a) Proof that
∫𝑋𝑛𝑑𝑥= 𝑋𝑛+1 /(𝑛 + 1) + 𝐶 𝑓𝑟𝑜𝑚 𝑑(𝑋𝑛)𝑑𝑥 = 𝑛𝑋𝑛−1
b)Evaluate
∫ (𝑥3−2𝑥2−4𝑥−4)/ (𝑥2+𝑥−2) 𝑑𝑥 limit from 2 to 3
Correct to 4 significant figures
a) 𝑚 = 𝑛 + 1
𝑑(𝑋𝑚)/𝑑𝑥 = 𝑚𝑋𝑚−1
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒 ∫𝑑(𝑋𝑚)/𝑑𝑥 = ∫𝑚𝑋𝑚−1 𝑑𝑥
∫𝑚𝑋𝑚−1 𝑑𝑥 = 𝑚𝑋𝑚−1+1 /(𝑚−1+1) = 𝑋𝑚 + 𝑘
∫𝑚𝑋𝑛−1 𝑑𝑥= 𝑋𝑚+𝑘
𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒 𝑏𝑦 𝑚
∫𝑚(𝑋𝑚−1 𝑑𝑥)/ 𝑚 = 𝑋𝑚 /𝑚 + 𝑘/𝑚
∫𝑋𝑛−1 𝑑𝑥 = 𝑋𝑚/𝑚 + 𝑘/𝑚
𝑘/𝑚=𝐶 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑚 = 𝑛+1
Thus; ∫𝑋𝑛−1 𝑑𝑥 = 𝑋𝑛+1 /𝑛+1 + 𝐶
b)
∫ (𝑥 − 3) + (−𝑥 – 4)/(𝑥2 + 𝑥 – 2) dx
∫ (𝑥 − 3) – (𝑥 + 4)/(𝑥2 + 𝑥 – 2) dx
(𝑥 + 4) /(𝑥2 + 𝑥 – 2) = (𝑥 + 4)/ (𝑥 − 1)(𝑥 + 2)
𝑃𝑎𝑟𝑡𝑖𝑎𝑙 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛
𝐴/(𝑥 – 1) + 𝐵/(𝑥 + 2) = 𝑥 + 4/(𝑥 − 1)(𝑥 + 2)
𝐴(𝑥 + 2) + 𝐵(𝑥 − 1) = 𝑥 + 4
𝑊ℎ𝑒𝑛 𝑥 = 1
𝐴(3) = 5 𝐴 =5/3
𝑊ℎ𝑒𝑛 𝑥 = −2
𝐵(−3) = 2 𝐵 = −2/3
Now we have [5/3/(𝑥 – 1)] – 2/3/(𝑥 + 2)]
Inserting this value into the integration we have;
∫ {(𝑥 − 3) − [5/3/(𝑥 – 1)] – [2/3/(𝑥 + 2)]} dx
∫ (𝑥 − 3) − ∫ [5/3/(𝑥 – 1)] – 2/3/(𝑥 + 2)]dx
∫ (𝑥 − 3)dx − ∫ 5/3/(𝑥 – 1)dx + ∫2/3/(𝑥 + 2)dx
∫ (𝑥 − 3)dx − 5/3∫ 1/(𝑥 – 1)dx + 2/3∫1/(𝑥 + 2)dx
[x2/2 – 3x] – [5/3 In(x -1)] + [2/3 In(x+2)]
= −0.5 − 1.1552 + 0.1487 = −1.5064
a) Show directly from definition (i.e from first principle) that derivative of f(x) = 2x3 is 6x2
b) Find the differential coefficient of y = 5x using first principle
a) f(x) = 2x3 is 6x2
Let x be increased by change in x
𝑓(𝑥 + Δ𝑥) = 2(𝑥 + Δ𝑥)3
= 2(𝑥3 + 3𝑥2Δ𝑥 + 3𝑥Δ𝑥2 + Δ𝑥3 )
= 2𝑥3 + 6𝑥2Δ𝑥 + 6𝑥Δ𝑥2 + 2Δ𝑥3
𝑎𝑠 𝑓(𝑥)𝑐ℎ𝑎𝑛𝑔𝑒𝑠 𝑏𝑦 𝑎 𝑟𝑎𝑡𝑒 𝑜𝑓 Δ𝑥
𝑓(𝑥 + Δ𝑥)/Δ𝑥 = (2𝑥3 + 6𝑥2Δ𝑥 + 6𝑥Δ𝑥2 + 2Δ𝑥3 )/Δ𝑥
𝑓(𝑥 + Δ𝑥)/Δ𝑥 = 2𝑥3/Δ𝑥 + 6𝑥2 + 6𝑥Δ𝑥 + 2Δ𝑥
𝑎𝑠 Δ𝑥 → 0
𝑓(𝑥 + Δ𝑥)/Δ𝑥 = 6𝑥2
b) y = 5x
f(x) = y = 5x
𝑓(𝑥+Δ𝑥) = 5(𝑥+Δ𝑥)
𝑓(𝑥+Δ𝑥)=5𝑥+5Δ𝑥
𝑓(𝑥+Δ𝑥)/Δ𝑥=5𝑥/Δ𝑥 + 5
Δ𝑥→0 𝑑𝑦/𝑑𝑥=5
a) Differentiate the following by the general rule
- y= 3(t-2)3
- 𝑦 = 3𝑥 – 1/√𝑥 + 1/𝑥
- 𝑦=6 + 1/𝑥3
b) An alternating Voltage is given by V=100Sin200t Volts, where ‘t’ is the time in seconds. Calculate the rate of change of voltage when
- t=0.005s
- t=0.01s
c) Find the rate of change of ‘y’ with respect to ‘x’ of the following
- y = 3x In2x
- y = 3x2 Sin x
a)
i) y = 3(t-2)2
y = f(x) xn
dy/dx = nXn-1
dy/dt = 2(3)(t-2)2-1
y = 6(t-2)
ii) 𝑦 = 3𝑥 – 1/√𝑥 + 1/𝑥
y = 3x – x-1/2 + x-1
dy/dx = 3 – (-1/2x-1/2-1 + (-1x-1-1)
= 3 + ½ x-3/2 – x-2
𝑑𝑦/𝑑𝑥 = 3 + 1/2𝑥3/2 −1/𝑥2
iii) 𝑦=6 + 1/𝑥3
y = 6 + 1/x3
y = 6 + x-3
dy/dx = 0 -3x-3-1
dy/dx = -3x-4
𝑑𝑦/𝑑𝑥 = −3/𝑥4
b)
V = 100 Sin200t volts
dv/dt = (100)200 Cos 200t
= 200 Cos 200t
i) when t =0.005s
dv/dt = 20000 Cos 200 (0.005)
=20000 Cos 1 = 19996.95 v/s
ii) when t = 0.01s
dv/dt = 2000 Cos 200 (0.10)
= 19987.8 v/s
b)
i) y = 3x In2x
let u=3x and v = In2x
du/dx =3x
v can also be written as v = In2 + Inx
dv/dx = 1/x
dy/dx = udv + vdu
dy/dx = 3x (1/x) + In2x (3x)
dy/dx = 3 + 3x In2x
ii) y = 3x2 Sin x
Let u = 3x2 and v = Sinx
du/dx = 6x dv/dx = Cosx
dy/dx = udv + vdu
dy/dx = 3x2 Cosx + 6x Sinx
a) Differentiate the following using
i) y = 4sin5x/5x2
ii) y = 3√θ3 /2sin2θ
b) Differentiate the following using chain rule
i) y = 3 cos(5x3 + 2)
ii) y = √(3x2 + 4x – 1)
iii) y = (2x3 + 5x)5
i) y = 4sin5x/5x2
Let U = 4sin 5x and v = 5x4
du/dx = 4 × 5 cos5x
dv/dx = 20x3
dy/dx = (Vdu/dx – Udv/dx)/V2
dy/dx = (5x4 × 20cos5x – 4sin5x × 20x3)/(5x4)2
dy/dx = (100x4 cos5x – 80x3sin5x)/25x8
ii)
y = 3√θ3 /2sin2θ = 3θ3/2 /2sin2θ
u = 3θ3/2 and v=2sin2θ
du/dθ =3/2 × 3θ3/2 -1 =9/2θ1/2 =9√θ
dv/dθ = 4cos2θ
Using dy/dθ = (Vdu/dθ – Udv/dθ)/V2
= (2sin2 θ × 9/2 √θ – 3√θ3 × 4cos2θ)/(2sin2θ)2
=(9 √θ sin2 θ – 12√θ3 cos2θ)/(4sin2 2θ)
b)
i) y = (2x3 + 5x)5
let u = 2x3 + 5x and y = u5
dy/dx = dy/du × du/dx
= 5u4 × 6x2 + 5 = 5(6x2 + 5)u4
=(30x2 + 25)u4
=(30x2 + 25)(2x3 + 5x)4
ii) y = √(3x2 + 4x -1)
let u = 3x2 + 4x -1
y = √u = u1/2
dy/dx = dy/du × du/dx
=1/2 u-1/2 × 6x + 4
=1/2(6x +4) × 1/√u
now inserting the value
= 2(3x + 2)/2√(3x2 + 4x – 1)
= (3x + 2)/√(3x2 + 4x – 1)
a)
i) Determine
∫ (11 -3x)/(x2 + 2x -3) dx
ii) Integrate the following indefinite integrals
∫3x4 dx and ∫ 2/x2 dx
b) Evaluate the definite integrals ( where necessary to 4 significant figures)
- ∫41 5x2 dx
- ∫20 3sin t dt
- ∫32 2/3x dx
a)
i) ∫(11 -3x)/(x2 + 2x -3) dx
x2 + 2x – 3 = (x + 3)(x -1)
∫(11 – 3x)/ (x + 3)(x -1) dx
Splitting the equation to partial fraction
A/x+3 + B/x -1 = (11 – 3x)/(x +3)(x – 1)
Multiply through by (x + 3)(x – 1)
A(x – 1) + B(x+3) = 11 -3x
when x = +1
A(0) + B(4) = 8
B = 8/4 = 2
when x = -3
A(-4) = 20
A = -20/4 = -5
Substituting the value of A and B
2/(x+3) + -3/(x -1)
∫2/(x+3) dx – ∫3/(x -1)
2 ∫1/(x+3) – 3∫1/(x-1)
u = x+3 du/dx =1 du=dx
v x+1 dv/dx = 1 dv=dx
2∫1/u du – 3∫1/v dv
2Inu – 3Inv
Inserting the value of U and V
2In(x+3) – 3In(x-1)
b)
i) ∫41 5x2 dx = [5x3 /3] limit from 1 to 4
=5/3[x3] Now inserting the limit;
5/3[43 – 13 ] =105
ii) ∫20 3sint dt = 3[-cos t] limit from 0 to 2
Now inserting the limit;
-3[cos(2) –cos(0) ] =1.827 ×10-3|
Answer to 4 dp is 0.0018
iii)
∫32 2/3x dx = 2/3 ∫32 (1/x) dx
=2/3 [Inx]
inserting the limit 2/3[In3 – In2]
2/3[1.09861 – 0.6931] =0.2703
a) Determine the following
- ∫ 2sin(4x +9)dx
- ∫1/2(5x – 3)6 dx
- ∫10 (3x + 1)5dx
b) If ∫ 8x3 -3x2 + 4x – 5 dx
Determine the value of “I” where x = 3, given that at x = 2, I = 26
a)
i) ∫ 2sin(4x +9)dx
Let u = 4x+9
du/dx= 4 dx = du/4
∫ 2sinu du/4 =1/2[-cosu]
= -1/2cos(4x + 9) + C
ii) ∫1/2(5x – 3)6 dx
let u= 5x-3, du/dx = 5 therefore dx= du/5
∫1/2 u6 du/5 =1/10 ∫u6 du
1/10 ×u7 = u7/70
Inserting the value of u (5x-30)/70 + C
iii) ∫10 (3x + 1)5dx
let u= 3x+1, du/dx= 3 therefore dx=du/3
∫10 u5 du/3 = 1/3[u6/6]
Inserting the value of u 1/18 [(3x -1)6]
Putting the limit of 0 to 1
1/18[(3(1) + 1)6 – (3(0) + 1)6 ]
1/18(46 – 16) = 227.5
b) I = ∫8x3 -3x2 +4x -5 dx
=8x4/4 -3x3/3 + 4x2/2 – 5x + C
=2x4 – x + 2x2 -5x +c
When x= 2 and = 26
26 = 2(2)4 – (2)3 + 2(2)2 – 5(2) +C
C+22 = 26
C =4
When x= 3
I = 2(2)4 – (2)3 + 2(2)2 – 5(2) +4 =142
a. Sketch the graph y = x3 + 2x2 – 5x – 6 between x = -3 and x = 2 and determine the area enclosed by the curve and the x – axis.
b. The velocity v of a body t seconds after certain instance is (2t2 + 5 ) m/s. Find by integration how far its moves in the interval from t = 0 to t = 4s and produce a graph.
c. Determine the area enclosed by y = 2x + 3, the x – axis and ordinate x = 1 and x = 4. Produce the graph
a.
y = x3 + 2x2 – 5x – 6 x= -3, x= 2
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| y | -18 | 0 | 4 | 0 | -6 | -8 | 0 | 24 |
Graph of y = x3 + 2x2 – 5x – 6 x= -3, x= 2
∫-1-3 ydx – ∫2-1ydx
∫ydx = ∫ x3 +2x2 -5x -6
=[x4/4 + 2x3/3 – 5x2/2 – 6x]
Inserting the two different limit to the equation -1 to -3 and -1 to 2
Area = 5.33 – ( -15.75) = 21.08
b) Recall v = du/dx and dx =vdt or ∫tt0 vdt
x = ∫40 (2t2 +5)dt = (2t3/3 + 5t)
Inserting the limit we have [2(4)3/3 + 5(4)] – 0 =62.67m
| t | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| v | 7 | 5 | 7 | 13 | 23 | 37 | 55 |
c. y = 2x + 3 x=1, x=4
Area =∫41 (2x + 3)dx = [2x3/3 + 3x]
Inserting the limit from 1 to 4
[2(4)3 /3 + 3(4)] – [2(1)3/3 + 3] = 42
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 3 | 5 | 7 | 9 | 11 | 13 |
Differentiate the following
- y = 5x4. Sin x
- y = cos(2x – 3x)
a.
y = x3 + 2x2 – 5x – 6 x= -3, x= 2
| x | -4 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| y | -18 | 0 | 4 | 0 | -6 | -8 | 0 | 24 |
Graph of y = x3 + 2x2 – 5x – 6 x= -3, x= 2
∫-1-3 ydx – ∫2-1ydx
∫ydx = ∫ x3 +2x2 -5x -6
=[x4/4 + 2x3/3 – 5x2/2 – 6x]
Inserting the two different limit to the equation -1 to -3 and -1 to 2
Area = 5.33 – ( -15.75) = 21.08
b) Recall v = du/dx and dx =vdt or ∫tt0 vdt
x = ∫40 (2t2 +5)dt = (2t3/3 + 5t)
Inserting the limit we have [2(4)3/3 + 5(4)] – 0 =62.67m
| t | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
| v | 7 | 5 | 7 | 13 | 23 | 37 | 55 |
c. y = 2x + 3 x=1, x=4
Area =∫41 (2x + 3)dx = [2x3/3 + 3x]
Inserting the limit from 1 to 4
[2(4)3 /3 + 3(4)] – [2(1)3/3 + 3] = 42
| x | 0 | 1 | 2 | 3 | 4 | 5 |
| y | 3 | 5 | 7 | 9 | 11 | 13 |
Integration can be applied in the calculation of areas, volumes, central points and many more
a) false b) true c) none of the above d) all of the above
True [b]
The differential of y = 3x2 . ex is ?
a) 3xex(x + 3) b) x2ex(x + 3) c) 3x3ex(x + 3) d) 3x2ex(x + 3)
d[(3x2.ex)/dx] = d/dx((3x2).ex) + d(ex). 3x3
9x2ex + 3x3 ex = 3x2 ex (3 + x ) [d]
lim (1 – √x )/(1 -x) by rationalization as x approaches 1 is?
a) 1 b) 1/2 c) 2 d) 0
I = ∫ 4x2 dx = 4x3/3 + C
I = 25 when x = 3 we have 4(3)3/3 + C = 25
(4 x 27)/3 + C = 25 => 36 + C = 25
C = 25 – 36 = – 11 [b]
the integral of ∫ cos x dx is?
a. Sin x b. – Sin x c. – Sin x + C d. Sin x + C
∫ cos x dx = sin x + C [d]
The slope of a constant value (like 1000) is always?
a. 1000 b. 0 c. undefined d. none of the above
slope of constant is always ‘0’ [b]
if f(x) = In(Inx), then (dy/dx) =
a. 1/x b. 1/Inx c. 1/xInx d. Inx/x
derivative of In (In x ) using chain rule, let In x = u such that y = In u => (dy/dx) = 1/x , (du/dx) = 1/x
=> (dy/dx) = (dy/du). (du/dx) = (1/x) x (1/In x) = (1/x Inx) [c]
They equation Xn = nXn-1 is an expression of ?
a. The product rule b. The chain rule c. The power rule d. The substitution rule
(d/dx)[xn] = n xn-1 is an experiment of power rule [c]
What do you understand about calculus?
a. Motion b. speed c. change d. development
calculus is approximately change [c]
what is (u/dx)[(5z2 + z3 – 7z4)]?
a. 10z2 + 3z3 – 14z4 b. 10z + 3z2 – 28z3 c. 10z +3z2 + 28z4 d. 10z +3z – 28z3
d/dx (5z2 + z3 – 7z4) =
10z + 3z2 – 28z3 [b]
∫ 1/x dx =
a. undefined because you cannot divide by zero b. loge(x) c. In(x) d. In(x) + C
∫(1/x) dx = In x + C [d]
which of the following is the best differential technique to use for (x2 + 1)5?
a. the product rule b. The chain rule c. The power rule d. The substitution rule
the best techniques for differential (X2 + 1)5 is a power rule d/dx((g(x))n
= ( n.a.(g1(x)) . (g(x)n-1) [c]
∫10 (x2 – 2x + 1) dx is
a. – 1 b. – 7/3 c. 7/3 d. 1
∫10 ( x2 – 2x + 1 )dx = [ x3/3 – 2x2/2 + x ]10
= [(13/3) – (1)2 + 1 ] – [(03/3) – (0)2 + (0)] = 1/3 – 1 + 1 = 1/3
if y = 3x2 + 2x + 1, then the value od dy/dx at X = 1 is
a. – 2 b. 0 c. 12 d. 8
y = 3x2 + 2x + 1
dy/dx = 6x + 2 , when x = 1 , dy/dx = 6(1) + 2 = 8 [d]
A function might be continuous or not, depends on its?
a. equation b. x – axis c. y- axis d. Domain
. A function might be continuous or not depending on its domain [d]
. Differentiate
- y = 5x2 . sinx
- Y = cos ( 2x – 3x )
- Y = 5x4. Sinx => u = 5x4 v = sin x
(du/dx) 20x3 (dv/dx) = cos x
(dy/dx) = u(dv/dx) + v(du/dx) = 5x4.cos x + 20 x3 sin x ( product rule )
- y = cos ( 2x – 3x ) = cos ( – x )
Let ( – x) = u such that y = cos u
Chain rule: dy/dx = (dy/du) . (du/dx)
du/dx = – 1 , dy/du = – sin u = – sin ( – x )
dy/dx = – 1 x -sin(-x) = sin(-x)
integrates by partial fraction on the given integral
∫ [(3x + 2) / (1 + – 2x3)] dx
[( 5x + 2 )/ ( 1 + x – 2x2] = [ (5x + 2)/(1 – x)(1 + 2x) ] = [ A/(1 – x) ] + [ B/(1 + 2x) ]
multiply through by (1 – x)(1 + 2x)
5x + 2 = A (1 + 2x ) + B (1 – x)
Put (x.1) ; 5(1) + 2 = A (1 + 2(1) + B (1 – 1 )
5 + 2 = 3A
A = 7/3
Put x = (-1/2) ; 5(-1/2) + 2 = A ( 1 + 2(-1/2) + B (1 – (– ½) )
– 5/2 + 2 = 3/2 B
– 1/2 = 3/2 B => B = (– 1/2) x (2/3) = – 1/3
Therefore (5x + 2 )/ ( 1 – x) (1 + 2x2) = [7/3(1 – x )] + ( – 1/ 3(1 + 2x) )
[7/3(1- x)] – [1/3( 1 + 2x )]
∫ (5x + 2 )/ ( 1 + x – 2x2 )dx = -7/3 In(x) – (1/6) In ( 1 + 2x)
= In(x)-7/3 – In (1 + 2x)1/6
= In(x)-7/3/(1 + 2x)1/6)
- use the first principles to find the derivative of f(x) = x3 + 3x
- find dy/dx of 2ex
- evaluate ∫ [(x3 – 25x + 1) / (x3 – 25)]dx
- f(x) = x3 + 3x
From first principle, f1(x) = f(x + dx) – f(x)/dx
f(x + dx ) = ( x + dx )2 + 3(x + dx)
f(x) = x2 + 3x
f1(x) = (x + dx )3 + 3(x + dx) – ( x3 + 3x) / dx
x2 + 2xdx + dx3 + 3x + 3dx = x2 – 3x / dx
2xdx + dx2 + 3dx / dx = dx(2x + dx + 3 ) / dx
f(x) = 2x + dx + 3
lim dx -> 0
f(x) = 2x + (0) + 3 = 2x + 3
- . dy/dx (2ex) = 2. dy/dx (ex)
dex/dx = ex therefore dy/dx (2ex) = 2ex
(x3 – 25x + 1)/ (x2 – 25) =
- [(x3 – 25x + 1)/ (x2 – 25)] . x + 1/x2 – 25
∫ [(x3 – 25x + 1)/ (x2 – 25)] dx = ∫ x dx + ∫ 1/ x2 – 25 dx
∫x dx = x2/2 + C
∫( 1/ x2 – 25 ) dx = ∫ 1/(x – 5 ) (x + 5 ) dx
1/(x – 5)(x + 5) = (A/x-5) + (B/x+5)
1. = A(x + 5) + B(x-5)
For x = 5 1 = A(5 + 5) + B(0) => 10A = 1 => A = 1/10
For x = – 5 1 = A(- 5 + 5) + B(-5 -5) => 1 = – 10B therefore B = – 1/ 10
Therefore ∫ [1/(x – 5)(x + 5)] dx = ∫ [1/10(x + 5)] dx
1/10 In (x – 5 ) = 1 /10 In (x + 5)
In (x – 5)1/10 = In (x + 5)1/10 => In [(x – 5)1/10] / (x + 5)1/10
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