Tasks 06-05 - Economic Applications & Integration by Parts

Section 06: Integral Calculus

A: Skills

Problem 1: Integration by Parts (x)

Evaluate each integral with integration by parts, \(\int u \, dv = uv - \int v \, du\). Write down \(u\), \(du\), \(dv\) and \(v\) first. Check the results of a) and c) by differentiating.

  1. \(\int x \cdot e^{4x} \, dx\)

  2. \(\int (3x - 2) \cdot e^x \, dx\)

  3. \(\int (2x + 5) \cdot e^{-x} \, dx\)

  4. \(\int 8x \cdot e^{0.5x} \, dx\)

  1. \(u = x\), \(du = dx\), \(dv = e^{4x} \, dx\), \(v = \frac{1}{4}e^{4x}\):

    \[\int x \cdot e^{4x} \, dx = \frac{x}{4}e^{4x} - \int \frac{1}{4}e^{4x} \, dx = \frac{x}{4}e^{4x} - \frac{1}{16}e^{4x} + C = \frac{1}{16}e^{4x}(4x - 1) + C\]

    Check (product rule, chain rule for \(e^{4x}\)): \(\frac{d}{dx}\left[\frac{1}{16}e^{4x}(4x - 1)\right] = \frac{4}{16}e^{4x}(4x - 1) + \frac{1}{16}e^{4x} \cdot 4 = \frac{1}{4}e^{4x}(4x - 1 + 1) = x \cdot e^{4x}\) ✓

  2. \(u = 3x - 2\), \(du = 3 \, dx\), \(dv = e^x \, dx\), \(v = e^x\):

    \[\int (3x - 2) \cdot e^x \, dx = (3x - 2)e^x - \int 3e^x \, dx = (3x - 2)e^x - 3e^x + C = (3x - 5)e^x + C\]

  3. \(u = 2x + 5\), \(du = 2 \, dx\), \(dv = e^{-x} \, dx\), \(v = -e^{-x}\):

    \[\int (2x + 5) \cdot e^{-x} \, dx = -(2x + 5)e^{-x} - \int \left(-2e^{-x}\right) dx = -(2x + 5)e^{-x} - 2e^{-x} + C = -(2x + 7)e^{-x} + C\]

    Check: \(\frac{d}{dx}\left[-(2x + 7)e^{-x}\right] = -2e^{-x} + (2x + 7)e^{-x} = (2x + 5)e^{-x}\) ✓

  4. \(u = 8x\), \(du = 8 \, dx\), \(dv = e^{0.5x} \, dx\), \(v = \frac{1}{0.5}e^{0.5x} = 2e^{0.5x}\):

    \[\int 8x \cdot e^{0.5x} \, dx = 16x \cdot e^{0.5x} - \int 16e^{0.5x} \, dx = 16x \cdot e^{0.5x} - 32e^{0.5x} + C = 16(x - 2)e^{0.5x} + C\]

Typical mistake: in c), writing \(v = e^{-x}\). The integral of \(e^{-x}\) is \(-e^{-x}\), and the two minus signs in \(-\int v \, du\) then turn into a plus.

Problem 2: Logarithms and the LIATE Rule (x)

Use the LIATE rule to decide which factor is \(u\). Then evaluate the integral (\(x > 0\)).

  1. \(\int x^3 \cdot \ln(x) \, dx\)

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

  3. \(\int (4x - 1) \cdot \ln(x) \, dx\)

Hint: In b), write \(\frac{\ln(x)}{x^2} = x^{-2} \cdot \ln(x)\).

In all three integrals, L comes before A, so \(u = \ln(x)\) and \(du = \frac{1}{x} \, dx\).

  1. \(dv = x^3 \, dx\), \(v = \frac{x^4}{4}\):

    \[\int x^3 \cdot \ln(x) \, dx = \frac{x^4}{4}\ln(x) - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx = \frac{x^4}{4}\ln(x) - \int \frac{x^3}{4} \, dx = \frac{x^4}{4}\ln(x) - \frac{x^4}{16} + C = \frac{x^4}{16}\left(4\ln(x) - 1\right) + C\]

  2. \(dv = x^{-2} \, dx\), \(v = \frac{x^{-1}}{-1} = -\frac{1}{x}\):

    \[\int x^{-2} \cdot \ln(x) \, dx = -\frac{\ln(x)}{x} - \int \left(-\frac{1}{x}\right) \cdot \frac{1}{x} \, dx = -\frac{\ln(x)}{x} + \int x^{-2} \, dx = -\frac{\ln(x)}{x} - \frac{1}{x} + C = -\frac{\ln(x) + 1}{x} + C\]

  3. \(dv = (4x - 1) \, dx\), \(v = 2x^2 - x\):

    \[\int (4x - 1) \cdot \ln(x) \, dx = \left(2x^2 - x\right)\ln(x) - \int \left(2x^2 - x\right) \cdot \frac{1}{x} \, dx = \left(2x^2 - x\right)\ln(x) - \int (2x - 1) \, dx\]

    \[= \left(2x^2 - x\right)\ln(x) - x^2 + x + C\]

Typical mistake: choosing \(u = x^3\) in a). Then \(dv = \ln(x) \, dx\), and you would need an antiderivative of \(\ln(x)\) to get \(v\). With \(u = \ln(x)\), the logarithm disappears after one step, because \(\frac{1}{x}\) cancels against the power of \(x\).

Problem 3: Repeated Integration by Parts (xx)

Evaluate each integral. After the first step, a new integral by parts remains: apply integration by parts again until no power of \(x\) is left.

  1. \(\int x^2 \cdot e^{3x} \, dx\)

  2. \(\int \left(x^2 - 4x\right) \cdot e^x \, dx\)

  3. \(\int x^3 \cdot e^x \, dx\) (three steps)

  1. First step: \(u = x^2\), \(du = 2x \, dx\), \(dv = e^{3x} \, dx\), \(v = \frac{1}{3}e^{3x}\):

    \[\int x^2 \cdot e^{3x} \, dx = \frac{x^2}{3}e^{3x} - \int \frac{2x}{3}e^{3x} \, dx\]

    Second step: \(u = \frac{2x}{3}\), \(du = \frac{2}{3} \, dx\), \(v = \frac{1}{3}e^{3x}\):

    \[\int \frac{2x}{3}e^{3x} \, dx = \frac{2x}{9}e^{3x} - \int \frac{2}{9}e^{3x} \, dx = \frac{2x}{9}e^{3x} - \frac{2}{27}e^{3x}\]

    Together:

    \[\int x^2 \cdot e^{3x} \, dx = \frac{x^2}{3}e^{3x} - \frac{2x}{9}e^{3x} + \frac{2}{27}e^{3x} + C = \frac{1}{27}e^{3x}\left(9x^2 - 6x + 2\right) + C\]

  2. First step: \(u = x^2 - 4x\), \(du = (2x - 4) \, dx\), \(v = e^x\):

    \[\int \left(x^2 - 4x\right)e^x \, dx = \left(x^2 - 4x\right)e^x - \int (2x - 4)e^x \, dx\]

    Second step: \(u = 2x - 4\), \(du = 2 \, dx\), \(v = e^x\): \(\int (2x - 4)e^x \, dx = (2x - 4)e^x - 2e^x = (2x - 6)e^x\).

    Together:

    \[\int \left(x^2 - 4x\right)e^x \, dx = \left(x^2 - 4x\right)e^x - (2x - 6)e^x + C = e^x\left(x^2 - 6x + 6\right) + C\]

    Check: \(\frac{d}{dx}\left[e^x\left(x^2 - 6x + 6\right)\right] = e^x\left(x^2 - 6x + 6\right) + e^x(2x - 6) = e^x\left(x^2 - 4x\right)\) ✓

  3. Each step lowers the power of \(x\) by one, and \(v = e^x\) every time:

    \[\int x^3 e^x \, dx = x^3 e^x - \int 3x^2 e^x \, dx\]

    \[\int 3x^2 e^x \, dx = 3x^2 e^x - \int 6x e^x \, dx, \qquad \int 6x e^x \, dx = 6x e^x - \int 6e^x \, dx = 6x e^x - 6e^x\]

    Together:

    \[\int x^3 e^x \, dx = x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x + C = e^x\left(x^3 - 3x^2 + 6x - 6\right) + C\]

Typical mistake: losing the minus sign in front of the second integral. Put the whole result of the second step in brackets, \(x^3 e^x - \left(3x^2 e^x - \ldots\right)\), and only then remove the brackets. The signs alternate: \(+, -, +, -\).

Problem 4: Definite Integrals by Parts (xx)

Compute each integral. Give the exact value and a decimal rounded to two decimal places.

  1. \(\int_0^2 (6 - x) \cdot e^x \, dx\)

  2. \(\int_1^4 \frac{\ln(x)}{\sqrt{x}} \, dx\)

  3. \(\int_0^3 t \cdot e^{-t/3} \, dt\)

  4. \(\int_1^2 (x - 1) \cdot e^{2x} \, dx\)

Hint: In b), \(\frac{1}{\sqrt{x}} = x^{-1/2}\) and \(\ln(4) = 2\ln(2)\). In c), \(e^{-t/3} = e^{-\frac{1}{3}t}\).

Find an antiderivative by parts first, then use the evaluation bar.

  1. \(u = 6 - x\), \(du = -dx\), \(v = e^x\): \(\int (6 - x)e^x \, dx = (6 - x)e^x - \int (-e^x) \, dx = (6 - x)e^x + e^x = (7 - x)e^x\).

    \[\int_0^2 (6 - x) \cdot e^x \, dx = \left[(7 - x)e^x\right]_0^2 = 5e^2 - 7 \approx 29.95\]

  2. \(u = \ln(x)\), \(du = \frac{1}{x} \, dx\), \(dv = x^{-1/2} \, dx\), \(v = \frac{x^{1/2}}{1/2} = 2\sqrt{x}\):

    \[\int x^{-1/2}\ln(x) \, dx = 2\sqrt{x}\ln(x) - \int 2x^{1/2} \cdot \frac{1}{x} \, dx = 2\sqrt{x}\ln(x) - \int 2x^{-1/2} \, dx = 2\sqrt{x}\ln(x) - 4\sqrt{x}\]

    \[\int_1^4 \frac{\ln(x)}{\sqrt{x}} \, dx = \left[2\sqrt{x}\ln(x) - 4\sqrt{x}\right]_1^4 = (4\ln(4) - 8) - (0 - 4) = 8\ln(2) - 4 \approx 1.55\]

  3. \(u = t\), \(du = dt\), \(dv = e^{-t/3} \, dt\), \(v = \frac{1}{-1/3}e^{-t/3} = -3e^{-t/3}\):

    \[\int t \cdot e^{-t/3} \, dt = -3t \cdot e^{-t/3} - \int \left(-3e^{-t/3}\right) dt = -3t \cdot e^{-t/3} - 9e^{-t/3}\]

    \[\int_0^3 t \cdot e^{-t/3} \, dt = \left[-3t \cdot e^{-t/3} - 9e^{-t/3}\right]_0^3 = \left(-9e^{-1} - 9e^{-1}\right) - (0 - 9) = 9 - \frac{18}{e} \approx 2.38\]

  4. \(u = x - 1\), \(du = dx\), \(v = \frac{1}{2}e^{2x}\): \(\int (x - 1)e^{2x} \, dx = \frac{x - 1}{2}e^{2x} - \frac{1}{4}e^{2x} = \frac{1}{4}e^{2x}(2x - 3)\).

    \[\int_1^2 (x - 1) \cdot e^{2x} \, dx = \left[\frac{1}{4}e^{2x}(2x - 3)\right]_1^2 = \frac{e^4}{4} - \left(-\frac{e^2}{4}\right) = \frac{e^4 + e^2}{4} \approx 15.50\]

Typical mistake: in c), taking the value at the lower limit as 0 because the first term vanishes at \(t = 0\). The second term \(-9e^{-t/3}\) is \(-9\) at \(t = 0\), so you subtract \(-9\).

Problem 5: Average Value (xx)

The average value of \(f\) on \([a; b]\) is \(f_{avg} = \frac{1}{b - a}\int_a^b f(x) \, dx\). Compute it. Give exact values and decimals rounded to two decimal places.

  1. \(f(x) = 6x^2 - 11x + 7\) on \([1; 3]\)

  2. \(f(x) = 4e^{0.5x}\) on \([0; 2]\)

  3. \(f(x) = \frac{15}{x}\) on \([3; 9]\)

  4. \(f(x) = (x + 2) \cdot e^{-x}\) on \([0; 2]\)

  5. The average value of \(f(x) = 8 - 0.3x\) on \([0; b]\) is 5. Determine \(b > 0\).

  1. \[f_{avg} = \frac{1}{3 - 1}\int_1^3 \left(6x^2 - 11x + 7\right) dx = \frac{1}{2}\left[2x^3 - 5.5x^2 + 7x\right]_1^3 = \frac{1}{2}\left((54 - 49.5 + 21) - (2 - 5.5 + 7)\right) = \frac{1}{2}(25.5 - 3.5) = 11\]

    A rectangle of height 11 over \([1; 3]\) has the same area as the region under the graph.

  2. \[f_{avg} = \frac{1}{2}\int_0^2 4e^{0.5x} \, dx = \frac{1}{2}\left[8e^{0.5x}\right]_0^2 = \frac{1}{2}(8e - 8) = 4(e - 1) \approx 6.87\]

  3. \[f_{avg} = \frac{1}{6}\int_3^9 \frac{15}{x} \, dx = \frac{1}{6}\left[15\ln|x|\right]_3^9 = \frac{15}{6}\left(\ln(9) - \ln(3)\right) = \frac{5}{2}\ln(3) \approx 2.75\]

  4. By parts with \(u = x + 2\), \(v = -e^{-x}\): \(\int (x + 2)e^{-x} \, dx = -(x + 2)e^{-x} - e^{-x} = -(x + 3)e^{-x}\).

    \[f_{avg} = \frac{1}{2}\left[-(x + 3)e^{-x}\right]_0^2 = \frac{1}{2}\left(-5e^{-2} + 3\right) = \frac{3 - 5e^{-2}}{2} \approx 1.16\]

  5. \[\frac{1}{b}\int_0^b (8 - 0.3x) \, dx = \frac{1}{b}\left(8b - 0.15b^2\right) = 8 - 0.15b = 5 \iff 0.15b = 3 \iff b = 20\]

Typical mistake: forgetting the factor \(\frac{1}{b - a}\). Then the result is the area (the integral), not the average height. In c), the interval has length \(9 - 3 = 6\), not 9.

B: Set Up & Check

Problem 6: From Words to Integrals (xx)

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

  1. A company launches a smart ring (a ring that measures sleep and pulse): “\(t\) months after the launch, the company sells \(s(t) = 250 + 19t - 0.5t^2\) rings per month.” Write the average monthly sales within the first 24 months as a formula and compute it.

    Check: A classmate gets 9,168. Explain what the classmate has computed and what is missing.

  2. “Prove that \(F(x) = (3x + 1) \cdot e^{-2x}\) is an antiderivative of \(f(x) = (1 - 6x) \cdot e^{-2x}\).” Write down the equation you have to show and show it.

    Check: A classmate differentiates each factor on its own, \(3 \cdot \left(-2e^{-2x}\right) = -6e^{-2x}\), and concludes that \(F\) is not an antiderivative of \(f\). Find the mistake.

  3. A company makes floor lamps: “At a sales volume of \(x\) lamps per month, the marginal revenue is \(MR(x) = 180 - 1.6x\) EUR per lamp.” Write the revenue from selling units 20 to 50 (the extra revenue when the monthly sales rise from 20 to 50 lamps) as a definite integral and compute it.

    Check: A classmate computes \(MR(50) - MR(20) = -48\) and says: “The revenue falls by 48 EUR.” Explain what this number really measures.

  4. A new burger restaurant in a shopping street: “\(t\) days after the opening, the restaurant sells \(b(t) = 160 + 300e^{-0.4t}\) burgers per day.” Compute the average number of burgers per day within the first 15 days and the long-term level \(\lim_{t \to \infty} b(t)\). Explain why the average lies above the long-term level.

    Check: A classmate says: “In the long run, the restaurant sells 460 burgers per day.” Find the mistake.

  5. “Evaluate \(\int (6x + 1) \cdot e^{2x} \, dx\).” Decide which factor is \(u\) and evaluate the integral.

    Check: A classmate chooses \(u = e^{2x}\) and \(dv = (6x + 1) \, dx\). Write down the integral \(\int v \, du\) that the classmate gets and explain why this choice does not help.

  1. The average over \([0; 24]\) divides the total by the length of the interval:

    \[s_{avg} = \frac{1}{24 - 0}\int_0^{24} \left(250 + 19t - 0.5t^2\right) dt = \frac{1}{24}\left[250t + 9.5t^2 - \frac{t^3}{6}\right]_0^{24} = \frac{1}{24}(6000 + 5472 - 2304) = \frac{9168}{24} = 382\]

    On average, the company sells 382 rings per month in the first 24 months.

    Check: 9,168 is the integral alone, the total number of rings sold in the 24 months. The division by the length \(24 - 0 = 24\) is missing.

    Typical mistake: averaging only the two end values, \(\frac{s(0) + s(24)}{2} = \frac{250 + 418}{2} = 334\). This ignores all the months in between, where the sales are higher.

  2. \(F\) is an antiderivative of \(f\) if \(F'(x) = f(x)\). Product rule with \(u = 3x + 1\), \(u' = 3\), \(v = e^{-2x}\), \(v' = -2e^{-2x}\) (chain rule):

    \[F'(x) = 3e^{-2x} + (3x + 1) \cdot \left(-2e^{-2x}\right) = (3 - 6x - 2)e^{-2x} = (1 - 6x)e^{-2x} = f(x) \; \checkmark\]

    No integration is needed: differentiating \(F\) is the proof.

    Check: the derivative of a product is not the product of the derivatives. The classmate needs the product rule, \(u'v + uv'\), as above.

    Typical mistake: integrating \(f\) by parts instead. This also works, but it takes longer and gives \(F\) only up to a constant, which you then have to compare.

  3. The integral of the marginal revenue is the change in revenue:

    \[\int_{20}^{50} (180 - 1.6x) \, dx = \left[180x - 0.8x^2\right]_{20}^{50} = (9000 - 2000) - (3600 - 320) = 7000 - 3280 = 3720\]

    Selling lamps 20 to 50 brings 3,720 EUR more revenue per month.

    Check: \(MR(50) - MR(20) = -48\) is the change of the marginal revenue: the 50th lamp brings 48 EUR less than the 20th lamp. The revenue itself still rises, because \(MR(x) > 0\) on \([20; 50]\).

    Typical mistake: subtracting the values of \(MR\) instead of integrating. \(MR\) is a rate (EUR per lamp), and only its integral gives EUR.

  4. Average over \([0; 15]\) with \(\int e^{-0.4t} \, dt = \frac{1}{-0.4}e^{-0.4t} = -2.5e^{-0.4t}\):

    \[b_{avg} = \frac{1}{15}\int_0^{15} \left(160 + 300e^{-0.4t}\right) dt = \frac{1}{15}\left[160t - 750e^{-0.4t}\right]_0^{15} = \frac{1}{15}\left(2400 - 750e^{-6} + 750\right) = 210 - 50e^{-6} \approx 209.88\]

    Long-term level: \(e^{-0.4t} \to 0\) for \(t \to \infty\), so \(\lim_{t \to \infty} b(t) = 160\) burgers per day.

    The term \(300e^{-0.4t}\) is positive, so \(b(t) > 160\) on every day. The opening effect adds extra burgers, most of them in the first days. The average of the first 15 days therefore lies above 160.

    Check: 460 is the start value \(b(0) = 160 + 300\). The classmate has used \(e^{0} = 1\) instead of the limit \(e^{-0.4t} \to 0\).

    Typical mistake: confusing the start value with the long-term level. For \(t \to \infty\), only the constant 160 remains.

  5. LIATE: A (the polynomial \(6x + 1\)) comes before E, so \(u = 6x + 1\), \(du = 6 \, dx\), \(dv = e^{2x} \, dx\), \(v = \frac{1}{2}e^{2x}\):

    \[\int (6x + 1) \cdot e^{2x} \, dx = \frac{6x + 1}{2}e^{2x} - \int 3e^{2x} \, dx = \frac{6x + 1}{2}e^{2x} - \frac{3}{2}e^{2x} + C = (3x - 1)e^{2x} + C\]

    Check: with \(u = e^{2x}\), \(du = 2e^{2x} \, dx\) and \(v = 3x^2 + x\). The new integral is \(\int \left(3x^2 + x\right) \cdot 2e^{2x} \, dx\). The power of \(x\) has gone up from 1 to 2, so this integral is harder than the one you started with.

    Typical mistake: choosing \(u\) as the factor that comes first in the product. Choose \(u\) so that it gets simpler when you differentiate it: the polynomial, not the exponential.

Problem 7: One Paragraph, All Conditions (xx)

An electronics company launches a handheld game console (a small games device with its own screen). The monthly sales volume is modelled by \(f(t) = a + b \cdot t \cdot e^{-kt}\) QU per month (1 QU = 1,000 consoles), \(t\) months after the launch, with positive constants \(a\), \(b\) and \(k\). In the long run, the company expects to sell 9 QU per month. The sales are highest 4 months after the launch, at 15 QU per month. The management wants to know the total sales volume within the first year, the average monthly sales volume within the first 6 months and the month in which the sales fall fastest. The stock of 150 QU in the warehouse should last for the whole first year.

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

Check: An assistant writes the condition “the sales are highest after 4 months, at 15 QU” as \(f'(4) = 15\). Explain what is wrong and give the correct conditions.

  1. Model: \(f(t) = a + b \cdot t \cdot e^{-kt}\) for \(t \geq 0\), with \(a, b, k > 0\)
  2. Long-term level: \(\lim_{t \to \infty} f(t) = 9\). Since \(t \cdot e^{-kt} \to 0\), this means \(a = 9\)
  3. Highest sales after 4 months (necessary condition): \(f'(4) = 0\)
  4. The highest value: \(f(4) = 15\)
  5. Total sales volume within the first year: \(\int_0^{12} f(t) \, dt\)
  6. Average monthly sales volume within the first 6 months: \(\frac{1}{6 - 0}\int_0^{6} f(t) \, dt\)
  7. Fastest fall: the time \(t_w\) where \(f'\) has its minimum, so \(f''(t_w) = 0\) with a sign change of \(f''\) (an inflection point of \(f\))
  8. The stock lasts for the first year: \(\int_0^{12} f(t) \, dt \leq 150\)

Check: the assistant has mixed up the position and the value. “Highest after 4 months” is a statement about the slope: \(f'(4) = 0\). “At 15 QU” is a statement about the value: \(f(4) = 15\). So there are two conditions, and \(f'(4) = 15\) is neither of them.

Typical mistake: writing “sales fall fastest” as \(f'(t) = 0\). At \(f'(t) = 0\), the sales do not change at all. The fastest fall is the most negative slope, i.e. the minimum of \(f'\).

C: Applied

Problem 8: Mechanical Keyboards (xx)

A small company assembles mechanical keyboards for gamers. At an output of \(x\) keyboards per week, the marginal revenue and the marginal costs are

\[MR(x) = 140 - 0.5x \qquad \text{and} \qquad MC(x) = 0.005x^2 - 1.1x + 90\]

EUR per keyboard. The fixed costs are 3,500 EUR per week. At the moment, the company makes and sells 120 keyboards per week. Its capacity (the maximum possible output) is 160 keyboards per week. Round to two decimal places.

  1. Compute the revenue from selling units 60 to 120. Without sales, there is no revenue: determine the revenue function \(R(x)\) and the weekly revenue at 120 keyboards.

(for verification: \(R(120) = 13{,}200\) EUR)

  1. Show that the average value of \(MR\) on \([0; 120]\) equals the price per keyboard at 120 keyboards, \(p(120) = \frac{R(120)}{120}\). Explain why the average value of \(MR\) on \([0; x]\) is always the price \(p(x)\).

  2. The variable costs are the costs that grow with the output, i.e. the costs without the fixed costs. Compute the variable costs per keyboard at 120 keyboards as the average value of \(MC\) on \([0; 120]\). Compute the weekly costs and the weekly profit at 120 keyboards.

(for verification: variable costs of 5,760 EUR per week)

  1. The owner estimates both average values with the value in the middle of the interval, \(MR(60)\) and \(MC(60)\). Compare the estimates with b) and c). Explain why the estimate is exact for the linear \(MR\), and decide whether the estimate for \(MC\) is too high or too low.

  2. Compute the change in the weekly profit when the company raises its output from 120 keyboards to its capacity, using one definite integral. Decide whether the step pays off and give the new weekly profit.

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

\[\int_{60}^{120} (140 - 0.5x) \, dx = \left[140x - 0.25x^2\right]_{60}^{120} = (16800 - 3600) - (8400 - 900) = 13200 - 7500 = 5700\]

Selling keyboards 60 to 120 brings 5,700 EUR per week.

Revenue function: \(R(x) = \int (140 - 0.5x) \, dx = 140x - 0.25x^2 + K\) with \(R(0) = K = 0\), so \(R(x) = 140x - 0.25x^2\) and \(R(120) = 16800 - 3600 = 13200\) EUR per week.

b) Average value of \(MR\):

\[MR_{avg} = \frac{1}{120 - 0}\int_0^{120} (140 - 0.5x) \, dx = \frac{1}{120}\left(R(120) - R(0)\right) = \frac{13200}{120} = 110\]

The price is \(p(120) = \frac{R(120)}{120} = \frac{13200}{120} = 110\) EUR per keyboard ✓.

In general: by the Fundamental Theorem of Calculus and \(R(0) = 0\),

\[\frac{1}{x - 0}\int_0^x MR(s) \, ds = \frac{R(x) - R(0)}{x} = \frac{R(x)}{x} = p(x)\]

The price is the revenue per keyboard, so it is the average of what each keyboard adds to the revenue.

c) The variable costs of 120 keyboards are \(C(120) - C(0) = \int_0^{120} MC(x) \, dx\):

\[\int_0^{120} \left(0.005x^2 - 1.1x + 90\right) dx = \left[\frac{0.005}{3}x^3 - 0.55x^2 + 90x\right]_0^{120} = 2880 - 7920 + 10800 = 5760\]

\[MC_{avg} = \frac{5760}{120} = 48 \text{ EUR per keyboard}\]

Weekly costs: \(C(120) = 3500 + 5760 = 9260\) EUR. Weekly profit: \(P(120) = 13200 - 9260 = 3940\) EUR.

d) \(MR(60) = 140 - 30 = 110\), exactly the average value from b). \(MC(60) = 18 - 66 + 90 = 42\), but the average value from c) is 48.

  • \(MR\) is linear. Its graph is a straight line, so on the left half of \([0; 120]\) it lies above \(MR(60)\) by exactly as much as it lies below on the right half. Both triangles have the same area, and the average is the value in the middle.
  • \(MC\) is not linear, so the value in the middle need not be the average. Here the estimate \(MC(60) = 42\) EUR is too low by \(48 - 42 = 6\) EUR per keyboard.

e) The change in profit is the integral of the marginal profit \(MR - MC\) (net change theorem):

\[\int_{120}^{160} \left(50 + 0.6x - 0.005x^2\right) dx = \left[50x + 0.3x^2 - \frac{0.005}{3}x^3\right]_{120}^{160} = 8853.33 - 7440 = 1413.33\]

The profit rises by about 1,413.33 EUR per week, so the step pays off. The new weekly profit is \(3940 + 1413.33 = 5353.33\) EUR.

Check: \(MR(160) - MC(160) = 60 - 42 = 18 > 0\). Even the last keyboard still adds more to the revenue than to the costs.

Typical mistake: in c), computing \(\frac{C(120)}{120} = \frac{9260}{120} \approx 77.17\) EUR. This is the average cost per keyboard including the fixed costs. The average value of \(MC\) only contains the variable costs, because the fixed costs disappear when you differentiate.

Problem 9: An Open-Air Cinema (xx)

An open-air cinema in a city park shows films on summer evenings. Before the opening, the operator pays 40,000 EUR for the screen, the projector and the chairs. \(t\) weeks after the opening, the cinema makes a profit at a rate of

\[P'(t) = 1.2t - 0.048t^2\]

thousand EUR per week. Round to two decimal places.

  1. Determine the time span after the opening in which the profit rate is positive. Explain what a negative profit rate means for the cinema.

  2. The operator ends the season when the profit rate becomes zero again. Compute the profit of the whole season. Give the net result of the season after the set-up costs.

(for verification: the profit of the season is 125 thousand EUR)

  1. Determine the week with the highest profit rate and prove that it is a maximum. Give the highest profit rate.

  2. Compute the average weekly profit rate over the season. Show that it is exactly \(\frac{2}{3}\) of the highest profit rate.

  3. Because of good weather forecasts, the operator thinks about staying open three weeks longer. Compute the change in profit for these three weeks with the same model and decide whether the operator should stay open. Compute the average weekly profit rate over the longer season as well.

a) Zeros of the profit rate: \(1.2t - 0.048t^2 = 0.048t(25 - t) = 0 \iff t = 0\) or \(t = 25\).

The parabola opens downward (\(-0.048 < 0\)), so it is positive between its zeros. Test value: \(P'(10) = 12 - 4.8 = 7.2 > 0\). The profit rate is positive for \(0 < t < 25\), i.e. during the first 25 weeks.

For \(t > 25\), the profit rate is negative, e.g. \(P'(26) = 31.2 - 32.448 < 0\). Each further week then costs more (staff, film rights) than it brings in, and the total profit falls.

b) The season runs from \(t = 0\) to \(t = 25\). The integral of the profit rate is the profit of the season (net change theorem):

\[\int_0^{25} \left(1.2t - 0.048t^2\right) dt = \left[0.6t^2 - 0.016t^3\right]_0^{25} = 375 - 250 = 125\]

The season brings a profit of 125 thousand EUR. After the set-up costs: \(125 - 40 = 85\), so the net result is 85,000 EUR.

c) \(P''(t) = 1.2 - 0.096t = 0 \iff t = 12.5\).

Proof: \(P'\) is a parabola that opens downward (\(-0.048 < 0\)), so its vertex is a maximum. (Equivalently, \(P'''(t) = -0.096 < 0\).)

\[P'(12.5) = 15 - 7.5 = 7.5\]

The profit rate is highest in the middle of week 13 (\(t = 12.5\)), at 7,500 EUR per week.

d) \[\text{average} = \frac{1}{25 - 0}\int_0^{25} P'(t) \, dt = \frac{125}{25} = 5 \text{ thousand EUR per week}\]

\(\frac{5}{7.5} = \frac{2}{3}\) ✓. On average, the cinema makes 5,000 EUR per week, two thirds of its best week.

e) \[\int_{25}^{28} \left(1.2t - 0.048t^2\right) dt = \left[0.6t^2 - 0.016t^3\right]_{25}^{28} = (470.4 - 351.232) - 125 = 119.168 - 125 = -5.832\]

The three extra weeks lower the profit by about 5,832 EUR, so the operator should not stay open. The profit rate is negative in all three weeks (a).

Average over the longer season: \(\frac{119.168}{28} \approx 4.26\) thousand EUR per week, lower than 5.

Typical mistake: closing the cinema after the week with the highest profit rate (c). The total profit keeps growing as long as the profit rate is positive, so the best time to close is when the rate reaches zero, at \(t = 25\).

Problem 10: A Photo-Editing App (xxx)

A software company sells a photo-editing app as a one-time purchase. The development of the app cost 30,000 EUR. After the launch, the sales fall slowly, while the costs for updates and customer support stay. \(t\) months after the launch, the company makes a profit at a rate of

\[P'(t) = (8 - 0.4t) \cdot e^{-0.04t}\]

thousand EUR per month. The accumulated profit \(P(t)\) (in thousand EUR) contains the development costs, so \(P(0) = -30\). Round to two decimal places.

  1. Find the accumulated profit \(P(t)\) with integration by parts and show your work. Check your result by differentiating.

(for verification: \(P(t) = 10(t + 5) \cdot e^{-0.04t} - 80\))

  1. Compute the profit that the app brings within the first 12 months and the accumulated profit \(P(12)\).

  2. Determine the time at which the profit rate is zero and explain what happens before and after this time. Determine the highest accumulated profit and prove that no other time \(t \geq 0\) gives a higher accumulated profit.

  3. Show that the app earns back its development costs in the fifth month after the launch, i.e. between \(t = 4\) and \(t = 5\).

Hint: \(P\) is continuous. If \(P\) is negative at \(t = 4\) and positive at \(t = 5\), its graph crosses the \(t\)-axis in between.

  1. Compute \(\lim_{t \to \infty} P(t)\) and interpret the result. A manager wants to keep the app on the market for four years. Compute \(P(48)\) and give a recommendation to the company.

a) LIATE: A before E, so \(u = 8 - 0.4t\), \(du = -0.4 \, dt\), \(dv = e^{-0.04t} \, dt\), \(v = \frac{1}{-0.04}e^{-0.04t} = -25e^{-0.04t}\):

\[\int (8 - 0.4t) \cdot e^{-0.04t} \, dt = -25(8 - 0.4t)e^{-0.04t} - \int \left(-25e^{-0.04t}\right) \cdot (-0.4) \, dt = -25(8 - 0.4t)e^{-0.04t} - \int 10e^{-0.04t} \, dt\]

\[= -25(8 - 0.4t)e^{-0.04t} + 250e^{-0.04t} + K = (10t + 50)e^{-0.04t} + K = 10(t + 5)e^{-0.04t} + K\]

Initial condition: \(P(0) = 10 \cdot 5 \cdot e^0 + K = 50 + K = -30\), so \(K = -80\):

\[P(t) = 10(t + 5) \cdot e^{-0.04t} - 80\]

Check (product rule, chain rule): \(P'(t) = 10e^{-0.04t} + 10(t + 5) \cdot (-0.04)e^{-0.04t} = (10 - 0.4t - 2)e^{-0.04t} = (8 - 0.4t)e^{-0.04t}\) ✓

b) Profit within the first 12 months (net change theorem):

\[\int_0^{12} P'(t) \, dt = P(12) - P(0) = \left(170e^{-0.48} - 80\right) - (-30) = 170e^{-0.48} - 50 \approx 55.19\]

The app brings about 55,190 EUR in the first year. The accumulated profit is \(P(12) = 170e^{-0.48} - 80 \approx 25.19\), so after one year the company is about 25,190 EUR in profit, including the development costs.

c) \(e^{-0.04t} > 0\), so \(P'(t) = 0 \iff 8 - 0.4t = 0 \iff t = 20\).

  • For \(0 \leq t < 20\): \(8 - 0.4t > 0\), so \(P'(t) > 0\). Each month adds profit, and \(P\) increases.
  • For \(t > 20\): \(8 - 0.4t < 0\), so \(P'(t) < 0\). The support costs are higher than the sales, and each month lowers the accumulated profit.

Highest accumulated profit: \(P'\) changes its sign from \(+\) to \(-\) at \(t = 20\), so \(P\) has a local maximum there. It is the only critical point for \(t \geq 0\): \(P\) increases on \([0; 20]\) and decreases for all \(t > 20\). So no other time gives more.

\[P(20) = 250e^{-0.8} - 80 \approx 32.33\]

The accumulated profit is highest after 20 months, at about 32,330 EUR.

d) \[P(4) = 90e^{-0.16} - 80 \approx -3.31 < 0, \qquad P(5) = 100e^{-0.2} - 80 \approx 1.87 > 0\]

\(P\) is continuous, so it crosses zero between \(t = 4\) and \(t = 5\). By c), \(P\) increases on \([0; 20]\), so this is the only zero in \([0; 20]\). The app earns back its development costs in the fifth month.

e) \(t \cdot e^{-0.04t} \to 0\) and \(e^{-0.04t} \to 0\) for \(t \to \infty\), so

\[\lim_{t \to \infty} P(t) = 0 + 0 - 80 = -80\]

Interpretation: if the company never takes the app off the market, the losses after month 20 use up the whole profit, and in the long run the company loses 80,000 EUR in total.

\(P(48) = 530e^{-1.92} - 80 \approx -2.30 < 0\). After four years, the losses since month 20 have already used up all the profit, and the app is in the red again.

Recommendation: the accumulated profit peaks after 20 months, so the company should take the app off the market (or stop the updates) then.

Typical mistake: in a), forgetting the initial condition and giving \(P(t) = 10(t + 5)e^{-0.04t}\). Then \(P(0) = 50\), which would mean a profit of 50,000 EUR on the launch day, instead of the development costs of \(-30\).

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: Action Cameras [22 pts. total]

A company launches a new action camera (a small, robust camera for filming sports). The function \(f\) given by

\[f(t) = 16 + 5t \cdot e^{-t/6}\]

describes the monthly sales volume in QU (1 QU = 1,000 cameras). \(t\) is the time in months after the launch. Round to two decimal places.

Part a)

Give the monthly sales volume at the launch. Compute the limit of \(f(t)\) at positive infinity and interpret its significance within this context. [3 pts.]

At the launch: \(f(0) = 16 + 0 = 16\), i.e. 16,000 cameras per month.

Limit: the exponential decay dominates the linear factor, so \(t \cdot e^{-t/6} \to 0\) for \(t \to \infty\):

\[\lim_{t \to \infty} f(t) = 16 + 5 \cdot 0 = 16\]

Interpretation: in the long run, the monthly sales volume settles at 16 QU, i.e. 16,000 cameras per month. The term \(5t \cdot e^{-t/6}\) is the extra demand after the launch, which fades out over time.

Grading: 1 Pt. \(f(0) = 16\), 1 Pt. limit 16 with the reason \(t \cdot e^{-t/6} \to 0\), 1 Pt. interpretation.

Part b)

Determine the month with the highest monthly sales volume and compute this sales volume. Prove that no other time \(t \geq 0\) has a higher monthly sales volume. [6 pts.]

First derivative (product rule on \(5t \cdot e^{-t/6}\), chain rule for \(e^{-t/6}\)):

\[f'(t) = 5e^{-t/6} + 5t \cdot \left(-\frac{1}{6}\right)e^{-t/6} = 5e^{-t/6}\left(1 - \frac{t}{6}\right)\]

Necessary condition: \(e^{-t/6} > 0\), so \(f'(t) = 0 \iff 1 - \frac{t}{6} = 0 \iff t = 6\).

Sufficient condition: for \(0 \leq t < 6\), \(1 - \frac{t}{6} > 0\), so \(f'(t) > 0\). For \(t > 6\), \(1 - \frac{t}{6} < 0\), so \(f'(t) < 0\). \(f'\) changes its sign from \(+\) to \(-\) at \(t = 6\): local maximum.

No other time is higher: \(t = 6\) is the only critical point for \(t \geq 0\). \(f\) increases on \([0; 6]\) and decreases for all \(t > 6\), so the local maximum is the highest value for \(t \geq 0\).

\[f(6) = 16 + 30e^{-1} \approx 27.04\]

The monthly sales volume is highest 6 months after the launch, at about 27.04 QU, i.e. about 27,040 cameras per month.

Grading: 2 Pts. \(f'(t)\) with product and chain rule, 1 Pt. \(t = 6\), 2 Pts. sign change and the argument for all \(t \geq 0\), 1 Pt. \(f(6) \approx 27.04\) QU.

Part c)

Prove that \(F\) given by

\[F(t) = 16t - 30t \cdot e^{-t/6} - 180 \cdot e^{-t/6} + C\]

is an antiderivative of \(f\). [4 pts.]

\(F\) is an antiderivative of \(f\) if \(F'(t) = f(t)\). Differentiate term by term:

  • \((16t)' = 16\) (power rule), and the constant \(C\) gives 0.

  • Product rule on \(-30t \cdot e^{-t/6}\) with \(u = -30t\), \(u' = -30\), \(v = e^{-t/6}\), \(v' = -\frac{1}{6}e^{-t/6}\) (chain rule):

    \[\left(-30t \cdot e^{-t/6}\right)' = -30e^{-t/6} + (-30t) \cdot \left(-\frac{1}{6}\right)e^{-t/6} = -30e^{-t/6} + 5t \cdot e^{-t/6}\]

  • Chain rule: \(\left(-180e^{-t/6}\right)' = -180 \cdot \left(-\frac{1}{6}\right)e^{-t/6} = 30e^{-t/6}\)

Collect:

\[F'(t) = 16 - 30e^{-t/6} + 5t \cdot e^{-t/6} + 30e^{-t/6} = 16 + 5t \cdot e^{-t/6} = f(t) \; \checkmark\]

So \(F\) is an antiderivative of \(f\). No integration by parts is needed for the proof.

Grading: 1 Pt. \(F' = f\) as the goal, 2 Pts. product rule on \(-30t \cdot e^{-t/6}\), 1 Pt. chain rule on \(-180e^{-t/6}\) and collecting to \(f(t)\).

Part d)

Compute the total sales volume within the first 12 months (1 year). [3 pts.]

With \(F\) from c) and \(C = 0\) (it cancels in \(F(12) - F(0)\)):

\[F(12) = 192 - 360e^{-2} - 180e^{-2} = 192 - 540e^{-2}, \qquad F(0) = 0 - 0 - 180 = -180\]

\[\int_0^{12} f(t) \, dt = F(12) - F(0) = 192 - 540e^{-2} + 180 = 372 - 540e^{-2} \approx 298.92\]

Within the first year, the company sells about 298.92 QU, i.e. about 298,920 cameras.

Grading: 1 Pt. \(\int_0^{12} f(t) \, dt = F(12) - F(0)\), 1 Pt. \(F(0) = -180\) (not 0), 1 Pt. result with unit.

Part e)

Compute the average monthly sales volume within the first 18 months. Compare it with the long-term level from a) and the highest monthly sales volume from b). [4 pts.]

Average value over \([0; 18]\):

\[f_{avg} = \frac{1}{18 - 0}\int_0^{18} f(t) \, dt = \frac{1}{18}\left[F(18) - F(0)\right]\]

\[F(18) = 288 - 540e^{-3} - 180e^{-3} = 288 - 720e^{-3}\]

\[f_{avg} = \frac{1}{18}\left(288 - 720e^{-3} + 180\right) = \frac{468 - 720e^{-3}}{18} = 26 - 40e^{-3} \approx 24.01\]

On average, the company sells about 24.01 QU (about 24,010 cameras) per month in the first 18 months.

Comparison: \(16 < 24.01 < 27.04\). The average lies clearly above the long-term level of 16 QU, because the first 18 months contain the whole launch effect. It lies below the highest value, because the sales are lower than 27.04 QU in all other months.

Grading: 1 Pt. formula with \(\frac{1}{18}\), 1 Pt. \(F(18)\), 1 Pt. \(f_{avg} \approx 24.01\) QU, 1 Pt. comparison with both values.

Part f)

The marketing manager claims: “No matter over how many months we compute the average monthly sales volume from the launch on, it always stays above 16 QU.” Decide whether the claim is true and give a reason. [2 pts.]

The claim is true. For every \(t > 0\), \(5t \cdot e^{-t/6} > 0\), so \(f(t) > 16\). Then the area under the graph of \(f\) on \([0; T]\) is larger than the area of the rectangle with height 16:

\[\int_0^T f(t) \, dt > 16T, \qquad \text{so} \qquad f_{avg} = \frac{1}{T}\int_0^T f(t) \, dt > 16 \quad \text{for every } T > 0\]

The average comes closer and closer to 16 when \(T\) grows, but it never falls below it.

Grading: 1 Pt. decision with \(f(t) > 16\) for \(t > 0\), 1 Pt. reason via the integral or the rectangle.