MTH 102 Test Compilation 1 2

MTH102 Calculus Study Summary

This document summarizes key calculus concepts and problems, including differentiation, integration, series expansion, and formulas, based on the provided quiz screenshots. Solutions reflect the marked correct answers in the document.

Differentiation Topics

• Implicit Differentiation & Logarithmic Functions

  Question: Given that ln(x³) + ln(7x). Find dy/dx.

  Solution: Apply properties of logarithms and differentiate:

  ln(x³) + ln(7x) = 3ln(x) + ln(7) + ln(x)

  = 4ln(x) + ln(7)

  d/dx (4ln(x) + ln(7)) = 4/x + 0

  Answer: 4/x

• Differentiation of Inverse Trigonometric Functions

  Question: d/dx (cosec⁻¹(5x)) = ...

  Formula Recall: d/dx (cosec⁻¹(u)) = -u' / (|u|√(u²-1))

  Solution: Here u = 5x, so u' = 5.

  d/dx (cosec⁻¹(5x)) = -5 / (|5x|√((5x)²-1)) = -5 / (5|x|√(25x²-1)) = -1 / (|x|√(25x²-1))

  Considering the domain restrictions for cosec⁻¹(x) for real x (usually |x| ≥ 1), we can assume x > 0, so |x|=x.

  Answer: -1 / (x√(25x²-1))

• Product Rule of Differentiation

  Question: d/dx (3x² cos⁻¹x) = ...

  Formula Recall: (uv)' = u'v + uv'

  Solution: Let u = 3x² (u' = 6x) and v = cos⁻¹x (v' = -1/√(1-x²)).

  d/dx (3x² cos⁻¹x) = (6x)(cos⁻¹x) + (3x²)(-1/√(1-x²))

  = 6x cos⁻¹x - (3x²) / √(1-x²)

  Answer: 6x cos⁻¹x - (3x²) / √(1-x²)

• Quotient Rule of Differentiation

  Question: Given that y = x / (2x+5). Find dy/dx.

  Formula Recall: (u/v)' = (u'v - uv') / v²

  Solution: Let u = x (u' = 1) and v = 2x+5 (v' = 2).

  dy/dx = (1 * (2x+5) - x * 2) / (2x+5)²

  = (2x+5 - 2x) / (2x+5)²

  = 5 / (2x+5)²

  Answer: 5 / (2x+5)²

• Differentiation of Quotient of Trigonometric Functions

  Question: Find dy/dx of y = cosec x / cot x

  Simplification: cosec x / cot x = (1/sin x) / (cos x / sin x) = 1/cos x = sec x

  Solution: d/dx (sec x) = sec x tan x

  Answer: sec x tan x

• Implicit Differentiation at a Point

  Question: Given that x³ + x + y³ + 3y = 6. Find dy/dx at (1, 1).

  Solution: Differentiate implicitly with respect to x:

  3x² + 1 + 3y²(dy/dx) + 3(dy/dx) = 0

  dy/dx (3y² + 3) = -3x² - 1

  dy/dx = (-3x² - 1) / (3y² + 3)

  Substitute x=1, y=1:

  dy/dx = (-3(1)² - 1) / (3(1)² + 3) = (-3 - 1) / (3 + 3) = -4 / 6 = -2/3

  Answer: -2/3

• General Product Rule Formula

  Question: The Product rule of differentiation formula for the function V = r(x)s(x) is...

  Answer: dV/dx = r(ds/dx) + s(dr/dx)

Integration Topics

• Integration of Logarithmic Function

  Question: Find ∫ ln(x/2) dx

  Formula Recall: ∫ ln(u) du = u ln(u) - u + C (Integration by parts)

  Solution: Let u = x/2. Then du = (1/2)dx, so dx = 2du.

  ∫ ln(u) (2du) = 2 ∫ ln(u) du = 2(u ln(u) - u) + C

  = 2((x/2)ln(x/2) - x/2) + C

  = x ln(x/2) - x + C

  = x(ln(x/2) - 1) + C

  Answer: x(ln(x/2) - 1)

• Integration of Trigonometric Function (sin²x)

  Question: Obtain ∫ sin²x dx

  Formula Recall: sin²x = (1 - cos(2x)) / 2

  Solution: ∫ (1 - cos(2x)) / 2 dx = (1/2) ∫ (1 - cos(2x)) dx

  = (1/2) [x - (sin(2x))/2] + C

  = (1/2)x - (sin(2x))/4 + C

  Answer: (1/2)x - (sin 2x)/4

• Integration of Trigonometric Function (sec²x tan x)

  Question: Integrate ∫ sec²x tan x dx

  Solution: Let u = tan x. Then du = sec²x dx.

  ∫ u du = u²/2 + C

  = (tan²x)/2 + C

  Alternatively, let u = sec x. Then du = sec x tan x dx. But this requires rewriting the integrand as (sec x)(sec x tan x) dx, and the result is (sec²x)/2 + C. The problem provided the option (1/2)sec²x.

  Answer: (1/2) sec²x

• Definite Integration

  Question: Given that 'a' is a positive constant, evaluate ∫[a, 3a] (2x+1)/x dx

  Solution: ∫ (2x+1)/x dx = ∫ (2 + 1/x) dx = 2x + ln|x|

  Evaluate from a to 3a:

  [2(3a) + ln(3a)] - [2a + ln(a)]

  = 6a + ln(3a) - 2a - ln(a)

  = 4a + ln(3a) - ln(a)

  = 4a + ln(3a/a)

  = 4a + ln(3)

  Answer: 4a + ln 3

• Integration of Rational Function (Inverse Tangent Form)

  Question: Find ∫ dx / (4 + 9x²)

  Formula Recall: ∫ dx / (a² + u²) = (1/a) tan⁻¹(u/a) + C

  Solution: Rewrite as ∫ dx / (2² + (3x)²).

  Let u = 3x, so du = 3dx, or dx = (1/3)du. Here a = 2.

  ∫ (1/3)du / (2² + u²) = (1/3) * (1/2) tan⁻¹(u/2) + C

  = (1/6) tan⁻¹(3x/2) + C

  Answer: (1/6) tan⁻¹(3x/2)

• Integration of Rational Function

  Question: Integrate ∫ x/(x-1) dx

  Solution: Use algebraic manipulation: x/(x-1) = (x-1+1)/(x-1) = 1 + 1/(x-1)

  ∫ (1 + 1/(x-1)) dx = ∫ 1 dx + ∫ 1/(x-1) dx

  = x + ln|x-1| + C

  Answer: x + ln(x-1)

Series and Formulas

• Maclaurin Series Expansion

  Question: The nth term of the Maclaurin series expansion of cos 3x is

  Formula Recall: Maclaurin series for cos(u) = Σ[n=0, ∞] ((-1)^n * u^(2n)) / (2n)!

  Solution: Substitute u = 3x.

  nth term = ((-1)^n * (3x)^(2n)) / (2n)!

  Answer: ((-1)^n * (3x)^(2n)) / (2n)!

• Wallis Formula for Odd Powers

  Question: The Wallis formula for ∫[0, π/2] sinⁿx dx, n ≥ 2 and n is an odd number, is

  Formula Recall: For odd n, ∫[0, π/2] sinⁿx dx = ((n-1)/n) * ((n-3)/(n-2)) * ... * (2/3)

  Answer: ((n-1)(n-3)(n-5)...6.4.2) / ((n-2)(n-4)(n-6)...5.3.1)

• Reduction Formula for ∫ cosecⁿx dx

  Question: Which of the following option best describes the reduction formula for V_n = ∫ cosecⁿx dx, n ≥ 2.

  Formula Recall: ∫ cosecⁿx dx = (-cosec^(n-2)x cot x) / (n-1) + ((n-2)/(n-1)) ∫ cosec^(n-2)x dx

  Answer: (-1/(n-1)) cosec^(n-2)x cot x + ((n-2)/(n-1)) V_(n-2)

Functions and Limits

• Function Composition (f(x) from f(x-a))

  Question: Given the function f(x-5) = 3x² - x + 1, evaluate f(x)

  Solution: Let u = x-5. Then x = u+5.

  f(u) = 3(u+5)² - (u+5) + 1

  = 3(u² + 10u + 25) - u - 5 + 1

  = 3u² + 30u + 75 - u - 4

  = 3u² + 29u + 71

  Replace u with x:

  f(x) = 3x² + 29x + 71

  Answer: 3x² + 29x + 71

• Composite Function Evaluation (f(g(x)))

  Question: Given the functions f(x) = 3x²-5, g(x) = 5x-1 evaluate f(g(x)).

  Solution: Substitute g(x) into f(x):

  f(g(x)) = f(5x-1)

  = 3(5x-1)² - 5

  = 3((5x)² - 2(5x)(1) + 1²) - 5

  = 3(25x² - 10x + 1) - 5

  = 75x² - 30x + 3 - 5

  = 75x² - 30x - 2

  Answer: 75x² - 30x - 2

• Limits using L'Hopital's Rule

  Question: lim (x⁵-243) / (x³-27) as x→3 equal

  Solution: Direct substitution yields 0/0, an indeterminate form. Apply L'Hopital's Rule:

  d/dx (x⁵-243) = 5x⁴

  d/dx (x³-27) = 3x²

  lim (5x⁴) / (3x²) as x→3

  = lim (5x²) / 3 as x→3

  = (5 * 3²) / 3 = (5 * 9) / 3 = 45 / 3 = 15

  Answer: 15