Tasks 06-01 - Antiderivatives & Indefinite Integrals

Section 06: Integral Calculus

Problem 1: Basic Power Rule (x)

Evaluate the following indefinite integrals:

  1. \(\int x^5 \, dx\)

  2. \(\int 3x^4 \, dx\)

  3. \(\int (x^2 + x + 1) \, dx\)

  4. \(\int (4x^3 - 2x^2 + 5) \, dx\)

  5. \(\int 7 \, dx\)

  6. \(\int (x^6 - x^3 + 2x) \, dx\)

  1. \(\int x^5 \, dx = \frac{x^6}{6} + C\)

    Verification: \(\frac{d}{dx}\left[\frac{x^6}{6}\right] = \frac{6x^5}{6} = x^5\) ✓

  2. \(\int 3x^4 \, dx = 3 \cdot \frac{x^5}{5} + C = \frac{3x^5}{5} + C\)

    Verification: \(\frac{d}{dx}\left[\frac{3x^5}{5}\right] = \frac{15x^4}{5} = 3x^4\) ✓

  3. \(\int (x^2 + x + 1) \, dx = \frac{x^3}{3} + \frac{x^2}{2} + x + C\)

  4. \(\int (4x^3 - 2x^2 + 5) \, dx = x^4 - \frac{2x^3}{3} + 5x + C\)

  5. \(\int 7 \, dx = 7x + C\)

  6. \(\int (x^6 - x^3 + 2x) \, dx = \frac{x^7}{7} - \frac{x^4}{4} + x^2 + C\)

Problem 2: Negative and Fractional Powers (x)

Rewrite each expression using exponents, then integrate:

  1. \(\int \frac{1}{x^2} \, dx\)

  2. \(\int \frac{3}{x^4} \, dx\)

  3. \(\int \sqrt{x} \, dx\)

  4. \(\int \frac{1}{\sqrt{x}} \, dx\)

  5. \(\int \left(x^2 + \frac{1}{x^3}\right) \, dx\)

  6. \(\int \left(\sqrt{x} - \frac{2}{x^2}\right) \, dx\)

  1. \(\int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C\)

  2. \(\int \frac{3}{x^4} \, dx = \int 3x^{-4} \, dx = 3 \cdot \frac{x^{-3}}{-3} + C = -\frac{1}{x^3} + C\)

  3. \(\int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} + C = \frac{2x^{3/2}}{3} + C = \frac{2\sqrt{x^3}}{3} + C\)

  4. \(\int \frac{1}{\sqrt{x}} \, dx = \int x^{-1/2} \, dx = \frac{x^{1/2}}{1/2} + C = 2\sqrt{x} + C\)

  5. \(\int \left(x^2 + \frac{1}{x^3}\right) \, dx = \int (x^2 + x^{-3}) \, dx = \frac{x^3}{3} + \frac{x^{-2}}{-2} + C = \frac{x^3}{3} - \frac{1}{2x^2} + C\)

  6. \(\int \left(\sqrt{x} - \frac{2}{x^2}\right) \, dx = \int (x^{1/2} - 2x^{-2}) \, dx = \frac{2x^{3/2}}{3} + \frac{2}{x} + C\)

Problem 3: Initial Value Problems (x)

Find the function \(f(x)\) that satisfies each condition:

  1. \(f'(x) = 3x^2\), \(f(1) = 5\)

  2. \(f'(x) = 2x + 4\), \(f(0) = 3\)

  3. \(f'(x) = 6x^2 - 4x + 1\), \(f(2) = 10\)

  4. \(f'(x) = x^3 - 1\), \(f(-1) = 2\)

  1. Step 1: \(f(x) = \int 3x^2 \, dx = x^3 + C\)

    Step 2: \(f(1) = 1 + C = 5 \implies C = 4\)

    Answer: \(f(x) = x^3 + 4\)

    Verification: \(f'(x) = 3x^2\) ✓, \(f(1) = 1 + 4 = 5\) ✓

  2. Step 1: \(f(x) = \int (2x + 4) \, dx = x^2 + 4x + C\)

    Step 2: \(f(0) = 0 + 0 + C = 3 \implies C = 3\)

    Answer: \(f(x) = x^2 + 4x + 3\)

  3. Step 1: \(f(x) = \int (6x^2 - 4x + 1) \, dx = 2x^3 - 2x^2 + x + C\)

    Step 2: \(f(2) = 2(8) - 2(4) + 2 + C = 16 - 8 + 2 + C = 10 + C = 10 \implies C = 0\)

    Answer: \(f(x) = 2x^3 - 2x^2 + x\)

  4. Step 1: \(f(x) = \int (x^3 - 1) \, dx = \frac{x^4}{4} - x + C\)

    Step 2: \(f(-1) = \frac{1}{4} - (-1) + C = \frac{1}{4} + 1 + C = \frac{5}{4} + C = 2 \implies C = \frac{3}{4}\)

    Answer: \(f(x) = \frac{x^4}{4} - x + \frac{3}{4}\)

Problem 4: Marginal Cost to Total Cost (xx)

A manufacturing company produces electronic components. The marginal cost function (cost of producing one additional unit) is given by:

\[MC(x) = C'(x) = 0.02x^2 - 2x + 50\]

where \(x\) is the number of units produced and cost is in euros.

  1. Find the general antiderivative of \(MC(x)\).

  2. If the fixed costs (costs when \(x = 0\)) are €1,200, find the specific total cost function \(C(x)\).

  3. Calculate the total cost of producing 100 units.

  4. Calculate the average cost per unit when producing 100 units.

  5. Interpret the meaning of the constant of integration in this context.

  1. General antiderivative: \[C(x) = \int (0.02x^2 - 2x + 50) \, dx = \frac{0.02x^3}{3} - x^2 + 50x + C\] \[C(x) = \frac{x^3}{150} - x^2 + 50x + C\]

  2. Using initial condition \(C(0) = 1200\): \[C(0) = 0 - 0 + 0 + C = 1200 \implies C = 1200\]

    Total cost function: \(C(x) = \frac{x^3}{150} - x^2 + 50x + 1200\)

  3. Cost of 100 units: \[C(100) = \frac{1000000}{150} - 10000 + 5000 + 1200\] \[C(100) = 6666.67 - 10000 + 5000 + 1200 = 2866.67\]

    Total cost: €2,866.67

  4. Average cost per unit: \[\overline{C}(100) = \frac{C(100)}{100} = \frac{2866.67}{100} = €28.67 \text{ per unit}\]

  5. Interpretation: The constant of integration \(C = 1200\) represents the fixed costs - expenses that must be paid regardless of production level (rent, equipment, salaries, etc.).

Problem 5: Revenue and Profit Analysis (xx)

A company sells premium coffee subscriptions. Their marginal revenue function is:

\[MR(x) = R'(x) = 60 - 0.4x\]

where \(x\) is the number of subscriptions and revenue is in euros.

The marginal cost function is:

\[MC(x) = C'(x) = 20 + 0.2x\]

Fixed costs are €500.

  1. Find the total revenue function \(R(x)\), given that \(R(0) = 0\).

  2. Find the total cost function \(C(x)\).

  3. Find the profit function \(P(x) = R(x) - C(x)\).

  4. Find the marginal profit function \(MP(x) = P'(x)\) and verify it equals \(MR(x) - MC(x)\).

  5. How many subscriptions maximize profit? What is the maximum profit?

  1. Revenue function: \[R(x) = \int (60 - 0.4x) \, dx = 60x - 0.2x^2 + C\]

    Using \(R(0) = 0\): \(C = 0\)

    Answer: \(R(x) = 60x - 0.2x^2\)

  2. Cost function: \[C(x) = \int (20 + 0.2x) \, dx = 20x + 0.1x^2 + C\]

    Using \(C(0) = 500\): \(C = 500\)

    Answer: \(C(x) = 20x + 0.1x^2 + 500\)

  3. Profit function: \[P(x) = R(x) - C(x) = (60x - 0.2x^2) - (20x + 0.1x^2 + 500)\] \[P(x) = 60x - 0.2x^2 - 20x - 0.1x^2 - 500\] \[P(x) = 40x - 0.3x^2 - 500\]

  4. Marginal profit: \[MP(x) = P'(x) = 40 - 0.6x\]

    Verification: \(MR(x) - MC(x) = (60 - 0.4x) - (20 + 0.2x) = 40 - 0.6x\) ✓

  5. Maximum profit: Set \(P'(x) = 0\): \(40 - 0.6x = 0 \implies x = \frac{40}{0.6} = 66.67\)

    Since we need whole subscriptions: \(x = 67\) subscriptions

    \(P(67) = 40(67) - 0.3(67)^2 - 500 = 2680 - 1346.7 - 500 = €833.30\)

    Maximum profit: Approximately €833 at 67 subscriptions

Problem 6: Velocity and Position (xx)

A delivery drone takes off from a warehouse roof (10 meters above ground). Its vertical velocity is given by:

\[v(t) = 12 - 4t \text{ meters per second}\]

where \(t\) is time in seconds after takeoff.

  1. Find the height function \(h(t)\), using the initial condition \(h(0) = 10\).

  2. At what time does the drone reach its maximum height?

  3. What is the maximum height reached?

  4. When does the drone return to the height of the warehouse roof (10 meters)?

  1. Height function: \[h(t) = \int (12 - 4t) \, dt = 12t - 2t^2 + C\]

    Using \(h(0) = 10\): \(C = 10\)

    Answer: \(h(t) = 12t - 2t^2 + 10\) meters

  2. Maximum height occurs when velocity = 0: \[v(t) = 12 - 4t = 0 \implies t = 3 \text{ seconds}\]

  3. Maximum height: \[h(3) = 12(3) - 2(9) + 10 = 36 - 18 + 10 = 28 \text{ meters}\]

  4. Return to roof height: Set \(h(t) = 10\): \[12t - 2t^2 + 10 = 10\] \[12t - 2t^2 = 0\] \[2t(6 - t) = 0\] \[t = 0 \text{ or } t = 6 \text{ seconds}\]

    The drone returns to roof height at \(t = 6\) seconds.

Problem 7: Compound Business Problem (xxx)

A startup company produces smart home devices. After analyzing production data, they model:

Marginal cost: \(MC(x) = 0.003x^2 - 0.6x + 80\) euros per unit

Marginal revenue: \(MR(x) = 120 - 0.2x\) euros per unit

where \(x\) is the number of units produced per month.

Fixed costs are €10,000 per month.

Part A: Function Determination

  1. Find the total cost function \(C(x)\).

  2. Find the total revenue function \(R(x)\) (assuming \(R(0) = 0\)).

  3. Find the profit function \(P(x)\).

Part B: Analysis

  1. Calculate the profit or loss when producing 100 units.

  2. Find the break-even point(s) where \(P(x) = 0\).

  3. Find the production level that maximizes profit.

  4. Calculate the maximum monthly profit.

Part C: Interpretation

  1. The company currently produces 80 units per month. Should they increase or decrease production? Justify your answer using marginal analysis.

Part A

  1. Cost function: \[C(x) = \int (0.003x^2 - 0.6x + 80) \, dx = 0.001x^3 - 0.3x^2 + 80x + C\]

    With \(C(0) = 10000\): \[C(x) = 0.001x^3 - 0.3x^2 + 80x + 10000\]

  2. Revenue function: \[R(x) = \int (120 - 0.2x) \, dx = 120x - 0.1x^2 + C\]

    With \(R(0) = 0\): \(C = 0\) \[R(x) = 120x - 0.1x^2\]

  3. Profit function: \[P(x) = R(x) - C(x)\] \[P(x) = (120x - 0.1x^2) - (0.001x^3 - 0.3x^2 + 80x + 10000)\] \[P(x) = -0.001x^3 + 0.2x^2 + 40x - 10000\]

Part B

  1. Profit at 100 units: \[P(100) = -0.001(1000000) + 0.2(10000) + 40(100) - 10000\] \[P(100) = -1000 + 2000 + 4000 - 10000 = -5000\]

    Loss of €5,000 at 100 units.

  2. Break-even points: Solve \(P(x) = 0\) \[-0.001x^3 + 0.2x^2 + 40x - 10000 = 0\]

    Multiply by \(-1000\): \(x^3 - 200x^2 - 40000x + 10000000 = 0\)

    Using numerical methods or graphing: approximately \(x \approx 180\) units

    (There may be additional break-even points at higher production levels.)

  3. Maximum profit: \[P'(x) = -0.003x^2 + 0.4x + 40 = 0\]

    Using quadratic formula: \[x = \frac{-0.4 \pm \sqrt{0.16 + 0.48}}{-0.006} = \frac{-0.4 \pm 0.8}{-0.006}\]

    \(x = \frac{-0.4 + 0.8}{-0.006} = -66.7\) (rejected, negative)

    \(x = \frac{-0.4 - 0.8}{-0.006} = 200\) units

    Verify: \(P''(x) = -0.006x + 0.4\); \(P''(200) = -0.8 < 0\) ✓ (maximum)

  4. Maximum profit: \[P(200) = -0.001(8000000) + 0.2(40000) + 40(200) - 10000\] \[P(200) = -8000 + 8000 + 8000 - 10000 = -2000\]

    The maximum “profit” is actually a minimum loss of €2,000.

    This business model isn’t profitable at any production level with current pricing!

Part C

  1. Marginal analysis at 80 units:

    • \(MR(80) = 120 - 0.2(80) = 120 - 16 = €104\)
    • \(MC(80) = 0.003(6400) - 0.6(80) + 80 = 19.2 - 48 + 80 = €51.20\)

    Since \(MR(80) > MC(80)\), marginal revenue exceeds marginal cost.

    Recommendation: The company should increase production because each additional unit adds more to revenue than to cost, reducing the overall loss.

    Note: Even though the company is operating at a loss at all levels, they should produce at the level that minimizes loss (200 units) rather than shutting down, as long as revenue covers variable costs.

Problem 8: Extended Application - Population Growth (xxx)

A city’s population growth rate is modeled by:

\[P'(t) = 0.02t^2 - 0.3t + 2\]

where \(P(t)\) is population in thousands and \(t\) is years since 2020.

In 2020 (\(t = 0\)), the population was 50,000 (or \(P(0) = 50\) in thousands).

  1. Find the population function \(P(t)\).

  2. What is the population in 2025 (\(t = 5\))?

  3. In which year was the growth rate at its minimum?

  4. What is the minimum growth rate (in thousands per year)?

  5. The city plans major infrastructure if population reaches 60,000. In what year will this occur?

  6. Graph both \(P'(t)\) and \(P(t)\) for \(0 \leq t \leq 15\) and interpret the relationship between them.

  1. Population function: \[P(t) = \int (0.02t^2 - 0.3t + 2) \, dt = \frac{0.02t^3}{3} - \frac{0.3t^2}{2} + 2t + C\] \[P(t) = \frac{t^3}{150} - 0.15t^2 + 2t + C\]

    Using \(P(0) = 50\): \(C = 50\)

    Answer: \(P(t) = \frac{t^3}{150} - 0.15t^2 + 2t + 50\) (thousands)

  2. Population in 2025 (\(t = 5\)): \[P(5) = \frac{125}{150} - 0.15(25) + 2(5) + 50\] \[P(5) = 0.833 - 3.75 + 10 + 50 = 57.08\]

    Population: approximately 57,080 people

  3. Minimum growth rate: \[P''(t) = 0.04t - 0.3 = 0 \implies t = 7.5 \text{ years}\]

    Year: 2020 + 7.5 = mid-2027

  4. Minimum growth rate: \[P'(7.5) = 0.02(56.25) - 0.3(7.5) + 2 = 1.125 - 2.25 + 2 = 0.875\]

    Minimum rate: 875 people per year (or 0.875 thousand per year)

  5. When population reaches 60,000 (\(P(t) = 60\)): \[\frac{t^3}{150} - 0.15t^2 + 2t + 50 = 60\] \[\frac{t^3}{150} - 0.15t^2 + 2t - 10 = 0\]

    Multiply by 150: \(t^3 - 22.5t^2 + 300t - 1500 = 0\)

    Using numerical methods: \(t \approx 6.8\) years

    Year: approximately late 2026 or early 2027

  6. Graph interpretation:

Two side-by-side graphs are shown. Left: The population growth rate P’(t) = 0.02t^2 - 0.3t + 2 is an upward-opening parabola that starts at 2, decreases to a minimum of 0.875 at t = 7.5 years (annotated with an arrow), then increases again. The rate remains positive throughout. Right: The total population P(t) is an increasing curve starting at 50 (thousands) at t = 0, crossing the target line of 60 (thousands) at approximately t = 6.8 years. The initial population of 50,000 is annotated.

Interpretation:

  • When \(P'(t) > 0\) (always, in this case), \(P(t)\) is increasing
  • When \(P'(t)\) decreases, \(P(t)\) increases more slowly (concave down)
  • When \(P'(t)\) increases, \(P(t)\) increases more rapidly (concave up)
  • The minimum of \(P'(t)\) at \(t = 7.5\) corresponds to an inflection point in \(P(t)\)