Integration

A curve with equation `y = f(x)` is such that `f′(x)` `=2x^(-1/3)-x^(1/3)`. It is given that `f (8) = 5`. Find `f(x)`

`int(2x^(-1/3) - x^(1/3))dx` (apply the formula `intx^alphadx=frac{1}{alpha+1}x^(alpha+1)+C`)

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

`f (8) = 5`

`frac{2x8^(2/3)}{2/3}-frac{8^(4/3)}{4/3}+C=5`

`C = 5` 

`f(x)=3x^(2/3)-3/4x^(4/3)+5` 

A curve is such that `(dy)/(dx)=frac{6}{(3x-2)^3` and `A (1, −3)` lies on the curve. A point is moving along the curve and at A the y-coordinate of the point is increasing at `3` units per second. 

a. Find the rate of increase at A of the x-coordinate of the point.
b. Find the equation of the curve. 

a At `x =1`, `(dy)/(dx)=6`[1]
`(dx)/(dt)=((dx)/(dy)xx(dy)/(dt))=1/6xx3=1/2` 

b. `y = (frac{6(2x-2)^(-2)}{-2})-:(3)[+C]` (Substitute `x = 1`, `y = -3`) 
`-3 = -1 + C`

`y = -(3x - 2)^-2 – 2` 

The diagram shows the curve with equation `y = 9 ( x^(-1/2 )− 4x^(-3/2))`. The curve crosses the x-axis at the point A. 

a.  Find the x-coordinate of A. 

b. Find the equation of the tangent to the curve at A.

c. Find the x-coordinate of the maximum point of the curve.

d. Find the area of the region bounded by the curve, the x-axis and the line `x=9`. 

a. x-coordinate of A leads to `y = 0`

`9 x (x^(-1/2) − 4x^(-3/2))  = 0`

`9x^(3/2) ( x – 4) = 0`

Solve out we have `x = 0` and `x = 4` (accept) 

b. `(dy)/(dx)=9(-1/2x^(-3/2)+6x^(-5/2))` 

At `x = 4` gradient `= 9(-1/16 + 6/32) = 9/8`

Equation is `y = 9/8 (x - 4)` 

c. `9x^(-5/2)(-1/2x+6)=0` 

`x = 12` 

d. `int9(x^(-1/2) - 4x^(-3/2))dx = 9(frac{x^(1/2)}{1/2}-frac{x^(-1/2)}{-1/2})`[2]

`= 9[(6 + 8/3) - (4+4)]`

`= 6` 

The equation of a curve is `y = 2sqrt( 3x +4) − x`.

a. Find the equation of the normal to the curve at the point `(4, 4)`, giving your answer in the form `y = mx + c`
b. Find the coordinates of the stationary point 
c. Determine the nature of the stationary point
d. Find the exact area of the region bounded by the curve, the x-axis and the lines `x = 0` and `x = 4`

a. `(dy)/(dx) = 3(3x +4)^(-0.5) – 1`
Gradient of tangent `= – 1/4` and the gradient of normal `= 4`

Equation of line is `(y – 4) = 4(x – 4)` or evaluate C 

So `y = 4x -12` 

b. `3(3x +4)^(-0.5) – 1 = 0`
Solving as far as `x = 5/3` 

`y = 2(3 xx 5/3 + 4)^0.5 – 5/3 = – 13/5` 

c. `(d^2y)/(dx^2)=-9/2(3x+4)^-1.5`
At `x = 5/3` `(d^2y)/(dx^2)`  is negative so the point is a maximum

d. `Area=int 2(3x + 4)^0.5 – xdx`
`= 4/9(3x + 4)^1.5 – 1/2 x^2`

`= (4/9 (16)^1.5 – 1/2 (4)^2) – 4/9(4)^1.5`

`= 256/9 – 8 – 32/9`

`= 152/9 = 16_9^8`

The diagram shows part of the curve with equation `y^2 = x − 2` and the lines `x = 5` and `y = 1`. The shaded region enclosed by the curve and the lines is rotated through `360°` about the x-axis. Find the volume obtained. 

Curve intersects `y = 1` at `(3, 1)`

Volume`= [pi]int(x – 2) d[x]`

`[pi][1/2 x^2 – 2x] or [pi][1/2(x –2)^2]`

`= [pi][(5^2/2 – 2 xx 5) – ((3^2)/2 – 2 xx 3)]`

`= [pi](5/2 + 3/2)`as a minimum requirement for their values 

Volume of cylinder `=  pi xx 1^2 x ( 5 – 3) = 2 pi`

Volume of solid `= 4 pi – 2pi = 2pi (~6.28)`

The gradient of a curve is given by `(dy)/dx = 6(3x – 5)^3 − kx^2` , where `k` is a constant. The curve has a stationary point at `(2, −3.5)`. 

a. Find the value of `k`
b. Find the equation of the curve 
c. Find `(d^2y)/dx^2`
d. Determine the nature of the stationary point `(2, -3.5)`

a. At stationary point `(dy)/dx = 0` so `6(3 xx 2 – 5)^3 – k xx 2^2 = 0` 
`k = 3/2` 

b. `y = 6/(4 xx 3) (3x  – 5)^4  – 1/3 kx^3 +C`
`-7/2 = 1/2( 3 xx 2 – 5)^4 – 1/3 xx 3/2 xx 2^3 + C` (leading to`–3.5 + C = –3.5`)

`y = 1/2(3x – 5)^4 – 1/2 x^3` 

c. `[3 x][18 (3x -5)^2][– 2kx]` 
`486x^2 1623x + 1350` or `– 1620x – 2kx`

d. At `x = 2`, `(d^2y)/dx^2 = 54(3 xx 3 – 5)^2 – 4k`
                        `= 48 > 0`, so it is minimum 

The diagram shows part of the curve with equation `y =  x^(1/2) + k^(2x^(-1/2))`, where `k` is a positive constant.

a. Find the coordinates of the minimum point of the curve, giving your answer in terms of `k`.

The tangent at the point on the curve where `x = 4k^2` intersects the y-axis at P. 

b. Find the y-coordinate of P in terms of `k`.

The shaded region is bounded by the curve, the x-axis and the lines `x = 9/4k^2` and `x = 4k^2`. 

c. Find the area of the shaded region in terms of `k`. 

a. `(dy)/dx = 1/2 x^(-1/2) – 1/2 k^2 x^(-3/2)`
`1/2 x^(-1/2) – 1/2 k^2 x^(-3/2) = 0` leading to `1/2 x^(-1/2) = 1/2 k^2 x^(-3/2)` 

`(k^2, 2k)` 

b. When `x = 4k^2`, `(dy)/dx = [1/(4k) – 1/(16k) ] = 3/(16k)`
`y = [2k + k^2 xx 1/(2k) ] = (5k)/ 2`

Equation of tangent is `y – (5k)/ 2 = 1/ (16k)(x – 4k^2)`

When `x = 0`, `y = [(5k)/2 – (3k)/4] = (7k)/4`

c. `int(x^(1/2) + k^(2x^(-1/2)))dx = (2x^(3/2))/3 + 2k^(2x^(1/2))`
                             `= ((16k^3)/3 + 4k^3) – ((9k^3)/4 + 3k^3)`

`= (49k^3)/12`

A curve has equation `y = f'(x)`, and it is given that `f'(x)  = 2x^2 – 7 – 4/(x^2)`. 

a. Given that `f(1) = –1/3` , find `f(x)`.
b. Find the coordinates of the stationary points on the curve.
c. Find `f''(x)` 
d. Hence, or otherwise, determine the nature of each of the stationary points. 

`a.``f(x) = 2/3 x^3 – 7x + 4x^(-1) [+C]`
`-1/3 = 2/3 – 7 + 4 + c` leading to  `C= 2`

`f(x) = 2/3x^3 – 7x + 4x^-1 + 2`

b. `2x^4 – 7x^2 – 4 = 0`
`(2x^2 +1)(x^2 – 4) = 0`

`x = +-  2`

`2/3(2)^3 – 7(2) + 4/2 + 2` leading to we have point A

`(2, –14/3)`

`2/3(-2)^3 – 7(-2) + 4/-2 + 2`  leading to we have point `B (- 2, 26/3)`

c. `f''(x) = 4x + 8x^-3`
`f''(2) = 9 > 0`, minimum at `x = 2`

`f''(– 2) = – 9 <0` maximum at `x = -2`

a. Find `int_1^oofrac{1}{(3x-2)^(3/2)}dx`

b.

The diagram shows the curve with equation `y = 1/(3x-2)^(3/2)`. The shaded region is bounded by the curve, the x-axis and the lines `x = 1` and `x = 2`. The shaded region is rotated through `360^o` about the x-axis. 

 Find the volume of revolution. 

The normal to the curve at the point `(1, 1)` crosses the y-axis at the point A. 

c. Find the y-coordinate of A. 

a. `((3x - 2)^(-1/2))/(-1/2) : 3`
`-2/3 [0 -1]`

`2/3`

b. `[π]inty^2dx = [π]int(3x -2)^-3dx = [pi]frac{(3x -2)^-2}{-2 xx 3}`
`[π][-1/6][1/16 -1]`

`(5π)/32`

c. `(dy)/dx = -3/2 xx 3(3x -2)^(-5/2)`
At `x = 1`, `(dy)/dx = -9/2`

`y -1 =  2/9(x-1)`

At A, `y = 7/9`

A curve is such that `(dy)/dx= 8/((3x + 2)^2`. The curve passes through the point `(2, 5_3^2)`. Find the equation of the curve.

`int8/((3x + 2)^2)dx`

`y =  frac{8/3}{(3x+2)}+C`

`5_3^2 = - frac{8/3}{(3x+2+2)} + C`

`C = 6`

So `y=frac{8/3}{3(3x+2)}+6`

The diagram shows the line `x=5/2`, part of the curve `y = 1/2 x + 7/10 − 1/((x -2)^(1/3))`  and the normal to the curve at the point A `(3, 6/5)`.

a. Find the x-coordinate of the point where the normal to the curve meets the x-axis.
b. Find the area of the shaded region, giving your answer correct to `2` decimal places.

a. `(dy)/dx = 1/2 + frac{1}{2(x- 2)^(4/3)}`
`(dy)/dx` at `x = 3 [1/2 + frac{1}{3(3 -2)^(4/3) }= 5/6]`

Gradient of normal `= frac{-1}{(dy)/dx} [ = -6/5]`

Equation of normal `y - 65`(their normal gradient) `(x-3)`

`[ y=-6/5x + 4.8 ⟹ 5y = 6x + 24]`

When `y = 0`, `x = 4`

`b.`Area under curve `= int(1/2 x + 7/10 -frac{1}{(x -2)^(1/3)})dx`
`= 1/4 x^2 + 7/10x -  frac{3(x-2)^(2/3)}{2}`

`= (9/4 + 2.1 -3/2) -(6.25/4 + 1.75 - frac{3 xx 0.5^(2/3)}{4})`

`= 0.48`

Area of triangle `= 0.6`

Total area `= 0.6 + 0.48 = 1.08`

The diagram shows the curves with equations `y = y = x^(-1/2)` and `y = 5/2 - x^(1/2)` . The curves intersect at the points A `(1/4 , 2)` and B `(4, 1/2)`. 

a. Find the area of the region between the two curves.

b. The normal to the curve `y = y = x^(-1/2)` at the point `(1, 1)` intersects the y-axis at the point `(0, p)`. 

Find the value of `p`.

a. `int(5/2 - x^(-1/2)-x^(-1/2))dx`
`{5/2x- 2/3x^(3/2)}{-}{2x^(1/2)}`

`(10 - 16/3 - 4)-(5/8 -1/12 -1)`

`9/8` or `1.125`

b. `(dy)/dx = -1/2 x^(-3/2)`

When `x =1`, `m=- 1/2`

`y – 1 = 2(x - 1)`

Through `(1,1)` with gradient `-1/m` or `(1 -p)/1 = 2`

When `x = 0`, `p = - 1`

A curve has equation`y = f(x)` and it is given that `f'(x) = (1/2x + k)^-2 – (1 + k)^-2`, where `k` is a constant. The curve has a minimum point at `x =2`. 

a. Find `f′′(x)` in terms of `k` and `x`, and hence find the set of possible values.

b. It is now given that `k = −3` and the minimum point is at `(2, 3 1/2)`. Find `f(x)`.

c. Find the coordinates of the other stationary point and determine its nature.

a. `f′′(x) = -(1/2x + k)^-3`
`f′′(2) > 0 ⟹ (1 + k)^-3 > 0`

`k < -1`

b. `[ f(x) = int(1/2x - 3)^-2 -(-2)^-2)dx] {((1/2x -3)^-1)/(-1 x 1/2)}{-x/4}`
`3_2^1 = 1 - 1/2 + C`

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

c. `(1/2 x -3)^-2 - (-2)^-2 = 0`

leading to `(1/2 x -3)^-2 = 4`, from that `x = 10`

`(10, - 1/2)`

`f′′(10) [ = (5-3)^-3] <0` so Maximum 

The diagram shows part of the curve with equation`y = x^2 + 1`. The shaded region enclosed by the curve, the y-axis and the line `y = 5` is rotated through `360^0` about the y-axis. Find the volume obtained.

`(π) ∫ (y -1)dy`

`(π)[(y^2)/2 – y]`

`(π)[(25/2 – 5) – (1/2 -1)] = 8 π (25.1)`

A curve has equation `y = x^2 − 2x − 3`. A point is moving along the curve in such a way that at P the y-coordinate is increasing at `4` units per second and the x-coordinate is increasing at `6` units per second. Find the x-coordinate of P.

`(dy)/dx = 2x -2`

`(dy)/dx = 4/6`

`(2x -2) = 4/6`

`x = 4/3`

The gradient of a curve at the point `(x, y)` is given by `(dy)/dx = 2(x + 3)^(1/2) − x`. The curve has a stationary point at `(a, 14)`, where `a` is a positive constant. 

a. Find the value a. 
b. Determine the nature of the stationary point.
c. Find the equation of the curve. 

a. `2(a + 3)^(1/2) – a = 0`
`4( a + 3) = a^2->   a^2 – 4a – 12 = 0`

`(a-6)(a+2) ->  a = 6`

b. `(d^2y)/dx^2 = (x + 3)^(1/2) – 1`
Sub `a-> (d^2y)/dx^2 = 1/3  – 1 = 2/3  < 0` Maximum

c. `y =frac{ 2(x +3)^(3/2)}{3/2} - 1/2x^2 + C`
Sub `x = a` and `y = 14-> 14 = 4/3(9)^(3/2) – 18 + C`

d. `y = 4/3(x + 3)^(3/2) - 1/2x^2 + 4`

The diagram shows part of the curve `y = 8/(x +2)` and the line `2y + x = 8`, intersecting at points A and B. The point C lies on the curve and the tangent to the curve at C is parallel to AB. 

a.Find, by calculation, the coordinates of A, B and C. [6]
b.Find the volume generated when the shaded region, bounded by the curve and the line, is rotated through `360^o`about the x-axis. [6]

a.Simultaneous equations `8/(x + 2) = 4 - 1/2x`
`x = 0` or `x = 6` `A(0,4)` and `B(6,1)`

At C `-8/((x + 2)^2) = - 1/2 ->` `C(2,2)`

b.Volume under line `= int(-1/2x +4)^2dx = pi [(x^3)/12 - 3x^2 + 16x] = 42pi`
 Volume under curve `=  pi(8/(x+2))^2dx =  pi[-64/(x+2)] = 24pi`

Subtracts and uses `0` to `6-> 18pi`

The diagram shows part of the curve `y = 6/x`. The points `(1, 6)` and `(3, 2)` lie on the curve. The shaded region is bounded by the curve and the lines `y = 2` and `x = 1`. 

a. Find the volume generated when the shaded region is rotated through `360^o` about the y-axis. [5]
b.The tangent to the curve at a point X is parallel to the line `y + 2x = 0`. Show that X lies on the line `y = 2x`. [3]

a.Volume `=piintx^2dy =piint36/(y^2)dy`
                `= pi[-36/y]`

Uses limits `2` to `6`correctly`->12pi`

Volume of cylinder `=  pi xx1^2 xx 4`or `int1^2dy = [y]` from `2` to `6`

Volume `= 12pi - 4pi = 8 π`

b.`(dy)/dx = -6/(x^2)`
`-6/(x^2) = -2 ->x =sqrt 3`

`y = 6/sqrt3 = 2sqrt3`  Lies on `y = 2x`

The equation of a curve is such that `(dy)/dx = 3x^(1/2) −3x^(-1/2)`. It is given that the point `(4, 7)` lies on the curve. Find the equation of the curve. [4]

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

 `7 = 16 - 12 + c`

`y = 2x^(3/2) - 6x^(1/2) + 3`

The diagram shows part of the curve with equation `y = x^3 − 2bx^2 + b^2x` and the line OA, where A is the maximum point on the curve. The x-coordinate of A is a and the curve has a minimum point at `(b, 0)`, where a and b are positive constants.

a.Show that `b = 3a`. [4]
b.Show that the area of the shaded region between the line and the curve is `ka^4`, where k is a fraction to be found. [7]

a.`(dy)/dx = 3x^2 – 4bx + b^2`
`3x^2 – 4bx + b^2 = 0 (3x – b)(x – b) = 0`

`x = b/3 or b`

`a = b/3-> b = 3a`

b.Area under curve `=int (x^3– 6ax^2 + 9a^2x)dx`
`(x^2)/4 – 2ax^3 + (9a^2x^2)/2`

`(a^2)/4 – 2a^4 + (9a^4)/2 = (11a^4)/4`

When `x = a`, `y = a^3 – 6a^3 + 9a^3 = 3a^3`

Area under line `= 1/2a x` `their 4a^3`

Shaded area `= (11a^4)/4 – 2a^4 = (3a^4)/4`