Tasks 06-02 - Definite Integrals & The Fundamental Theorem

Section 06: Integral Calculus

A: Skills

Problem 1: The Fundamental Theorem of Calculus (x)

If \(F\) is an antiderivative of \(f\), the Fundamental Theorem of Calculus says \(\int_a^b f(x) \, dx = F(x) \Big|_a^b = F(b) - F(a)\). Compute the definite integrals. Write the step with the evaluation bar each time.

  1. \(\int_0^2 (5x^4 - 4x) \, dx\)

  2. \(\int_1^3 (6x - 1) \, dx\)

  3. \(\int_{-1}^{2} (4x^3 + 2x) \, dx\)

  4. \(\int_2^5 7 \, dt\)

  5. \(\int_1^3 (4x + 1)(x - 2) \, dx\)

  6. \(\int_0^3 (x^3 - 2x^2 + 4x - 6) \, dx\)

Two steps from the lecture: find an antiderivative \(F\) (no \(+C\), the constants cancel), then compute \(F(\text{upper limit}) - F(\text{lower limit})\).

  1. \(\int_0^2 (5x^4 - 4x) \, dx = \left[x^5 - 2x^2\right]_0^2 = (32 - 8) - 0 = 24\)

  2. \(\int_1^3 (6x - 1) \, dx = \left[3x^2 - x\right]_1^3 = (27 - 3) - (3 - 1) = 24 - 2 = 22\)

  3. \(\int_{-1}^{2} (4x^3 + 2x) \, dx = \left[x^4 + x^2\right]_{-1}^{2} = (16 + 4) - \left((-1)^4 + (-1)^2\right) = 20 - 2 = 18\)

  4. The variable is \(t\): \(\int_2^5 7 \, dt = 7t \Big|_2^5 = 35 - 14 = 21\)

  5. Expand first: \((4x + 1)(x - 2) = 4x^2 - 8x + x - 2 = 4x^2 - 7x - 2\).

    \[\int_1^3 (4x^2 - 7x - 2) \, dx = \left[\frac{4x^3}{3} - 3.5x^2 - 2x\right]_1^3 = (36 - 31.5 - 6) - \left(\frac{4}{3} - 3.5 - 2\right) = -1.5 - \left(-\frac{25}{6}\right) = \frac{8}{3}\]

  6. \(\int_0^3 (x^3 - 2x^2 + 4x - 6) \, dx = \left[\frac{x^4}{4} - \frac{2x^3}{3} + 2x^2 - 6x\right]_0^3 = \left(\frac{81}{4} - 18 + 18 - 18\right) - 0 = \frac{9}{4}\)

Typical mistake: dropping the brackets at a negative lower limit. In c), \(F(-1) = (-1)^4 + (-1)^2 = 1 + 1 = 2\). Whoever writes \(-1^4 - 1^2 = -2\) gets \(20 + 2 = 22\). Put a negative limit into brackets before you raise it to a power.

Problem 2: Integrals with Roots and Powers (xx)

Rewrite each integrand as a sum of powers of \(x\). Then compute the definite integral.

  1. \(\int_1^4 (6\sqrt{x} - x) \, dx\)

  2. \(\int_1^8 x^{2/3} \, dx\)

  3. \(\int_1^4 \frac{1}{x\sqrt{x}} \, dx\)

  4. \(\int_1^3 \frac{6}{x^3} \, dx\)

  5. \(\int_4^9 \frac{x + 1}{\sqrt{x}} \, dx\)

  1. \(6\sqrt{x} - x = 6x^{1/2} - x\). With \(4^{3/2} = \left(\sqrt{4}\right)^3 = 8\):

    \[\int_1^4 \left(6x^{1/2} - x\right) dx = \left[6 \cdot \frac{x^{3/2}}{3/2} - \frac{x^2}{2}\right]_1^4 = \left[4x^{3/2} - \frac{x^2}{2}\right]_1^4 = (32 - 8) - \left(4 - \frac{1}{2}\right) = 24 - 3.5 = 20.5\]

  2. With \(8^{5/3} = \left(\sqrt[3]{8}\right)^5 = 2^5 = 32\):

    \[\int_1^8 x^{2/3} \, dx = \left[\frac{x^{5/3}}{5/3}\right]_1^8 = \frac{3}{5}\left[x^{5/3}\right]_1^8 = \frac{3}{5}(32 - 1) = \frac{93}{5} = 18.6\]

  3. \(x\sqrt{x} = x^1 \cdot x^{1/2} = x^{3/2}\), so \(\frac{1}{x\sqrt{x}} = x^{-3/2}\):

    \[\int_1^4 x^{-3/2} \, dx = \left[\frac{x^{-1/2}}{-1/2}\right]_1^4 = \left[-\frac{2}{\sqrt{x}}\right]_1^4 = -\frac{2}{2} - \left(-\frac{2}{1}\right) = -1 + 2 = 1\]

  4. \(\frac{6}{x^3} = 6x^{-3}\):

    \[\int_1^3 6x^{-3} \, dx = \left[6 \cdot \frac{x^{-2}}{-2}\right]_1^3 = \left[-\frac{3}{x^2}\right]_1^3 = -\frac{3}{9} - (-3) = -\frac{1}{3} + 3 = \frac{8}{3}\]

  5. Divide each term by \(\sqrt{x}\): \(\frac{x + 1}{\sqrt{x}} = x^{1/2} + x^{-1/2}\). With \(9^{3/2} = 27\) and \(4^{3/2} = 8\):

    \[\int_4^9 \left(x^{1/2} + x^{-1/2}\right) dx = \left[\frac{2}{3}x^{3/2} + 2x^{1/2}\right]_4^9 = (18 + 6) - \left(\frac{16}{3} + 4\right) = 24 - \frac{28}{3} = \frac{44}{3}\]

Typical mistake: with a negative power, going the wrong way. In d), the new power is \(-3 + 1 = -2\), not \(-4\). The power rule always adds 1, also for negative powers.

Problem 3: Rectangles and the Riemann Sum (xx)

The definite integral is the limit of Riemann sums, \(\int_a^b f(x) \, dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \cdot \Delta x\) with \(\Delta x = \frac{b - a}{n}\). Let \(f(x) = 2x^2 + x + 2\) on the interval \([0; 4]\).

  1. Compute the Riemann sum with \(n = 4\) rectangles and right endpoints (each rectangle is as high as the graph at the right edge of its subinterval). Give \(\Delta x\) and the four right endpoints first.

  2. Compute the Riemann sum with \(n = 8\) rectangles and right endpoints.

  3. Compute the exact value \(\int_0^4 f(x) \, dx\) with the Fundamental Theorem of Calculus. Compare it with a) and b). Explain why both sums are too large and what happens when \(n\) grows.

  1. \(\Delta x = \frac{4 - 0}{4} = 1\). The subintervals are \([0; 1]\), \([1; 2]\), \([2; 3]\) and \([3; 4]\), so the right endpoints are \(1\), \(2\), \(3\) and \(4\).
\(x_i^*\) 1 2 3 4
\(f(x_i^*)\) 5 12 23 38

\[\sum_{i=1}^{4} f(x_i^*) \cdot \Delta x = (5 + 12 + 23 + 38) \cdot 1 = 78\]

  1. \(\Delta x = \frac{4 - 0}{8} = 0.5\), right endpoints \(0.5, 1, 1.5, \ldots, 4\):
\(x_i^*\) 0.5 1 1.5 2 2.5 3 3.5 4
\(f(x_i^*)\) 3 5 8 12 17 23 30 38

\[\sum_{i=1}^{8} f(x_i^*) \cdot \Delta x = (3 + 5 + 8 + 12 + 17 + 23 + 30 + 38) \cdot 0.5 = 136 \cdot 0.5 = 68\]

  1. \[\int_0^4 (2x^2 + x + 2) \, dx = \left[\frac{2x^3}{3} + \frac{x^2}{2} + 2x\right]_0^4 = \frac{128}{3} + 8 + 8 = \frac{176}{3} \approx 58.67\]

    The sum with 4 rectangles is about \(19.33\) too large, the sum with 8 rectangles only about \(9.33\). On \([0; 4]\), \(f\) increases (\(f'(x) = 4x + 1 > 0\)). So on every subinterval, the right endpoint gives the largest function value, and every rectangle sticks out above the graph. With more and narrower rectangles, the parts that stick out get smaller. As \(n \to \infty\), the Riemann sum approaches the exact value \(\frac{176}{3}\): that is how the definite integral is defined.

Typical mistake: forgetting to multiply by \(\Delta x\) in b). The sum of the eight heights is 136, but each rectangle is only 0.5 wide.

Problem 4: Properties of Definite Integrals (xx)

Use the properties of definite integrals (reversing limits, splitting intervals, constant multiple, sum and difference) to answer a) to e). You do not know the functions, only the given integrals.

  1. Given \(\int_2^6 f(x) \, dx = 11\), find \(\int_6^2 f(x) \, dx\).

  2. Given \(\int_0^5 g(x) \, dx = 6\) and \(\int_0^5 h(x) \, dx = -4\), find \(\int_0^5 [3g(x) - 2h(x)] \, dx\).

  3. Given \(\int_{-1}^{2} f(x) \, dx = 4\) and \(\int_2^7 f(x) \, dx = 10\), find \(\int_{-1}^{7} f(x) \, dx\).

  4. Given \(\int_1^{10} f(x) \, dx = 30\) and \(\int_1^4 f(x) \, dx = 12\), find \(\int_4^{10} f(x) \, dx\).

  5. Given \(\int_2^5 f(x) \, dx = 3\), find \(\int_5^2 4f(x) \, dx\).

  6. A student claims that \(\int_0^3 x \cdot x \, dx = \left(\int_0^3 x \, dx\right)^2\). Decide by computing both sides whether the student is right.

  1. Reversing limits changes the sign: \(\int_6^2 f(x) \, dx = -\int_2^6 f(x) \, dx = -11\)

  2. Sum/difference and constant multiple:

    \[\int_0^5 [3g(x) - 2h(x)] \, dx = 3\int_0^5 g(x) \, dx - 2\int_0^5 h(x) \, dx = 3 \cdot 6 - 2 \cdot (-4) = 18 + 8 = 26\]

  3. Splitting intervals at \(x = 2\): \(\int_{-1}^{7} f(x) \, dx = \int_{-1}^{2} f(x) \, dx + \int_2^7 f(x) \, dx = 4 + 10 = 14\)

  4. Splitting intervals at \(x = 4\): \(\int_1^4 f(x) \, dx + \int_4^{10} f(x) \, dx = \int_1^{10} f(x) \, dx\), so

    \[12 + \int_4^{10} f(x) \, dx = 30 \iff \int_4^{10} f(x) \, dx = 18\]

  5. Constant multiple, then reversing limits:

    \[\int_5^2 4f(x) \, dx = 4\int_5^2 f(x) \, dx = 4 \cdot \left(-\int_2^5 f(x) \, dx\right) = 4 \cdot (-3) = -12\]

  6. Left side: \(\int_0^3 x^2 \, dx = \frac{x^3}{3} \Big|_0^3 = 9\). Right side: \(\int_0^3 x \, dx = \frac{x^2}{2} \Big|_0^3 = 4.5\), and \(4.5^2 = 20.25\). Since \(9 \neq 20.25\), the student is wrong. There is no product rule for integrals: multiply out the integrand first, then integrate.

Typical mistake: in b), \(-2 \cdot (-4) = -8\) instead of \(+8\), which gives 10. Put a negative given value into brackets before you multiply.

Problem 5: Signed Area and Total Area (xx)

For each function, compute (1) the signed area \(\int_a^b f(x) \, dx\) and (2) the total area between the graph and the \(x\)-axis on the given interval.

Hint: Find the zeros of \(f\) in the interval first. For the total area, compute the integral on each piece between the zeros separately and add up the absolute values.

  1. \(f(x) = 6 - 2x\) from \(x = 0\) to \(x = 5\)

  2. \(f(x) = 2x^2 - 11x + 9\) from \(x = 0\) to \(x = 3\)

  3. \(f(x) = x^3 - 9x\) from \(x = -1\) to \(x = 4\)

  1. Zero: \(6 - 2x = 0 \iff x = 3\). For \(x < 3\), \(f(x) > 0\) (e.g. \(f(0) = 6\)), for \(x > 3\), \(f(x) < 0\). Antiderivative \(F(x) = 6x - x^2\).

    1. Signed area: \(\int_0^5 (6 - 2x) \, dx = \left[6x - x^2\right]_0^5 = 30 - 25 = 5\)

    2. Pieces: \(\int_0^3 (6 - 2x) \, dx = 18 - 9 = 9\) (above the \(x\)-axis) and \(\int_3^5 (6 - 2x) \, dx = 5 - 9 = -4\) (below the \(x\)-axis). Total area: \(9 + |-4| = 13\).

  2. Zeros: \(2x^2 - 11x + 9 = (x - 1)(2x - 9) = 0 \iff x = 1\) or \(x = 4.5\). Only \(x = 1\) lies in \([0; 3]\). \(f(0) = 9 > 0\) and \(f(2) = -5 < 0\), so the graph is above the \(x\)-axis on \([0; 1]\) and below it on \([1; 3]\). Antiderivative \(F(x) = \frac{2x^3}{3} - 5.5x^2 + 9x\) with \(F(0) = 0\), \(F(1) = \frac{25}{6}\) and \(F(3) = 18 - 49.5 + 27 = -4.5\).

    1. Signed area: \(\int_0^3 f(x) \, dx = F(3) - F(0) = -4.5 = -\frac{9}{2}\)

    2. Pieces: \(\int_0^1 f(x) \, dx = \frac{25}{6}\) and \(\int_1^3 f(x) \, dx = -4.5 - \frac{25}{6} = -\frac{26}{3}\). Total area: \(\frac{25}{6} + \frac{26}{3} = \frac{25}{6} + \frac{52}{6} = \frac{77}{6} \approx 12.83\).

  3. Zeros: \(x^3 - 9x = x(x^2 - 9) = x(x - 3)(x + 3) = 0 \iff x = 0\), \(x = 3\) or \(x = -3\). In \([-1; 4]\) lie \(x = 0\) and \(x = 3\). Test values: \(f(-1) = 8 > 0\), \(f(1) = -8 < 0\), \(f(4) = 28 > 0\). Antiderivative \(F(x) = \frac{x^4}{4} - 4.5x^2\) with \(F(-1) = -4.25\), \(F(0) = 0\), \(F(3) = 20.25 - 40.5 = -20.25\) and \(F(4) = 64 - 72 = -8\).

    1. Signed area: \(\int_{-1}^{4} f(x) \, dx = F(4) - F(-1) = -8 + 4.25 = -3.75 = -\frac{15}{4}\)

    2. Pieces: \(\int_{-1}^{0} f(x) \, dx = 4.25\), \(\int_0^3 f(x) \, dx = -20.25\) and \(\int_3^4 f(x) \, dx = -8 + 20.25 = 12.25\). Total area: \(4.25 + 20.25 + 12.25 = 36.75 = \frac{147}{4}\).

Typical mistake: taking the absolute value of the whole integral as the total area. In a), \(|5| = 5\), but the total area is 13: the piece below the \(x\)-axis cancels part of the piece above it in the signed area. An area is never negative, a signed area can be.

B: Set Up & Check

Problem 6: From Rates to Totals (xx)

Translate each statement into a definite integral. Then answer the check question: it tells you whether your translation works.

  1. An online bank: “The bank gains new accounts at a rate of \(k(t) = 80 + 12t\) accounts per month, \(t\) months after the start of an advertising campaign.” Write the number of new accounts from the end of month 3 to the end of month 6 as a definite integral and compute it.

    Check: Which unit does your result have? A colleague computes \(k(6) - k(3) = 36\). What does this number measure?

  2. A carpentry workshop makes wall shelves: “At an output of \(x\) shelves per week, the marginal costs are \(C'(x) = 14 + 0.06x\) EUR per shelf. The fixed costs are 880 EUR per week.” Write the rise of the weekly costs when the output rises from 50 to 150 shelves as a definite integral and compute it.

    Check: A trainee adds the fixed costs and gets 2,880 EUR. Determine the cost function \(C(x)\) and decide with \(C(150) - C(50)\) who is right.

  3. A furniture maker installs a CNC milling machine (a computer-controlled machine that cuts wood): “As the machine gets older, it causes maintenance costs at a rate of \(m(t) = 150 + 25t\) EUR per month, \(t\) months after the installation.” Write the maintenance costs within the first year as a definite integral and compute them.

    Check: A colleague uses the limits 1 and 12. Compute the colleague’s value. Which period is missing?

  4. A new café: “In the first 6 months, the profit rate is \(P'(t) = 3t - 6\) thousand EUR per month.” Write the net profit and the total cash flow (all money that moved, \(\int_a^b |P'(t)| \, dt\)) of the first 6 months as definite integrals and compute both.

    Check: A colleague computes the total cash flow as \(\left|\int_0^6 P'(t) \, dt\right|\). Compare the colleague’s value with yours. When do both values agree?

  5. A software company sells a project-management tool as a monthly subscription: “The number of subscribers changes at a rate of \(b(t) = 60 - 15t\) subscribers per month, \(t\) months after the launch of a new version.” Compute \(\int_0^{10} b(t) \, dt\) and state its meaning.

    Check: A colleague writes: “After 10 months, the company has \(-150\) subscribers, that is the area between the graph and the \(t\)-axis.” Correct both parts of the sentence.

  1. The integral of a rate gives the net change (net change theorem). The end of month 3 is \(t = 3\), the end of month 6 is \(t = 6\):

    \[\int_3^6 (80 + 12t) \, dt = \left[80t + 6t^2\right]_3^6 = (480 + 216) - (240 + 54) = 696 - 294 = 402\]

    Check: accounts per month times months gives accounts: 402 new accounts. \(k(6) - k(3) = 152 - 116 = 36\) accounts per month: by this much, the rate grew between \(t = 3\) and \(t = 6\). It is not a number of accounts.

    Typical mistake: subtracting the rates at the limits. The Fundamental Theorem subtracts the values of the antiderivative \(K(t) = 80t + 6t^2\), not the values of the rate.

  2. The integral of the marginal costs is the change in costs:

    \[C(150) - C(50) = \int_{50}^{150} (14 + 0.06x) \, dx = \left[14x + 0.03x^2\right]_{50}^{150} = (2100 + 675) - (700 + 75) = 2000 \text{ EUR}\]

    Check: \(C(x) = 0.03x^2 + 14x + 880\) (antiderivative with \(C(0) = 880\)). \(C(150) = 3655\) and \(C(50) = 1655\), so \(C(150) - C(50) = 2000\) EUR. The fixed costs appear in both values and cancel. You are right, the trainee is not.

    Typical mistake: adding the fixed costs to a change in costs. They are paid at every output, so they do not change when the output rises.

  3. The first year runs from \(t = 0\) to \(t = 12\):

    \[\int_0^{12} (150 + 25t) \, dt = \left[150t + 12.5t^2\right]_0^{12} = 1800 + 1800 = 3600 \text{ EUR}\]

    Check: \(\int_1^{12} (150 + 25t) \, dt = 3600 - (150 + 12.5) = 3437.50\) EUR. The colleague misses the first month, from \(t = 0\) to \(t = 1\), with maintenance costs of 162.50 EUR.

    Typical mistake: starting at \(t = 1\) because “the first month” sounds like \(t = 1\). The variable \(t\) counts the time since the installation, so the first month is the interval \([0; 1]\).

  4. Net profit:

    \[\int_0^6 (3t - 6) \, dt = \left[1.5t^2 - 6t\right]_0^6 = 54 - 36 = 18 \text{ thousand EUR}\]

    For the total cash flow, split at the sign change: \(3t - 6 = 0 \iff t = 2\). For \(t < 2\), the café makes a loss, for \(t > 2\) a profit.

    \[\int_0^2 (3t - 6) \, dt = 6 - 12 = -6, \qquad \int_2^6 (3t - 6) \, dt = 18 - (-6) = 24\]

    \[\int_0^6 |3t - 6| \, dt = |-6| + 24 = 30 \text{ thousand EUR}\]

    Check: the colleague gets \(|18| = 18\), i.e. the net profit again. The losses of the first two months cancel part of the later profits. Both values only agree if the profit rate does not change its sign in the interval.

    Typical mistake: putting the absolute value around the whole integral. For the total cash flow, find the sign change first, integrate each piece and add the absolute values.

  5. \[\int_0^{10} (60 - 15t) \, dt = \left[60t - 7.5t^2\right]_0^{10} = 600 - 750 = -150\]

    Meaning: within the first 10 months after the new version, the number of subscribers falls by 150. At first, more companies sign up than cancel (\(b(t) > 0\) for \(t < 4\)), later more cancel.

    Check: the integral is the change in the number of subscribers, not the number itself. The company had some subscribers before the new version and now has 150 fewer. And an area is never negative: \(-150\) is a signed area. The area between the graph and the \(t\)-axis on \([0; 10]\) is \(\int_0^4 b(t) \, dt + \left|\int_4^{10} b(t) \, dt\right| = 120 + 270 = 390\).

    Typical mistake: reading the integral of a rate as the level of the quantity. It only tells you by how much the level changed.

Problem 7: One Paragraph, All Conditions (xx)

A company builds wooden pergolas (garden roofs on posts) for garden centres. At an output of \(x\) QU per month, its marginal costs are \(C'(x) = 0.3x^2 - 5x + 32\) MU per QU. The rent for the hall and the salaries of the permanent staff come to 90 MU per month, even without production. The company sells every QU at the same price of 40 MU. At the moment, it builds 10 QU per month. The management wants to know by how much the costs rise if the output rises to 15 QU, the profit at the current output, and whether the step from 10 to 15 QU raises the profit.

(QU = quantity units, MU = monetary units)

List every condition and every quantity in this paragraph as a mathematical statement. Do not solve anything.

Check: An assistant writes the change in profit of the step as \(P'(15) - P'(10)\). Explain what this difference measures and give the correct integral.

  1. Marginal costs: \(C'(x) = 0.3x^2 - 5x + 32\), the derivative of the cost function \(C\)
  2. Fixed costs: \(C(0) = 90\)
  3. Revenue with a fixed price: \(R(x) = 40x\), so \(R'(x) = 40\)
  4. Profit: \(P(x) = R(x) - C(x)\), so \(P'(x) = 40 - C'(x)\)
  5. Rise of the costs: \(C(15) - C(10) = \int_{10}^{15} C'(x) \, dx\) (the fixed costs cancel)
  6. Costs at the current output: \(C(10) = C(0) + \int_0^{10} C'(x) \, dx = 90 + \int_0^{10} C'(x) \, dx\)
  7. Profit at the current output: \(P(10) = R(10) - C(10) = 400 - C(10)\)
  8. Change in profit: \(P(15) - P(10) = \int_{10}^{15} \left(40 - C'(x)\right) dx\). The step raises the profit if this integral is positive.

Check: \(P'(15) - P'(10)\) is the change of the marginal profit, i.e. of the rate. It says how much the profit per extra QU changes, not how much the profit changes. The change in profit is the integral of the rate, item 8.

Typical mistake: putting the 90 MU into the rise of the costs, \(90 + \int_{10}^{15} C'(x) \, dx\). The fixed costs belong to \(C(10)\) and to \(C(15)\) alike, so they cancel in the difference.

C: Applied

Problem 8: Registrations for a City Marathon (xx)

A city organizes a marathon with 5,000 start places. The registration opens 10 weeks before the race. \(t\) weeks after the opening, runners register at a rate of

\[s(t) = 6t^2 + 50t + 180\]

registrations per week (\(0 \leq t \leq 10\)). Round to whole registrations where necessary.

  1. Compute the number of registrations within the first 8 weeks. Explain why the result is a number of registrations and not a rate.

  2. The weekly report of the organizers only lists the rate at the end of weeks 2, 4, 6 and 8. Estimate the registrations of the first 8 weeks with a Riemann sum with 4 rectangles and right endpoints. Compare the estimate with a) and explain why it is too high.

  3. Some runners withdraw their registration. The withdrawals come at a rate of \(w(t) = 0.75t^2 + 4t\) withdrawals per week. Compute the number of withdrawals within the first 8 weeks and the number of start places that are taken after 8 weeks. Name the property of definite integrals that you use.

(for verification: 3,808 start places are taken after 8 weeks)

  1. The registration closes as soon as all 5,000 start places are taken. Decide whether this happens in week 9 or in week 10. Use c) and the splitting property, so that you only need to integrate from \(t = 8\) on.

a) By the net change theorem, the integral of the rate gives the number of registrations:

\[\int_0^8 (6t^2 + 50t + 180) \, dt = \left[2t^3 + 25t^2 + 180t\right]_0^8 = 1024 + 1600 + 1440 = 4064\]

Within the first 8 weeks, 4,064 runners register. Unit: registrations per week times weeks gives registrations. The integral adds up the rate over the time, so it is a number of registrations, not a rate.

b) \(\Delta t = \frac{8 - 0}{4} = 2\), right endpoints \(2, 4, 6, 8\):

\(t\) 2 4 6 8
\(s(t)\) 304 476 696 964

\[(304 + 476 + 696 + 964) \cdot 2 = 2440 \cdot 2 = 4880\]

The estimate is \(4880 - 4064 = 816\) registrations (about 20 %) too high. The rate grows over the whole period (\(s'(t) = 12t + 50 > 0\) for \(t \geq 0\)). Each rectangle uses the rate at the end of its two weeks, which is the highest rate in these two weeks. So every rectangle sticks out above the graph of \(s\).

c) Withdrawals:

\[\int_0^8 (0.75t^2 + 4t) \, dt = \left[0.25t^3 + 2t^2\right]_0^8 = 128 + 128 = 256\]

The taken start places change at the rate \(s(t) - w(t)\). With the difference property:

\[\int_0^8 \left(s(t) - w(t)\right) dt = \int_0^8 s(t) \, dt - \int_0^8 w(t) \, dt = 4064 - 256 = 3808\]

After 8 weeks, 3,808 start places are taken (property: sum/difference).

d) Rate of taken places: \(s(t) - w(t) = 5.25t^2 + 46t + 180\) with the antiderivative \(G(t) = 1.75t^3 + 23t^2 + 180t\). Splitting at \(t = 8\): \(\int_0^9 = \int_0^8 + \int_8^9\).

Week 9 is the interval \([8; 9]\):

\[\int_8^9 (5.25t^2 + 46t + 180) \, dt = G(9) - G(8) = 4758.75 - 3808 = 950.75\]

After 9 weeks: \(3808 + 950.75 = 4758.75\), i.e. about 4,759 < 5,000 places. The registration is still open.

Week 10 is the interval \([9; 10]\): \(G(10) - G(9) = 5850 - 4758.75 = 1091.25\). After 10 weeks, \(4758.75 + 1091.25 = 5850 > 5000\). So the last start places are taken in week 10, and the registration closes before the race.

Problem 9: Hand-Knitted Wool Sweaters (xx)

A small company sells hand-knitted wool sweaters online. For \(x\) sweaters per month, the marginal revenue is \(MR(x) = 170 - 0.5x\) EUR per sweater and the marginal costs are \(MC(x) = 26 + 0.4x\) EUR per sweater. The fixed costs (studio rent, yarn storage, website) are 4,900 EUR per month. Without sales, there is no revenue.

  1. Compute the revenue \(R(100)\) and the costs \(C(100)\) of 100 sweaters per month with definite integrals, using \(R(100) = R(0) + \int_0^{100} MR(x) \, dx\) and the same idea for \(C\). Compute the profit of 100 sweaters.

(for verification: \(C(100) = 9500\) EUR)

  1. Determine the output at which the marginal revenue equals the marginal costs. Prove that the profit is highest there.

  2. Compute the extra profit when the output rises from 100 sweaters to the output from b) as a definite integral. Give the maximum profit and the price per sweater at this output (price = revenue per sweater).

  3. The owner says: “If 160 sweaters are good, 200 sweaters are even better.” Compute \(\int_{160}^{200} \left(MR(x) - MC(x)\right) dx\) and explain the result to the owner.

a) Revenue with \(R(0) = 0\):

\[R(100) = 0 + \int_0^{100} (170 - 0.5x) \, dx = \left[170x - 0.25x^2\right]_0^{100} = 17000 - 2500 = 14500 \text{ EUR}\]

Costs with the fixed costs \(C(0) = 4900\):

\[C(100) = 4900 + \int_0^{100} (26 + 0.4x) \, dx = 4900 + \left[26x + 0.2x^2\right]_0^{100} = 4900 + 2600 + 2000 = 9500 \text{ EUR}\]

Profit: \(P(100) = 14500 - 9500 = 5000\) EUR per month.

b) \(MR(x) = MC(x)\):

\[170 - 0.5x = 26 + 0.4x \iff 144 = 0.9x \iff x = 160\]

Proof: \(P'(x) = MR(x) - MC(x) = 144 - 0.9x\). For \(x < 160\), \(P'(x) > 0\) (e.g. \(P'(100) = 54\)), and for \(x > 160\), \(P'(x) < 0\) (e.g. \(P'(200) = -36\)). \(P'\) changes its sign from \(+\) to \(-\) at \(x = 160\): maximum. It is the only critical point, so no other output brings more profit. (Also: \(P''(x) = -0.9 < 0\).)

c) The integral of the marginal profit is the change in profit:

\[P(160) - P(100) = \int_{100}^{160} (144 - 0.9x) \, dx = \left[144x - 0.45x^2\right]_{100}^{160} = (23040 - 11520) - (14400 - 4500) = 11520 - 9900 = 1620 \text{ EUR}\]

Maximum profit: \(P(160) = 5000 + 1620 = 6620\) EUR per month.

Price: \(R(160) = 170 \cdot 160 - 0.25 \cdot 160^2 = 27200 - 6400 = 20800\) EUR, so the price is \(\frac{20800}{160} = 130\) EUR per sweater.

d) \[\int_{160}^{200} (144 - 0.9x) \, dx = \left[144x - 0.45x^2\right]_{160}^{200} = (28800 - 18000) - 11520 = -720 \text{ EUR}\]

The integral is negative: from 160 to 200 sweaters, the profit falls by 720 EUR per month, to \(6620 - 720 = 5900\) EUR. Above 160 sweaters, each further sweater costs more than it brings in (\(MC > MR\)). More sweaters mean more revenue, but not more profit.

Problem 10: Balance Bikes for Children (xxx)

A company produces balance bikes (small bikes without pedals for children) and sells them to toy shops for 110 EUR each. At an output of \(x\) balance bikes per month, the marginal costs are \(MC(x) = C'(x) = 30 + 0.05x\) EUR per bike. The fixed costs are 28,000 EUR per month. The capacity (the maximum possible output) is 1,200 balance bikes per month.

  1. A chain of toy shops orders 400 more balance bikes per month, so the output would rise from 700 to 1,100 bikes per month. Compute the rise of the costs with a definite integral and explain why the fixed costs do not appear. Compute the extra profit of the 400 additional bikes.

  2. Determine the cost function \(C(x)\). Show that the monthly profit is \(P(x) = -0.025x^2 + 80x - 28000\).

(for verification: \(C(x) = 0.025x^2 + 30x + 28000\))

  1. Determine the break-even point and the profit limit (upper break-even point). Explain what the profit limit means for this company.

  2. Determine the output with the highest profit within the capacity and prove that no other possible output brings more. Compute this profit.

  3. A second shift would raise the capacity to 2,000 balance bikes per month. It adds 4,700 EUR per month to the fixed costs, the marginal costs stay the same. Determine the output with the highest profit with the second shift and prove that it is a maximum. Decide with a definite integral whether the second shift pays off.

a) The integral of the marginal costs is the change in costs:

\[C(1100) - C(700) = \int_{700}^{1100} (30 + 0.05x) \, dx = \left[30x + 0.025x^2\right]_{700}^{1100} = (33000 + 30250) - (21000 + 12250) = 30000 \text{ EUR}\]

The fixed costs of 28,000 EUR are paid at 700 bikes and at 1,100 bikes alike, so they cancel in the difference \(C(1100) - C(700)\).

Extra revenue: \(400 \cdot 110 = 44000\) EUR. Extra profit: \(44000 - 30000 = 14000\) EUR per month. (As one integral: \(\int_{700}^{1100} (110 - 30 - 0.05x) \, dx = \left[80x - 0.025x^2\right]_{700}^{1100} = 57750 - 43750 = 14000\).)

b) Antiderivative of the marginal costs with \(C(0) = 28000\):

\[C(x) = \int (30 + 0.05x) \, dx = 30x + 0.025x^2 + k, \qquad C(0) = k = 28000\]

\[C(x) = 0.025x^2 + 30x + 28000\]

Revenue at a fixed price: \(R(x) = 110x\).

\[P(x) = R(x) - C(x) = 110x - 0.025x^2 - 30x - 28000 = -0.025x^2 + 80x - 28000 \quad \checkmark\]

c) \(P(x) = 0 \iff -0.025x^2 + 80x - 28000 = 0 \iff x^2 - 3200x + 1120000 = 0\). Quadratic formula:

\[x = \frac{3200 \pm \sqrt{3200^2 - 4 \cdot 1120000}}{2} = \frac{3200 \pm \sqrt{5760000}}{2} = \frac{3200 \pm 2400}{2}\]

\(x = 400\) or \(x = 2800\). Break-even point: 400 balance bikes per month. Profit limit: 2,800 balance bikes per month.

Meaning: only above 2,800 bikes per month would the company make a loss again. The capacity of 1,200 bikes lies far below this limit, so every output from 401 up to the capacity brings a profit.

d) \(P'(x) = 80 - 0.05x\). The only critical point is \(P'(x) = 0 \iff x = 1600\), which is above the capacity. On the closed interval \([0; 1200]\), \(P'(x) \geq P'(1200) = 20 > 0\): \(P\) increases on the whole interval. The global maximum on \([0; 1200]\) therefore lies at the right end of the interval. Compare the ends: \(P(0) = -28000\) and

\[P(1200) = -36000 + 96000 - 28000 = 32000 \text{ EUR}\]

The highest profit is 32,000 EUR per month at 1,200 balance bikes, the capacity. Here \(P'(1200) \neq 0\): the maximum lies at the end of the interval, not at a critical point.

e) Now the interval is \([0; 2000]\), and \(x = 1600\) lies inside. \(P'(1600) = 0\) and \(P''(x) = -0.05 < 0\): maximum at 1,600 bikes. It is the only critical point, and the ends of the interval give less: \(P(1600) = 36000\), but \(P(0) = -28000\) and \(P(2000) = 32000\) (all before the costs of the second shift). So no other output brings more. (The extra fixed costs only move \(P\) down by 4,700, \(P'\) stays the same.)

Extra profit from the extra output, before the costs of the second shift:

\[\int_{1200}^{1600} (80 - 0.05x) \, dx = \left[80x - 0.025x^2\right]_{1200}^{1600} = (128000 - 64000) - (96000 - 36000) = 4000 \text{ EUR}\]

The second shift costs 4,700 EUR per month but brings only 4,000 EUR: the profit falls by 700 EUR to \(32000 + 4000 - 4700 = 31300\) EUR per month. The second shift does not pay off.

D: FSP Problem

Work on this problem as in the exam: write down every step. The points show what each part is worth.

FSP Problem: Heat Pumps in the Warehouse [21 pts. total]

A company produces heat pumps (heating systems that take warmth from the outside air) and keeps them in its own warehouse until they are sold. \(t\) months after the start of a new production line, the stock in the warehouse changes at the rate

\[r(t) = -t^3 - 13t^2 + 264t\]

heat pumps per month during the first 12 months. The rate is production minus sales: where \(r(t) > 0\), the company produces more heat pumps than it sells. At \(t = 0\), 1,000 heat pumps are in stock. Round to whole heat pumps where necessary.

Part a)

Determine the zeros of \(r\). Decide in which period the stock grows and in which period it falls. [4 pts.]

(for verification: \(r(t) = -t(t - 11)(t + 24)\))

Factor out \(-t\):

\[r(t) = -t\left(t^2 + 13t - 264\right) = 0 \iff t = 0 \text{ or } t^2 + 13t - 264 = 0\]

Quadratic formula:

\[t = \frac{-13 \pm \sqrt{13^2 + 4 \cdot 264}}{2} = \frac{-13 \pm \sqrt{1225}}{2} = \frac{-13 \pm 35}{2}\]

\(t = 11\) or \(t = -24\). Since \(0 \leq t \leq 12\), \(t = -24\) is not a possible time. So \(r(t) = -t(t - 11)(t + 24)\) with the zeros \(t = 0\) and \(t = 11\) in the domain.

Sign: test values \(r(1) = -1 - 13 + 264 = 250 > 0\) and \(r(12) = -1728 - 1872 + 3168 = -432 < 0\).

  • \(0 < t < 11\): \(r(t) > 0\), the company produces more than it sells, the stock grows.
  • \(11 < t \leq 12\): \(r(t) < 0\), the company sells more than it produces, the stock falls.

Grading: 1 Pt. factoring out \(-t\), 1 Pt. quadratic formula, 1 Pt. zeros 0 and 11 with \(t = -24\) excluded, 1 Pt. signs with meaning.

Part b)

Determine the time at which the stock grows fastest. Prove that the rate is highest there within the 12 months and give this rate. [5 pts.]

Necessary condition: \(r'(t) = -3t^2 - 26t + 264 = 0 \iff 3t^2 + 26t - 264 = 0\)

\[t = \frac{-26 \pm \sqrt{26^2 + 12 \cdot 264}}{6} = \frac{-26 \pm \sqrt{3844}}{6} = \frac{-26 \pm 62}{6}\]

\(t = 6\) or \(t = -\frac{44}{3}\) (not in the domain).

Sufficient condition: \(r''(t) = -6t - 26\), so \(r''(6) = -62 < 0\): \(r\) has a maximum at \(t = 6\).

Global: \(t = 6\) is the only critical point in \([0; 12]\). Compare with the ends of the interval:

\[r(0) = 0, \qquad r(6) = -216 - 468 + 1584 = 900, \qquad r(12) = -432\]

The stock grows fastest 6 months after the start, with 900 heat pumps per month. No other time in the 12 months has a higher rate.

Grading: 1 Pt. \(r'(t)\), 1 Pt. \(t = 6\) with \(t = -\frac{44}{3}\) excluded, 1 Pt. \(r''(6) < 0\), 1 Pt. only critical point and ends of the interval, 1 Pt. 900 heat pumps per month.

Part c)

Compute the total change of the stock within the first 12 months. Give the stock at the end of the 12 months. [4 pts.]

(for verification: the stock changes by 6,336 heat pumps)

Approach: by the net change theorem, the integral of the rate is the change of the stock.

Antiderivative:

\[F(t) = -\frac{t^4}{4} - \frac{13t^3}{3} + 132t^2\]

Fundamental Theorem of Calculus:

\[\int_0^{12} r(t) \, dt = F(12) - F(0) = (-5184 - 7488 + 19008) - 0 = 6336\]

Within the first 12 months, the stock grows by 6,336 heat pumps. At the end, \(1000 + 6336 = 7336\) heat pumps are in stock.

Grading: 1 Pt. approach with the integral from 0 to 12, 1 Pt. antiderivative, 1 Pt. 6,336, 1 Pt. stock 7,336.

Part d)

The warehouse has room for 7,400 heat pumps. Determine the time with the highest stock and decide whether the warehouse is large enough. Explain why the stock at the end of the 12 months does not answer this question. [4 pts.]

Highest stock: the rate \(r\) is the derivative of the stock. By part a), \(r\) changes its sign from \(+\) to \(-\) at \(t = 11\), and it is the only sign change in \((0; 12)\). The stock grows up to \(t = 11\) and falls afterwards, so it is highest after 11 months.

Stock after 11 months:

\[\int_0^{11} r(t) \, dt = F(11) - F(0) = -\frac{14641}{4} - \frac{17303}{3} + 15972 \approx 6544.08\]

Stock: \(1000 + 6544.08 \approx 7544\) heat pumps \(> 7400\). The warehouse is not large enough: around month 11, the stock is about 144 heat pumps larger than the warehouse can hold.

Explanation: the stock at the end, 7,336 heat pumps, fits into the warehouse. But the stock is higher before, because it falls in the last month. The end value only shows the net change, not the peak on the way.

Grading: 1 Pt. highest stock at \(t = 11\) with the sign change of \(r\), 1 Pt. \(\int_0^{11} r(t) \, dt \approx 6544\), 1 Pt. 7,544 > 7,400 with decision, 1 Pt. explanation.

Part e)

Compute the change of the stock in the last month from your results in c) and d), without integrating again. Name the property you use. The total movement of the stock, \(\int_0^{12} |r(t)| \, dt\), counts every change of the stock, up and down. Compute it and explain the difference between the net change and the total movement for the company. [4 pts.]

Splitting intervals: \(\int_0^{11} r(t) \, dt + \int_{11}^{12} r(t) \, dt = \int_0^{12} r(t) \, dt\), so

\[\int_{11}^{12} r(t) \, dt = 6336 - 6544.08 = -208.08\]

In the last month, the stock falls by about 208 heat pumps.

Total movement: split at the sign change \(t = 11\):

\[\int_0^{12} |r(t)| \, dt = \int_0^{11} r(t) \, dt + \left|\int_{11}^{12} r(t) \, dt\right| = 6544.08 + 208.08 \approx 6752\]

Meaning: the net change of 6,336 heat pumps only compares the end with the start. The total movement of about 6,752 heat pumps shows that the stock first grew by about 6,544 and then fell by about 208. The net change hides the fall at the end and the peak above the capacity. For the warehouse, the path of the stock matters, not just the result after 12 months.

Grading: 1 Pt. splitting property, 1 Pt. \(-208\), 1 Pt. total movement 6,752, 1 Pt. meaning.