Engineering Mathematics: cram pack

Formula sheets, the key method steps for every topic and 20 practice problems, with full solutions at the back.

New problemsBack to the course

In the print window, choose “Save as PDF” to keep a copy on your device.

1Formulas and methods

Limits

Direct substitution
try x = a first; 0/0 means simplify
Factor and cancel
xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + a·xⁿ⁻² + … + aⁿ⁻¹)
Conjugate
(√x − √a)(√x + √a) = x − a
Limits at infinity
divide by the highest power of x; equal degrees give the ratio of leading coefficients
Special limit
limx→∞ (1 + a/x)bx = eab
L'Hôpital's rule
lim f/g = lim f′/g′ (0/0 or ∞/∞)

Derivatives

Power rule
(xⁿ)′ = n·xⁿ⁻¹
Product rule
(uv)′ = u′v + uv′
Quotient rule
(u/v)′ = (u′v − uv′)/v²
Chain rule
f(g(x))′ = f′(g(x))·g′(x)
Common derivatives
(eˣ)′ = eˣ, (ln x)′ = 1/x, (sin x)′ = cos x, (cos x)′ = −sin x

Method

  1. Identify the structure: sum, product, quotient or composition.
  2. Apply the power rule to each power of x.
  3. Use the product, quotient or chain rule where needed.
  4. Simplify, and evaluate at the point if asked.

Integrals

Power rule
∫xⁿ dx = xⁿ⁺¹/(n + 1) + C
Definite integral
∫ab f dx = F(b) − F(a)
Common integrals
∫eˣ dx = eˣ, ∫1/x dx = ln|x|, ∫sin x dx = −cos x
Area between curves
∫ab (f − g) dx

Method

  1. Rewrite the integrand as a sum of simple terms where possible.
  2. Apply the power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1), and standard integrals for ex, 1/x, sin and cos.
  3. Use substitution when you see a function and its derivative together.
  4. For a definite integral, evaluate F(b) − F(a).

Complex numbers

Rectangular form
z = a + jb, j² = −1, z̄ = a − jb
Modulus and argument
|z| = √(a² + b²), tan θ = b/a (check the quadrant)
Polar and exponential form
z = r∠θ = r(cos θ + j sin θ) = r·ejθ
Division
z₁/z₂ = z₁z̄₂/|z₂|²
Product and quotient
r₁∠θ₁ · r₂∠θ₂ = r₁r₂∠(θ₁ + θ₂), r₁∠θ₁ / r₂∠θ₂ = (r₁/r₂)∠(θ₁ − θ₂)
De Moivre
(r∠θ)ⁿ = rⁿ∠nθ
n-th roots
zk = R1/n∠((φ + 360°·k)/n), k = 0 … n − 1
Quadratic with p² < 4q
z = −p/2 ± j√(q − p²/4)

Method

  1. Decide which form suits the operation: rectangular a + jb for adding and subtracting, polar r∠θ for multiplying, dividing, powers and roots.
  2. Multiply out with j² = −1; divide by multiplying the top and bottom by the conjugate of the denominator.
  3. Convert with r = √(a² + b²) and θ from a sketch of the point (check the quadrant), and back with a = r cos θ, b = r sin θ.
  4. For powers use De Moivre, (r∠θ)ⁿ = rⁿ∠nθ; for zⁿ = R∠φ list all n roots R1/n∠((φ + 360°k)/n).

Differential equations

Exponential growth/decay
dy/dt = ky → y = y0ekt
Half-life
k = ln 2 / t½
Newton's law of cooling
T(t) = Ta + (T0 − Ta)e−kt
Integrating factor
y′ + p(x)y = q(x): μ = e∫p dx, (μy)′ = μq
Mixing tank
dm/dt = r·c − (r/V)·m

Matrices & determinants

2×2 determinant
det = ad − bc
2×2 inverse
A⁻¹ = (1/det)·[d −b; −c a]
Product
(AB)ij = Σk AikBkj
Eigenvalues
det(A − λI) = 0
Cramer’s rule
xi = det(Ai)/det(A)

Vectors: dot & cross product

Magnitude
|a| = √(ax² + ay² + az²)
Unit vector
û = a/|a|
Components from an angle
Fx = F·cos θ, Fy = F·sin θ
Dot product
a·b = axbx + ayby + azbz = |a||b|cos θ
Scalar projection of a onto b
a·b/|b|
Work
W = F·d
Cross product
a × b = (aybz − azby, azbx − axbz, axby − aybx)
Areas
parallelogram |a × b|, triangle ½|a × b|
Moment about O
MO = r × F, r = A − O
Volume of a parallelepiped
V = |a·(b × c)|

Method

  1. Write every vector in components; for a force given by a magnitude and an angle, use F·cos θ and F·sin θ.
  2. For lengths and directions use the magnitude √(x² + y² + z²) and the unit vector a/|a|.
  3. Use the dot product for angles, projections, perpendicularity and work: a·b = |a||b|cos θ.
  4. Use the cross product for perpendicular vectors, areas and moments: write the three components carefully, keeping the order a × b.

Probability

Complement
P(not A) = 1 − P(A)
Addition rule
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Independent events
P(A ∩ B) = P(A)·P(B)
Conditional probability
P(A|B) = P(A ∩ B)/P(B)
Binomial
P(X = k) = C(n,k)·pk(1 − p)n−k

Descriptive statistics

Mean
x̄ = Σx/n
Sample standard deviation
s = √(Σ(x − x̄)²/(n − 1))
Median
middle value of the sorted data
z-score
z = (x − μ)/σ
Normal distribution
68% within ±1σ, 95% within ±2σ

Normal distribution

z-score
z = (x − μ)/σ
Below x
P(X < x) = Φ(z)
Above x
P(X > x) = 1 − Φ(z)
Between
Φ(z2) − Φ(z1)
Percentile
x = μ + z·σ

Method

  1. Convert the value to a z-score: z = (x − μ)/σ.
  2. Look up Φ(z), the area to the left of z.
  3. For the area to the right use 1 − Φ(z); for an interval subtract the two left areas.
  4. For a percentile, work backwards: find z for the area, then x = μ + z·σ.

2Practice problems

1.LimitsMedium
xyy = L

Find the limit L = limx→∞ (3x2 − 2x + 6)/(2x2 + x + 3).

f(x) = (3x2 − 2x + 6)/(2x2 + x + 3)x → ∞
L =
2.LimitsHard
xyx = 0

Find the limit L = limx→0 (1 − cos 4x)/(x²), with x in radians.

f(x) = (1 − cos 4x)/(x²)x → 0
L =
3.DerivativesMedium
xyx = 0

Let f(x) = (3x + 1)3. Find f′(0).

f(x) = (3x + 1)3x = 0
f′(0) =
4.DerivativesHard
xyx = 3

The tangent line to f(x) = 2x2 + 4x − 2 at x = 3 crosses the y-axis at y = b. Find b.

f(x) = 2x2 + 4x − 2x = 3
b =
5.IntegralsMedium
xy11.5y = x ln x

Evaluate I = ∫₁1.5 x·ln x dx.

∫₁1.5 x ln x dx
I =
6.IntegralsHard
xy1.5y = 3xy = 2x²

The region between the line y = 3x and the parabola y = 2x² is rotated about the x-axis. Find the volume V of the solid.

y = 3xy = 2x²
V =
7.Complex numbersMedium
ReIm55−5−5θz

Write z = −8 + j8 in exponential form r·ejθ. Give θ in radians, between −π and π.

z = −8 + j8
θ = rad
8.Complex numbersHard
ReIm22−2−244−4−4z₁z₂

Solve z² + 5z + 31.25 = 0. The roots are a ± jb; find the real part a.

p = 5q = 31.25
a =
9.Differential equationsMedium
tyy(0) = 5y(0.5) = ?

Solve y″ + 7y′ + 12y = 0 with y(0) = 5 and y′(0) = -2. Find y(0.5).

y″ + 7y′ + 12y = 0y(0) = 5y′(0) = -2
y(0.5) =
10.Differential equationsHard
tyy(0) = 2y(0.25) = ?

Solve y″ + 3y′ + 11.25y = 0 with y(0) = 2 and y′(0) = 0 (released from rest). Find y(0.25); angles in radians.

y″ + 3y′ + 11.25y = 0y(0) = 2y′(0) = 0
y(0.25) =
11.Matrices & determinantsMedium
A =31−31B =−42−1−3

Let C = AB for the matrices shown. Find the entry c21 (row 2, column 1).

A = [3 1; −3 1]B = [−4 2; −1 −3]
c21 =
12.Matrices & determinantsHard
A =0−257

Find the larger eigenvalue λ of the matrix A shown.

A = [0 −2; 5 7]
λ =
13.Vectors: dot & cross productMedium
xyzab

Find the angle θ between the vectors a = (−1, 4, −2) and b = (2, 6, −5), in degrees.

a = (−1, 4, −2)b = (2, 6, −5)
θ = °
14.Vectors: dot & cross productHard
xyzABC

Find the area A of the triangle with corners A(−3, −4, −1), B(1, −2, 2) and C(1, −3, −2). Coordinates are in metres.

A(−3, −4, −1)B(1, −2, 2)C(1, −3, −2)
A = m²
15.ProbabilityMedium
xp(x)00.310.120.330.3μ = 1.6

For the distribution shown, find the standard deviation σ of X.

x: 0, 1, 2, 3p(x): 0.3, 0.1, 0.3, 0.3
σ =
16.ProbabilityHard
€p(x)−50.7200.25450.05

A game costs €5 to play. You win €25 with probability 0.25 and €50 with probability 0.05; otherwise you win nothing. The stake is not returned. Find the expected net gain E(G) per game.

stake = €5€25: p = 0.25€50: p = 0.05
E(G) = €
17.Descriptive statisticsMedium
10152025303540455055

Find the sample standard deviation s of the data: 42, 37, 10, 29, 57.

n = 5
s =
18.Descriptive statisticsHard
z −2−1012

Exam scores are normally distributed with mean 500 and standard deviation 40. Find the z-score z of a score of 560.

μ = 500σ = 40x = 560
z =
19.Normal distributionMedium
32 min35.1 minX + Y?

X is normal with mean 12 min and standard deviation 1.5 min; Y is independent and normal with mean 20 min and standard deviation 2 min. Find the probability P that X + Y is more than 35.1 min. Round z to two decimals.

μX = 12, σX = 1.5μY = 20, σY = 2X + Y > 35.1 min
P =
20.Normal distributionHard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =

3Solutions

1.Limits

Answer: L = 1.5

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

Divide by the highest power

(3 + (−2)/x + 6/x2)/(2 + 1/x + 3/x2)

Let x → ∞

L = 3/2 = 1.5

Answer

L = 1.5

For equal degrees the limit is the ratio of the leading coefficients.

2.Limits

Answer: L = 8

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

First application

4·sin 4x/(2x) → 0/0

Second application

16·cos 4x/2

Substitute x = 0

L = 16/2 = 8

Answer

L = 8
3.Derivatives

Answer: f′(0) = 9

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Chain rule

f′(x) = 3·(3x + 1)2·3

Substitute

f′(0) = 3·12·3 = 9

Answer

f′(0) = 9
4.Derivatives

Answer: b = −20

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Point and slope

f(3) = 28
f′(x) = 4x + 4 → f′(3) = 16

Tangent line at x = 0

b = f(x₀) − f′(x₀)·x₀ = 28 − 16·3 = −20

Answer

b = −20
5.Integrals

Answer: I = 0.1436

Integration by parts: ∫u·dv = u·v − ∫v·du, from the product rule. Choose u as the factor that gets simpler when you differentiate it (a power of x, or ln x), and dv as the factor you can integrate (eˣ, sin x, cos x, or x when u = ln x). For a definite integral, evaluate u·v at both limits: ∫ₐᵇ u dv = [u·v]ₐᵇ − ∫ₐᵇ v du. A factor x² needs two rounds.

Choose u and dv

u = ln x, du = dx/x; dv = x dx, v = x²/2

Antiderivative

∫x ln x dx = (x²/2)ln x − ∫(x/2) dx = (x²/2)ln x − x²/4

Limits

(1.125·ln 1.5 − 0.5625) − (0 − 0.25) = 0.1436

Answer

I = 0.1436
6.Integrals

Answer: V = 12.72

Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.

Intersection points

3x = 2x² → x = 0, x = 1.5

Washer method

V = π∫₀1.5 ((3x)² − (2x²)²) dx

Integrate

V = π(9·1.5³/3 − 4·1.5⁵/5) = π·4.05 = 12.72

Answer

V = 12.72
7.Complex numbers

Answer: θ = 2.356 rad

Polar form: z = r∠θ = r(cos θ + j sin θ) = r·ejθ, with r = |z| and θ = arg z. Back to rectangular form: a = r cos θ, b = r sin θ. The argument needs the quadrant: θ = atan(b/a) only when a > 0; for a < 0 add or subtract 180°. In polar form multiplication is easy: multiply the moduli and add the arguments; for division divide the moduli and subtract the arguments.

Modulus

r = √(64 + 64) = 11.31

Argument in radians

θ = 135° = 135·π/180 = 2.356 rad

Exponential form

z = 11.31·ej2.356

Answer

θ = 2.356 rad
8.Complex numbers

Answer: a = −2.5

De Moivre and roots: (r∠θ)ⁿ = rⁿ∠nθ. The equation zⁿ = R∠φ has exactly n roots, all with modulus R1/n, at the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1: they lie evenly spaced on a circle, 360°/n apart. A quadratic z² + pz + q = 0 with p² < 4q has the complex roots z = −p/2 ± j√(q − p²/4); a polynomial with real coefficients always has its complex roots in conjugate pairs a ± jb.

Discriminant

p² − 4q = 25 − 125 = −100 < 0

Quadratic formula

z = (−5 ± j√100)/2 = −2.5 + j5, −2.5 − j5

Answer

a = −2.5

Answer

a = −2.5
9.Differential equations

Answer: y(0.5) = 2.257

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.

Characteristic roots

(r + 3)(r + 4) = 0 → r = −3, −4

Constants from the initial values

A + B = 5, −3A − 4B = -2
A = 18, B = −13

Evaluate

y(0.5) = 18·e−1.5 + (−13)·e−2 = 2.257

Answer

y(0.5) = 2.257
10.Differential equations

Answer: y(0.25) = 1.474

Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.

Complex roots

r = −1.5 ± 3j

Constants

A = 2, B = 1.5·2/3 = 1

Evaluate

y(0.25) = e−0.375(2·cos 0.75 + 1·sin 0.75) = 1.474

Answer

y(0.25) = 1.474
11.Matrices & determinants

Answer: c21 = 11

Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.

Row times column

c21 = (−3)·(−4) + 1·(−1) = 11

Answer

c21 = 11
12.Matrices & determinants

Answer: λ = 5

Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.

Characteristic equation

λ² − (7)λ + (10) = 0

Solve

λ = (7 ± √(9))/2 → λ = 2, 5
λ = 5

Answer

λ = 5
13.Vectors: dot & cross product

Answer: θ = 29.99 °

Dot product: a·b = axbx + ayby + azbz = |a|·|b|·cos θ, a scalar. It gives the angle between two vectors, cos θ = a·b/(|a|·|b|), and it is zero when they are perpendicular. The scalar projection of a onto b is a·b/|b|, and a constant force F moving through a displacement d does work W = F·d.

Dot product

a·b = (−1)·2 + 4·6 + (−2)·(−5) = 32

Magnitudes

|a| = √((−1)² + 4² + (−2)²) = 4.583
|b| = √(2² + 6² + (−5)²) = 8.062

Angle

cos θ = 32/(4.583·8.062) = 0.8661
θ = arccos(0.8661) = 29.99°

Answer

θ = 29.99°
14.Vectors: dot & cross product

Answer: A = 8.617 m²

Cross product: a × b = (aybz − azby, azbx − axbz, axby − aybx) is perpendicular to both vectors, and |a × b| = |a|·|b|·sin θ is the area of the parallelogram they span (the triangle has half of it). Order matters: b × a = −(a × b). The moment of a force F applied at point A about point O is MO = r × F, with r = A − O.

Edge vectors

AB = B − A = (4, 2, 3)
AC = C − A = (4, 1, −1)

Cross product

(AB × AC)x = AByACz − ABzACy = 2·(−1) − 3·1 = −5
(AB × AC)y = ABzACx − ABxACz = 3·4 − 4·(−1) = 16
(AB × AC)z = ABxACy − AByACx = 4·1 − 2·4 = −4
|AB × AC| = √((−5)² + 16² + (−4)²) = 17.23

Area of the triangle

A = ½·17.23 = 8.617 m²

Answer

A = 8.617 m²
15.Probability

Answer: σ = 1.2

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Mean

μ = 0·0.3 + 1·0.1 + 2·0.3 + 3·0.3 = 1.6

Mean of the squares

E(X²) = 0·0.3 + 1·0.1 + 4·0.3 + 9·0.3 = 4

Variance

Var(X) = 4 − 1.6² = 1.44
σ = √1.44 = 1.2

Answer

σ = 1.2
16.Probability

Answer: E(G) = 3.75 €

Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.

Net outcomes

G: 20 (p = 0.25), 45 (p = 0.05), −5 (p = 0.7)

Expected value

E(G) = 20·0.25 + 45·0.05 − 5·0.7 = 3.75 €

Answer

E(G) = 3.75 €

A positive value means you win money on average.

17.Descriptive statistics

Answer: s = 17.31

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Mean

x̄ = 175/5 = 35

Squared deviations

Σ(x − x̄)² = 7² + 2² + (−25)² + (−6)² + 22² = 1198

Standard deviation

s = √(1198/4) = 17.31

Answer

s = 17.31
18.Descriptive statistics

Answer: z = 1.5

Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.

Standardise

z = (560 − 500)/40 = 1.5

Answer

z = 1.5
19.Normal distribution

Answer: P = 0.1075

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the sum

μ = 32 min, σ = √(1.5² + 2²) = 2.5 min

Standardise

z = (35.1 − 32)/2.5 = 1.24

Upper tail

P = 1 − Φ(1.24) = 1 − 0.8925 = 0.1075

Answer

P = 0.1075
20.Normal distribution

Answer: P = 0.8413

Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).

Distribution of the difference

μ = 500 − 490 = 10 g, σ = 10 g

Standardise

z = (0 − 10)/10 = −1

Probability

P = 1 − Φ(−1) = 0.8413

Answer

P = 0.8413

ExamProblems · examproblems.com/courses/engineering-mathematics