Step-by-Step Answers for Calculus Assignment

Part A: Multiple Choice Questions (20 Marks)

1. What is the limit of sin⁡(𝑥)/𝑥 as 𝑥 approaches 0?
   Answer: B) 1
   Explanation: This is a famous limit in calculus. As x approaches 0, both sin x and x approach 0, but their ratio approaches 1. This can be proven using L'Hôpital's rule or by considering the geometric interpretation of the sine function.
2. The derivative of 𝑓(𝑥)=𝑥^3 using the power rule is:
   Answer: A) 3𝑥^2
   Explanation: The power rule states that for f(x) = x^n, f'(x) = nx^(n-1). Here, n = 3, so f'(x) = 3x^2.
3. Evaluate the integral ∫𝑥^2 𝑑𝑥.
   Answer: A) 𝑥^3/3 + 𝐶
   Explanation: To integrate x^n, increase the power by 1 and divide by the new power. So, ∫𝑥^2 𝑑𝑥 = 𝑥^(2+1)/(2+1) + 𝐶 = 𝑥^3/3 + 𝐶.
4. What is the derivative of cos⁡(𝑥)?
   Answer: A) −sin⁡(𝑥)
   Explanation: The derivative of cosine is negative sine.
5. Which of the following is a technique for finding the integral of a function?
   Answer: D) Integration by parts
   Explanation: Integration by parts is a method used to find integrals, especially useful for products of functions.
6. If 𝑓(𝑥)=ln⁡(𝑥), then 𝑓′(𝑥) is:
   Answer: A) 1/𝑥
   Explanation: The derivative of the natural logarithm function is 1/x.
7. The area under the curve 𝑦=𝑥^2 from 𝑥=1 to 𝑥=3 is:
   Answer: C) 26/3
   Explanation: To find the area, integrate x^2 from 1 to 3: ∫𝑥^2 𝑑𝑥 = [𝑥^3/3]_1^3 = 27/3 - 1/3 = 26/3.
8. What is the derivative of 𝑒^(2𝑥)?
   Answer: A) 2𝑒^(2𝑥)
   Explanation: Using the chain rule, d/dx(e^u) = e^u * du/dx. Here, u = 2x, so du/dx = 2.
9. Which of the following statements about limits is true?
   Answer: C) Limits can be used to determine the behavior of a function as it approaches a specific value.
   Explanation: This is the fundamental purpose of limits in calculus.
10. The integral of 1/𝑥 is:
    Answer: A) ln⁡|𝑥| + 𝐶
    Explanation: The natural logarithm is the antiderivative of 1/x.

Part B: Problem Solving (40 Marks)

11. Determine the limit: lim_(𝑥→0) (𝑒^(𝑥)−1)/𝑥.
    Step-by-step solution:
    Recognize this as an indeterminate form 0/0 as x approaches 0.
    Apply L'Hôpital's rule: differentiate numerator and denominator separately.
    lim_(𝑥→0) (𝑒^(𝑥)−1)/𝑥 = 𝑒^(0) = 1
    Therefore, the limit is 1.
12. Differentiate 𝑓(𝑥)=𝑥^2+1/𝑥−1.
    Step-by-step solution:
    Use the quotient rule: d/dx [u/v] = (v * du/dx - u * dv/dx) / v^2
    Let u = x^2 + 1 and v = x - 1
    d/dx [𝑥^2+1]/[𝑥−1] = (2x(x-1) - (x^2+1))/ (x-1)^2 = (x^2 - 2x -1) / (x-1)^2
13. Find the derivative using the chain rule for 𝑔(𝑥)=sin(3𝑥^2).
    Step-by-step solution:
    Let u = 3x^2, so g(x) = sin(u)
    d/dx [sin(u)] = cos(u) * du/dx
    d/dx [3x^2] = 6x
    Combine: g'(x) = cos(3x^2) * 6x
14. Evaluate the definite integral ∫_(0 to 4) (2𝑥+3) 𝑑𝑥.
    Step-by-step solution:
    Integrate: ∫ (2x+3) dx = x^2 + 3x + C
    Apply limits:
    [x^2 + 3x]_0^4 = (16 + 12) - (0 + 0) = 28
    Therefore, the definite integral equals 28.
15. Using integration by parts, evaluate ∫𝑥 cos(𝑥) 𝑑𝑥.
    Step-by-step solution:
    Let u = x and dv = cos(x) dx
    Then du = dx and v = sin(x)
    Apply the formula: ∫ u dv = uv - ∫ v du
    ∫ x cos(x) dx = x sin(x) - ∫ sin(x) dx = x sin(x) + cos(x) + C

Part C: Theoretical Questions (40 Marks)

16. Explain the Fundamental Theorem of Calculus and its significance in connecting differentiation and integration.
    Explanation:
    The Fundamental Theorem of Calculus consists of two parts:
    Part 1: If f(x) is continuous on [a,b], then the function F(x) defined by F(x) = ∫[a to x] f(t)dt is continuous on [a,b] and differentiable on (a,b), and F'(x) = f(x).
    Part 2: If f is continuous on [a,b], then ∫[a to b] f(x)dx = F(b) - F(a), where F is any antiderivative of f.
    Significance:
    - It establishes the inverse relationship between differentiation and integration.
    - It provides a practical method for evaluating definite integrals.
    - It connects the concept of the area under a curve to the antiderivative of the function.
    - It allows for the development of computational methods in calculus.
    This theorem is crucial because it unifies the two main branches of calculus and provides a powerful tool for solving a wide range of mathematical and physical problems.
17. Discuss the concept of continuity and its importance in calculus, providing examples of functions that are not continuous and explaining why.
    Explanation:
    Continuity Definition: A function f(x) is continuous at a point x = a if:
    1. f(a) is defined
    2. lim[x→a] f(x) exists
    3. lim[x→a] f(x) = f(a)
    Importance in Calculus:
    - Differentiability: Continuous functions are more likely to be differentiable.
    - Intermediate Value Theorem: Guarantees that a continuous function takes on all values between any two of its function values.
    - Integration: Many integration techniques require continuity.
    - Limit evaluation: Continuity allows direct substitution in limit problems.
    Examples of Discontinuous Functions:
    - Step Function: f(x) = { 0 if x < 0, 1 if x ≥ 0 }
    - 1/x at x = 0
    - Floor Function: f(x) = ⌊x⌋
    - Dirichlet Function: f(x) = { 1 if x is rational, 0 if x is irrational }
    Understanding continuity is crucial for analyzing function behavior, solving limit problems, and applying many theorems in calculus. It forms the foundation for more advanced concepts like differentiability and integrability.