`

`int_0^5frac{5-3x}{(x+1)(3x+1)}dx=int_0^5 (-4/(x+1)+9/(3x+1)) dx`

`=[-4ln |x+1| +3ln| 3x+1|] _0^5`

`=-4ln 6 +3ln 16 +4ln 1 -3ln 1`

`=3ln 16 -4ln 6`

`=3ln 2^4 -4ln 6`

`=4ln 2^3 -4ln 6` 

`=4lnfrac{ 8}{6}` 

`=4ln frac{4}{3}` 

Let `I=int_0^(1/2)frac{4x^2}(sqrt(1-x^2))dx`.

(a) Using the substitution`x=sin theta`, show that `I=int_0^(pi/2)4 theta dθ`. 

(b) Hence show that `I=pi/3-sqrt3/2`. 

(a)

`x=sin theta` 

`(dx)/(dθ)=cos theta`

`dx=cos theta dθ`

`x=0→θ=0`

`x=1/2 -> sin theta =1/2→θ= pi/6`

`I= int_0^(pi/6) frac{4 theta}{sqrt( 1- theta)} cos theta dθ`

`I=int_0^(pi/6) frac{4theta}{sqrttheta} cos theta dθ`

`I=int_0^(pi/6)4 theta dθ`

(b)

`I=int_0^(pi/6) 4theta dθ`

`I=int_0^(pi/6)4(1/2-1/2cos 2θ) dθ` 

`I=int_0^(pi/6)(2-2cos 2θ) dθ`

`I=[2θ-sin 2θ] _0^(pi/6)`

`I=pi/3-sinfrac{pi}{ 3} -0+0`

`I=pi/3-sqrt3/2` 

It is given that `int_1^a xln x dx=30`, where `a>1`. Show that `a=sqrt(frac{119}{2ln a -1})`.

`int_1^axln x dx=30`

Using integration by parts:

`[1/2x^2ln x]_ 1^aint_-1^a((1/2x^2)(1/x))dx=30` 

`1/2a^2ln a -0int_-1^a1/2xdx=30`

`1/2a^2ln a -[1/4x^2]_1^a=30` 

`1/2a^2ln a -1/4a^2=30-1/4` 

`2a^2ln a -a^2=120-1` 

`a^2(2ln a -1)=119`

`a^2=frac{119}{2ln a -1}`

`a>1`

`a=sqrt(frac{119}{2ln a -1})` 

The diagram shows the curve 

`y=x^2e^(2-x)` and its maximum point `M`.

(a) Show that the x-coordinate of `M` is `2`. 

(b) Find the exact value of `int_0^2x^2e^(2-x)dx`. 

(a)

`y=x^2e^(2-x)`

Using the product rule:

`(dy)/dx=2xe^(2-x)-x^2e^(2-x)` 

`=e^(2-x)(2x-x^2)`

At `M`, `(dy)/dx=0`

`2x-x^2=0`

`x=0`or `x=2` 

`x>0`

`x=2`

(b)

Using integration by parts twice:

`int_0^2x^2e^(2-x)dx`

`=[-x^2e^(2-x)]_0^2-int_0^2-2xe^(2-x)dx` 

`=-4+0+2int_0^2xe^(2-x)dx` 

`=-4+2[-xe^(2-x)]_0^2-2int_0^2-e^(2-x)dx` 

`=-4-4+0+2[-e^(2-x)]_0^2` 

`=-8+2(-1+e^2)` 

`=2e^2-10`

Let `f(x)=frac{12+8x-x^2}{(2-x)(4+x^2)}`.

(a) Express `f(x)` in the form `A/(2-x)+(Bx+C)/(4+x^2)`.

(b) Show that `int_0^1f(x)dx=ln frac{25}{2}`. 

(a)

`f(x)=frac{12+8x-x^2}{(2-x)(4+x^2)}=A/(2-x)+(Bx+C)/(4+x^2)` 

`12+8x-x^2=A(4+x^2)+(Bx+C)(2-x)`

Letting `x=2`and `x=0` 

`A=3`and `C=0` 

Equating coefficients of `x^2`:

`-1=A-B`

`B=4` 

`f(x)=frac{12+8x-x^2}{(2-x)(4+x^2)}=3/(2-x)+(4x)/(4+x^2)`

(b)

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

`=[-3ln |2-x| +2ln| 4+x^2|]_ 0^1` 

`=-3ln 1 +2ln 5 +3ln 2 -2ln 4`

`=ln frac{(25)(8)}{16}`

`=lnfrac{ 25}{2}` 

It is given that `int_1^aln (2x) dx=1`, where `a>1`. Show that `a=1/2`exp `(1+(ln 2)/ a)` to where exp `x` denotes `e^x`. 

`int_1^a(1)ln (2x) dx=1`

Using integration by parts:

`[xln (2x)]_1^a-int_1^a(x)(1/x)dx=1` 

`aln 2a -ln 2 -int_1^a1dx=1` 

`aln 2a -ln 2 -a+1=1` 

`aln 2a =ln 2 +a` 

`ln 2a =1+(ln 2)/ a` 

`2a=exp( 1+(ln 2)/ a )`

`a=1/2exp (1+(ln 2)/ a)` 

Note: exp `(x)`means `e^x`.

(a) Express `frac{4+12x+x^2}{(3-x)(1+2x)^2}` in partial fraction. 

(b) Hence obtain the expansion of `frac{4+12x+x^2}{(3-x)(1+2x)^2}` in ascending powers of `x`, up to and including the term in `x^2`. 

(a)

`frac{4+12x+x^2}{(3-x)(1+2x)^2}=A/(3-x)+B/(1+2x)+C/((1+2x)^2)`

`4+12x+x^2=A(1+2x)^2+B(3-x)(1+2x)+C(3-x)`

Letting `x=3`, `x=-1/2`, `x=0`

`A=1, C=-1/2, B=3/2` 

`frac{4+12x+x^2}{(3-x)(1+2x)^2}=1/(3-x)+3/(2(1+2x))-1/(2(1+2x)^2)`

(b)

`1/(3-x)+3/(2(1+2x))-1/(2(1+2x)^2)=(3-x)^-1+3/2(1+2x)^-1-1/2(1+2x)^-2 (3-x)^-1`

`=3^-1(1-x/3)^-1` 

`=1/3{1+(-1)(-x/3)+frac{(-1)(-2)}{2!}(-x/3)^2+… }`

`=1/3+1/9x+1/27x^2+…` 

`3/2(1+2x)^-1`

`=3/2{1+(-1)(2x)+frac{(-1)(-2)}{2!}(2x)^2+… }`

`=3/2-3x+6x^2+…`

`1/2(1+2x)^-2`

`=1/2{1+(-2)(2x)+frac{(-2)(-3)}{2!}(2x)^2+… }`

`=1/2-2x+6x^2-…` 

`(3-x)^-1+3/2(1+2x)^-1-1/2(1+2x)^-2`

`=(1/3+1/9x+1/27x^2+… )+(3/2-3x+6x^2+…)-(1/2-2x+6x^2-… )`

`=(1/3+3/2-1/2)+(1/9-3+2)x+(1/27+6-6)x^2+…` 

`=4/3-8/9x+1/27x^2+…` 

Let `f(x)=frac{x^2-8x+9}{(1-x)(2-x)^2}`.

(a) Express `f(x)` in partial fractions. 

(b) Hence obtain the expansion of`f(x)` in ascending powers of `x`, up to and including the term in `x^2`. 

(a)

`f(x)=frac{x^2-8x+9}{(1-x)(2-x)^2}=A/(1-x)+B/(2-x)+C/((2-x)^2 )`

`x^2-8x+9=A(2-x)^2+B(1-x)(2-x)+C(1-x)` 

Letting `x=1, x=2, x=0`

`A=2, C=3, B=-1` 

`f(x)=frac{x^2-8x+9}{(1-x)(2-x)^2}=2/(1-x)-1/(2-x)+3/((2-x)^2 )`

(b)

`f(x)=frac{x^2-8x+9}{(1-x)(2-x)^2}=2/(1-x)-1/(2-x)+3/((2-x)^2 )`

`f(x)=2(1-x)^-1-(2-x)^-1+3(2-x)^-2`

`2(1-x)^-1`

`=2{1+(-1)(-x)+frac{(-1)(-2)}{2!}(-x)^2+… }`

`=2+2x+2x^2+…` 

`(2-x)^-1`

`=2^-1(1-x/2)^-1`

`=1/2{1+(-1)(x/2)+frac{(-1)(-2)}{2!}(x/2)^2+… }`

`=1/2+1/4x+1/8x^2+…`

`3(2-x)^-2`

`=3(2)^-2(1-x/2)^-2`

`=3/4{1+(-2)(-x/2)+frac{(-2)(-3)}{2!}(-x/2)^2+… }`

`=3/4+3/4x+9/16x^2+…`

`2(1-x)^-1-(2-x)^-1+3(2-x)^-2`

`=(2+2x+2x^2+… )-(1/2+1/4x+1/8x^2+… )+(3/4+3/4x+9/16x^2+…)` `=(2-1/2+3/4)+(2-1/4+3/4)x+(2-1/8+9/16)x^2+…`

`=9/4+5/2x+39/16x^2+…` 

(a) Given that `frac{x^3-2}{x^2(2x-1)}=A+B/x+C/(x^2)+D/(2x-1)`, find the values of the constant A, B, C and D. 

(b) Hence show that `int_1^2frac{x^3-2}{x^2(2x-1)}dx=3/2-2ln frac{9}{4} +1/4ln 3`. 

(a)

`frac{x^3-2}{x^2(2x-1)}=frac{x^3-2}{2x^3-x^2}=A+B/x+C/(x^2)+D/(2x-1)` 

`x^3-2=Ax^2(2x-1)+Bx(2x-1)+C(2x-1)+Dx^2`

Letting `x=1/2`and `x=0`

`D=-15/2` and`C=2` 

Equating coefficients of `x`:

`0=-B+2C`

`B=4`

Letting `x=1`:

`-1=A+B+C+D`

`A=1/2` 

(b)

`frac{x^3-2}{x^2(2x-1)}=1/2+4/x+2/(x^2)-15/(2(2x-1))` 

`int_1^2frac{x^3-2}{x^2(2x-1)}dx=int_1^2(1/2+4/x+2/(x^2)-15/(2(2x-1)))dx`

`=int_1^2(1/2+4/x+2x^-2-15/(2(2x-1)))dx`

`=[1/2x+4ln |x| -2x^-1-15/4ln |2x-1|]_1^2`

`=1+4ln 2 -1-15/4ln 3 -1/2-4ln 1 +2+15/4ln 1` 

`=3/2+4ln 2 -16/4ln 31 +1/4ln 3` 

`=3/2+2lnfrac{ 4}{9} +1/4ln 3`

`=3/2-2ln frac{9}{4} +1/4ln 3` 

The diagram shows the curve `y=x^2ln x` and its minimum point `M`.

(a) Find the exact values of the coordinates of `M`. 

(b) Find the exact value of the area of the shaded region bounded by the curve, the x-axis and the line `x=e`. 

(a)

`y=x^2ln x`

Using the product rule:

`(dy)/dx=2xln x +x`

At `M`, `(dy)/dx=0`

`x(2ln x +1)=0`

`x>0`at `M`

`x=e^(-1/2)=1/sqrte` 

`y=(e^(-1/2))^2ln e^(-1/2) =-1/(2e)` 

The coordinate of `M` are `(1/sqrte, -1/(2e))` 

(b)

Lower bound of region when `y=0`

`x^2ln x =0`

`x=0` or `x=e` 

Using integration by parts for the shaded region:

`int_0^e x^2ln x dx`

`=[1/3x^3ln x]_ 1^e-int_1^e1/3x^2dx`

`=1/3e^3-[1/9x^3]_1^e`

`=1/3e^3-1/9e^3+1/9`

`=2/9e^3+1/9` 

(a) Show that `int_2^4 4xln x dx=56ln 2 -12`. 

(b) Use the substitution `u=sin 4x` to find the exact value of `int_0^(1/24 pi)4x dx`. 

(a)

`int_2^4 4xln x dx=[2x^2ln x]_ 2^4-int_2^4 2xdx` 

`=32ln 4 -8ln 2 -[x^2]_2^4` 

`=64ln 2 -8ln 2 -(16-4)`

`=56ln 2 -12`

(b)

`u=sin 4x` 

`(du)/(dx)=4cos 4x dx`

`1-u^2=1-4x =4x`

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

`x=0→u=0` and `x=pi/24→u=1/2`

`int_0^(1/24 pi)4x dx`

`=int_0^(1/2)1/4( 1-u^2)du` 

`=[1/4u-1/12u^3]_0^(1/2)` 

`=11/96` 

(a) Express `4cos theta +3sin theta` in the form `Rcos (θ-α)`, where `R>0` and `0<α<1/2 pi`. Give the value of correct to `4` decimal places. 

(b) Hence solve the equation `4cos theta +3sin theta =2` for `0<θ<2π`. 

(c) Hence find `int frac{50}{(4cos theta +3sin theta)^2}dθ`. 

(a)

`4cos theta +3sin theta =Rcos theta cos alpha +Rsin theta sin alpha`

`Rsin alpha =3`

`Rcos alpha =4`

`R^2=3^2+4^2=25→R=5` 

`tan alpha =3/4→α=0.6435` 

`4cos theta +3sin theta =5cos (θ-0.6435)` 

(b)

Using the result from part a:

`5cos (θ-0.6435) =2`

`cos (θ-0.6435) =2/5` 

`θ-0.6435=1.1592…`or `θ-0.6435=5.1239…` 

`θ=1.80`or `θ=5.77`

(c)

Using the result from part a:

`int frac{50}{(4cos theta +3sin theta )^2}dθ` 

`=int frac{50}{25(θ-0.6435)} dθ` 

`=int 2(θ-0.6435) dθ`

`=2tan (θ-0.6435) +C`