Tasks 06-05 - Economic Applications & Integration by Parts

Section 06: Integral Calculus

Problem 1: Consumer Surplus (x)

For each demand function and equilibrium price, calculate the consumer surplus:

  1. \(D(q) = 100 - 2q\), equilibrium at \(q^* = 20\), \(p^* = 60\)

  2. \(D(q) = 80 - q\), equilibrium at \(q^* = 30\), \(p^* = 50\)

  3. \(D(q) = 200 - 5q\), equilibrium at \(q^* = 24\), \(p^* = 80\)

  4. \(D(q) = 150 - 3q\), equilibrium at \(q^* = 25\), \(p^* = 75\)

  1. \(D(q) = 100 - 2q\), \(q^* = 20\), \(p^* = 60\):

    \[CS = \int_0^{20} [(100 - 2q) - 60] \, dq = \int_0^{20} (40 - 2q) \, dq\] \[= \left[40q - q^2\right]_0^{20} = 800 - 400 = 400\]

  2. \(D(q) = 80 - q\), \(q^* = 30\), \(p^* = 50\):

    \[CS = \int_0^{30} [(80 - q) - 50] \, dq = \int_0^{30} (30 - q) \, dq\] \[= \left[30q - \frac{q^2}{2}\right]_0^{30} = 900 - 450 = 450\]

  3. \(D(q) = 200 - 5q\), \(q^* = 24\), \(p^* = 80\):

    \[CS = \int_0^{24} [(200 - 5q) - 80] \, dq = \int_0^{24} (120 - 5q) \, dq\] \[= \left[120q - \frac{5q^2}{2}\right]_0^{24} = 2880 - 1440 = 1440\]

  4. \(D(q) = 150 - 3q\), \(q^* = 25\), \(p^* = 75\):

    \[CS = \int_0^{25} [(150 - 3q) - 75] \, dq = \int_0^{25} (75 - 3q) \, dq\] \[= \left[75q - \frac{3q^2}{2}\right]_0^{25} = 1875 - 937.5 = 937.5\]

Problem 2: Producer Surplus (x)

For each supply function and equilibrium price, calculate the producer surplus:

  1. \(S(q) = 10 + q\), equilibrium at \(q^* = 20\), \(p^* = 30\)

  2. \(S(q) = 5 + 2q\), equilibrium at \(q^* = 15\), \(p^* = 35\)

  3. \(S(q) = 20 + 0.5q\), equilibrium at \(q^* = 40\), \(p^* = 40\)

  4. \(S(q) = 15 + 3q\), equilibrium at \(q^* = 10\), \(p^* = 45\)

  1. \(S(q) = 10 + q\), \(q^* = 20\), \(p^* = 30\):

    \[PS = \int_0^{20} [30 - (10 + q)] \, dq = \int_0^{20} (20 - q) \, dq\] \[= \left[20q - \frac{q^2}{2}\right]_0^{20} = 400 - 200 = 200\]

  2. \(S(q) = 5 + 2q\), \(q^* = 15\), \(p^* = 35\):

    \[PS = \int_0^{15} [35 - (5 + 2q)] \, dq = \int_0^{15} (30 - 2q) \, dq\] \[= \left[30q - q^2\right]_0^{15} = 450 - 225 = 225\]

  3. \(S(q) = 20 + 0.5q\), \(q^* = 40\), \(p^* = 40\):

    \[PS = \int_0^{40} [40 - (20 + 0.5q)] \, dq = \int_0^{40} (20 - 0.5q) \, dq\] \[= \left[20q - 0.25q^2\right]_0^{40} = 800 - 400 = 400\]

  4. \(S(q) = 15 + 3q\), \(q^* = 10\), \(p^* = 45\):

    \[PS = \int_0^{10} [45 - (15 + 3q)] \, dq = \int_0^{10} (30 - 3q) \, dq\] \[= \left[30q - \frac{3q^2}{2}\right]_0^{10} = 300 - 150 = 150\]

Problem 3: Complete Market Analysis (xx)

For each market, find equilibrium, then calculate CS, PS, and total surplus:

  1. \(D(q) = 120 - 4q\) and \(S(q) = 20 + 2q\)

  2. \(D(q) = 90 - 3q\) and \(S(q) = 30 + q\)

  3. \(D(q) = 200 - 2q\) and \(S(q) = 40 + 2q\)

  1. \(D(q) = 120 - 4q\) and \(S(q) = 20 + 2q\):

    Equilibrium: \(120 - 4q = 20 + 2q \Rightarrow 100 = 6q \Rightarrow q^* = \frac{50}{3}\) \(p^* = 20 + 2 \cdot \frac{50}{3} = 20 + \frac{100}{3} = \frac{160}{3}\)

    \[CS = \int_0^{50/3} \left[(120 - 4q) - \frac{160}{3}\right] dq = \int_0^{50/3} \left(\frac{200}{3} - 4q\right) dq\] \[= \left[\frac{200q}{3} - 2q^2\right]_0^{50/3} = \frac{200 \cdot 50}{9} - 2 \cdot \frac{2500}{9} = \frac{10000}{9} - \frac{5000}{9} = \frac{5000}{9} \approx 555.56\]

    \[PS = \int_0^{50/3} \left[\frac{160}{3} - (20 + 2q)\right] dq = \int_0^{50/3} \left(\frac{100}{3} - 2q\right) dq\] \[= \left[\frac{100q}{3} - q^2\right]_0^{50/3} = \frac{100 \cdot 50}{9} - \frac{2500}{9} = \frac{5000}{9} - \frac{2500}{9} = \frac{2500}{9} \approx 277.78\]

    Total surplus \(= \frac{5000}{9} + \frac{2500}{9} = \frac{7500}{9} = 833.33\)

  2. \(D(q) = 90 - 3q\) and \(S(q) = 30 + q\):

    Equilibrium: \(90 - 3q = 30 + q \Rightarrow 60 = 4q \Rightarrow q^* = 15\) \(p^* = 30 + 15 = 45\)

    \[CS = \int_0^{15} [(90 - 3q) - 45] \, dq = \int_0^{15} (45 - 3q) \, dq\] \[= \left[45q - \frac{3q^2}{2}\right]_0^{15} = 675 - 337.5 = 337.5\]

    \[PS = \int_0^{15} [45 - (30 + q)] \, dq = \int_0^{15} (15 - q) \, dq\] \[= \left[15q - \frac{q^2}{2}\right]_0^{15} = 225 - 112.5 = 112.5\]

    Total surplus \(= 337.5 + 112.5 = 450\)

  3. \(D(q) = 200 - 2q\) and \(S(q) = 40 + 2q\):

    Equilibrium: \(200 - 2q = 40 + 2q \Rightarrow 160 = 4q \Rightarrow q^* = 40\) \(p^* = 200 - 80 = 120\)

    \[CS = \int_0^{40} [(200 - 2q) - 120] \, dq = \int_0^{40} (80 - 2q) \, dq\] \[= \left[80q - q^2\right]_0^{40} = 3200 - 1600 = 1600\]

    \[PS = \int_0^{40} [120 - (40 + 2q)] \, dq = \int_0^{40} (80 - 2q) \, dq\] \[= \left[80q - q^2\right]_0^{40} = 3200 - 1600 = 1600\]

    Total surplus \(= 1600 + 1600 = 3200\)

Problem 4: Average Value (xx)

Find the average value of each function on the given interval:

  1. \(f(x) = 3x^2\) on \([0, 2]\)

  2. \(f(x) = x^3 - x\) on \([0, 2]\)

  3. \(f(x) = e^x\) on \([0, 3]\)

  4. \(f(x) = \frac{1}{x}\) on \([1, e]\)

  5. \(f(x) = 4 - x^2\) on \([-2, 2]\)

  1. \(f(x) = 3x^2\) on \([0, 2]\):

    \[f_{avg} = \frac{1}{2-0} \int_0^2 3x^2 \, dx = \frac{1}{2}[x^3]_0^2 = \frac{1}{2}(8) = 4\]

  2. \(f(x) = x^3 - x\) on \([0, 2]\):

    \[f_{avg} = \frac{1}{2} \int_0^2 (x^3 - x) \, dx = \frac{1}{2}\left[\frac{x^4}{4} - \frac{x^2}{2}\right]_0^2\] \[= \frac{1}{2}(4 - 2) = 1\]

  3. \(f(x) = e^x\) on \([0, 3]\):

    \[f_{avg} = \frac{1}{3} \int_0^3 e^x \, dx = \frac{1}{3}[e^x]_0^3 = \frac{1}{3}(e^3 - 1) \approx 6.36\]

  4. \(f(x) = \frac{1}{x}\) on \([1, e]\):

    \[f_{avg} = \frac{1}{e-1} \int_1^e \frac{1}{x} \, dx = \frac{1}{e-1}[\ln|x|]_1^e = \frac{1}{e-1}(1 - 0) = \frac{1}{e-1} \approx 0.582\]

  5. \(f(x) = 4 - x^2\) on \([-2, 2]\):

    \[f_{avg} = \frac{1}{4} \int_{-2}^2 (4 - x^2) \, dx = \frac{1}{4}\left[4x - \frac{x^3}{3}\right]_{-2}^2\] \[= \frac{1}{4}\left[\left(8 - \frac{8}{3}\right) - \left(-8 + \frac{8}{3}\right)\right] = \frac{1}{4}\left(\frac{16}{3} + \frac{16}{3}\right) = \frac{1}{4} \cdot \frac{32}{3} = \frac{8}{3}\]

Problem 5: Revenue and Cost Accumulation (xx)

  1. A company’s marginal revenue is \(MR(x) = 150 - 3x\) EUR/unit. Find the additional revenue from increasing production from 20 to 40 units.

  2. Marginal cost is \(MC(x) = 20 + 0.5x\) EUR/unit. Find the total cost of producing the first 50 units (assuming no fixed costs).

  3. If \(MR(x) = 100 - 2x\) and \(MC(x) = 10 + x\), find the profit-maximizing quantity and the total profit earned up to that quantity (starting from 0).

  1. Additional revenue from 20 to 40 units:

    \[\int_{20}^{40} (150 - 3x) \, dx = \left[150x - \frac{3x^2}{2}\right]_{20}^{40}\] \[= (6000 - 2400) - (3000 - 600) = 3600 - 2400 = 1200 \text{ EUR}\]

  2. Total cost for first 50 units:

    \[\int_0^{50} (20 + 0.5x) \, dx = \left[20x + 0.25x^2\right]_0^{50}\] \[= 1000 + 625 = 1625 \text{ EUR}\]

  3. Profit maximization:

    \(MR = MC\): \(100 - 2x = 10 + x \Rightarrow 90 = 3x \Rightarrow x = 30\)

    Total profit from 0 to 30: \[\int_0^{30} [MR(x) - MC(x)] \, dx = \int_0^{30} [(100 - 2x) - (10 + x)] \, dx\] \[= \int_0^{30} (90 - 3x) \, dx = \left[90x - \frac{3x^2}{2}\right]_0^{30}\] \[= 2700 - 1350 = 1350 \text{ EUR}\]

Problem 6: Business Applications (xx)

  1. A factory’s production rate is \(P(t) = 120 - 4t\) units per hour, where \(t\) is hours into the shift. Find the average production rate over an 8-hour shift.

  2. Daily sales revenue follows \(R(t) = 500 + 100\sin(\frac{\pi t}{12})\) EUR, where \(t\) is hours after midnight. Find the total revenue from 6 AM to 6 PM (12 hours).

  3. A machine depreciates at rate \(V'(t) = -5000e^{-0.2t}\) EUR per year. Find the total depreciation over the first 5 years.

  1. Average production over 8 hours:

    \[P_{avg} = \frac{1}{8} \int_0^8 (120 - 4t) \, dt = \frac{1}{8}\left[120t - 2t^2\right]_0^8\] \[= \frac{1}{8}(960 - 128) = \frac{832}{8} = 104 \text{ units/hour}\]

  2. Total revenue from \(t = 6\) to \(t = 18\):

    \[\int_6^{18} \left(500 + 100\sin\left(\frac{\pi t}{12}\right)\right) dt\] \[= \left[500t - 100 \cdot \frac{12}{\pi}\cos\left(\frac{\pi t}{12}\right)\right]_6^{18}\] \[= \left[500t - \frac{1200}{\pi}\cos\left(\frac{\pi t}{12}\right)\right]_6^{18}\]

    At \(t = 18\): \(500(18) - \frac{1200}{\pi}\cos(\frac{3\pi}{2}) = 9000 - 0 = 9000\) At \(t = 6\): \(500(6) - \frac{1200}{\pi}\cos(\frac{\pi}{2}) = 3000 - 0 = 3000\)

    Total revenue \(= 9000 - 3000 = 6000\) EUR

  3. Total depreciation over 5 years:

    \[\int_0^5 (-5000e^{-0.2t}) \, dt = -5000 \cdot \frac{1}{-0.2}[e^{-0.2t}]_0^5\] \[= 25000[e^{-0.2t}]_0^5 = 25000(e^{-1} - 1) = 25000(\frac{1}{e} - 1)\] \[\approx 25000(-0.632) = -15803 \text{ EUR}\]

    Total depreciation is approximately 15,803 EUR (value lost).

Problem 7: Profit Rate Analysis (xxx)

A startup’s monthly profit rate is \(P'(t) = 12t - t^2\) thousand EUR, where \(t\) is months after launch.

  1. During which months is the company profitable (positive profit rate)?

  2. Find the total profit accumulated from launch until the profit rate first becomes zero.

  3. Find the month when the instantaneous profit rate is highest, and what is that maximum rate?

  4. Calculate the average monthly profit rate over the first 12 months.

  1. When is \(P'(t) > 0\)?

    \(12t - t^2 > 0\) \(t(12 - t) > 0\)

    This is positive when \(0 < t < 12\).

    The company is profitable from month 0 to month 12.

  2. Total profit from \(t = 0\) to \(t = 12\):

    \[\int_0^{12} (12t - t^2) \, dt = \left[6t^2 - \frac{t^3}{3}\right]_0^{12}\] \[= 6(144) - \frac{1728}{3} = 864 - 576 = 288 \text{ thousand EUR}\]

  3. Maximum profit rate:

    \(P''(t) = 12 - 2t = 0 \Rightarrow t = 6\)

    Maximum rate: \(P'(6) = 12(6) - 36 = 72 - 36 = 36\) thousand EUR/month

    Maximum profit rate of 36 thousand EUR/month occurs at month 6.

  4. Average monthly profit rate over first 12 months:

    \[P'_{avg} = \frac{1}{12} \int_0^{12} (12t - t^2) \, dt = \frac{288}{12} = 24 \text{ thousand EUR/month}\]

Problem 8: Basic Integration by Parts (x)

Use integration by parts to evaluate the following integrals. Remember to verify your answers by differentiation.

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

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

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

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

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

  6. \(\int 5x \cdot e^{2x} \, dx\)

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

    Let \(u = x\), \(dv = e^x \, dx\), so \(du = dx\), \(v = e^x\)

    \[= x \cdot e^x - \int e^x \, dx = x \cdot e^x - e^x + C = e^x(x - 1) + C\]

    Verification: \(\frac{d}{dx}[e^x(x-1)] = e^x(x-1) + e^x = xe^x\) ✓

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

    Let \(u = 2x\), \(dv = e^x \, dx\), so \(du = 2 \, dx\), \(v = e^x\)

    \[= 2x \cdot e^x - \int 2e^x \, dx = 2xe^x - 2e^x + C = 2e^x(x - 1) + C\]

    Verification: \(\frac{d}{dx}[2e^x(x-1)] = 2e^x(x-1) + 2e^x = 2xe^x\) ✓

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

    Let \(u = x\), \(dv = e^{-x} \, dx\), so \(du = dx\), \(v = -e^{-x}\)

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

    Verification: \(\frac{d}{dx}[-e^{-x}(x+1)] = e^{-x}(x+1) - e^{-x} = xe^{-x}\) ✓

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

    Let \(u = x + 1\), \(dv = e^x \, dx\), so \(du = dx\), \(v = e^x\)

    \[= (x+1)e^x - \int e^x \, dx = (x+1)e^x - e^x + C = xe^x + C\]

    Verification: \(\frac{d}{dx}[xe^x] = e^x + xe^x = (x+1)e^x\) ✓

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

    Let \(u = x - 2\), \(dv = e^x \, dx\), so \(du = dx\), \(v = e^x\)

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

    Verification: \(\frac{d}{dx}[e^x(x-3)] = e^x(x-3) + e^x = (x-2)e^x\) ✓

  6. \(\int 5x \cdot e^{2x} \, dx\)

    Let \(u = 5x\), \(dv = e^{2x} \, dx\), so \(du = 5 \, dx\), \(v = \frac{1}{2}e^{2x}\)

    \[= 5x \cdot \frac{e^{2x}}{2} - \int \frac{5}{2}e^{2x} \, dx = \frac{5xe^{2x}}{2} - \frac{5e^{2x}}{4} + C\] \[= \frac{5e^{2x}}{4}(2x - 1) + C\]

    Verification: \(\frac{d}{dx}\left[\frac{5e^{2x}}{4}(2x-1)\right] = \frac{5 \cdot 2e^{2x}}{4}(2x-1) + \frac{5e^{2x}}{4} \cdot 2 = \frac{5e^{2x}}{2}(2x-1+1) = 5xe^{2x}\) ✓

Problem 9: Integration with Logarithms (x)

Use the LIATE rule to determine which function should be \(u\), then integrate.

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

  2. \(\int x^2 \cdot \ln(x) \, dx\)

  3. \(\int \ln(x) \, dx\) (Hint: write as \(\int 1 \cdot \ln(x) \, dx\))

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

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

    By LIATE: \(u = \ln(x)\) (L before A), \(dv = x \, dx\) So \(du = \frac{1}{x} \, dx\), \(v = \frac{x^2}{2}\)

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

    Verification: \(\frac{d}{dx}\left[\frac{x^2}{4}(2\ln x - 1)\right] = \frac{2x}{4}(2\ln x - 1) + \frac{x^2}{4} \cdot \frac{2}{x} = \frac{x(2\ln x - 1)}{2} + \frac{x}{2} = x\ln x\) ✓

  2. \(\int x^2 \cdot \ln(x) \, dx\)

    By LIATE: \(u = \ln(x)\), \(dv = x^2 \, dx\) So \(du = \frac{1}{x} \, dx\), \(v = \frac{x^3}{3}\)

    \[= \ln(x) \cdot \frac{x^3}{3} - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac{x^3 \ln(x)}{3} - \frac{1}{3}\int x^2 \, dx\] \[= \frac{x^3 \ln(x)}{3} - \frac{x^3}{9} + C = \frac{x^3}{9}(3\ln(x) - 1) + C\]

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

    Let \(u = \ln(x)\), \(dv = dx\), so \(du = \frac{1}{x} \, dx\), \(v = x\)

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

    Verification: \(\frac{d}{dx}[x(\ln x - 1)] = \ln x - 1 + x \cdot \frac{1}{x} = \ln x\) ✓

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

    By LIATE: \(u = \ln(x)\), \(dv = x^3 \, dx\) So \(du = \frac{1}{x} \, dx\), \(v = \frac{x^4}{4}\)

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

Problem 10: Repeated Integration by Parts (xx)

These integrals require applying integration by parts twice.

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

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

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

  4. \(\int (x^2 + x) \cdot e^x \, dx\)

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

    First application: \(u = x^2\), \(dv = e^x dx\), so \(du = 2x \, dx\), \(v = e^x\)

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

    Second application: For \(\int 2x \cdot e^x dx\), let \(u = 2x\), \(dv = e^x dx\)

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

    Combining: \[= x^2 e^x - (2xe^x - 2e^x) + C = x^2 e^x - 2xe^x + 2e^x + C\] \[= e^x(x^2 - 2x + 2) + C\]

    Verification: \(\frac{d}{dx}[e^x(x^2-2x+2)] = e^x(x^2-2x+2) + e^x(2x-2) = e^x(x^2-2x+2+2x-2) = x^2 e^x\) ✓

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

    First application: \(u = x^2\), \(dv = e^{-x} dx\), so \(du = 2x \, dx\), \(v = -e^{-x}\)

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

    Second application: For \(\int x \cdot e^{-x} dx\), let \(u = x\), \(dv = e^{-x} dx\), \(v = -e^{-x}\)

    \[\int x \cdot e^{-x} \, dx = -xe^{-x} + \int e^{-x} dx = -xe^{-x} - e^{-x}\]

    Combining: \[= -x^2 e^{-x} + 2(-xe^{-x} - e^{-x}) + C = -x^2 e^{-x} - 2xe^{-x} - 2e^{-x} + C\] \[= -e^{-x}(x^2 + 2x + 2) + C\]

    Verification: \(\frac{d}{dx}[-e^{-x}(x^2+2x+2)] = e^{-x}(x^2+2x+2) - e^{-x}(2x+2) = e^{-x}(x^2) = x^2 e^{-x}\) ✓

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

    First application: \(u = x^2\), \(dv = e^{2x} dx\), so \(du = 2x \, dx\), \(v = \frac{e^{2x}}{2}\)

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

    Second application: For \(\int x \cdot e^{2x} dx\), let \(u = x\), \(dv = e^{2x} dx\), \(v = \frac{e^{2x}}{2}\)

    \[\int x \cdot e^{2x} \, dx = \frac{xe^{2x}}{2} - \frac{e^{2x}}{4}\]

    Combining: \[= \frac{x^2 e^{2x}}{2} - \frac{xe^{2x}}{2} + \frac{e^{2x}}{4} + C = \frac{e^{2x}}{4}(2x^2 - 2x + 1) + C\]

  4. \(\int (x^2 + x) \cdot e^x \, dx\)

    Split the integral: \[= \int x^2 e^x \, dx + \int x e^x \, dx\]

    From part (a): \(\int x^2 e^x dx = e^x(x^2 - 2x + 2)\)

    From Problem 1(a): \(\int x e^x dx = e^x(x - 1)\)

    Combining: \[= e^x(x^2 - 2x + 2) + e^x(x - 1) + C = e^x(x^2 - 2x + 2 + x - 1) + C\] \[= e^x(x^2 - x + 1) + C\]

Problem 11: Definite Integrals by Parts (xx)

Evaluate the following definite integrals.

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

  2. \(\int_0^2 x \cdot e^{-x} \, dx\)

  3. \(\int_1^e x \cdot \ln(x) \, dx\)

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

  5. \(\int_0^2 (x + 1) \cdot e^x \, dx\)

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

    Using \(\int x e^x dx = e^x(x-1)\):

    \[= [e^x(x-1)]_0^1 = e^1(1-1) - e^0(0-1) = 0 - (-1) = 1\]

    Answer: \(1\)

  2. \(\int_0^2 x \cdot e^{-x} \, dx\)

    Using \(\int x e^{-x} dx = -e^{-x}(x+1)\):

    \[= [-e^{-x}(x+1)]_0^2 = -e^{-2}(3) - (-e^0)(1) = -3e^{-2} + 1\] \[= 1 - \frac{3}{e^2} \approx 1 - 0.406 = 0.594\]

    Answer: \(1 - 3e^{-2} \approx 0.594\)

  3. \(\int_1^e x \cdot \ln(x) \, dx\)

    Using \(\int x \ln(x) dx = \frac{x^2}{4}(2\ln x - 1)\):

    \[= \left[\frac{x^2}{4}(2\ln x - 1)\right]_1^e = \frac{e^2}{4}(2 \cdot 1 - 1) - \frac{1}{4}(2 \cdot 0 - 1)\] \[= \frac{e^2}{4}(1) - \frac{1}{4}(-1) = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4}\]

    Answer: \(\frac{e^2 + 1}{4} \approx 2.097\)

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

    Using \(\int x^2 e^x dx = e^x(x^2 - 2x + 2)\):

    \[= [e^x(x^2-2x+2)]_0^1 = e^1(1-2+2) - e^0(0-0+2) = e(1) - 2 = e - 2\]

    Answer: \(e - 2 \approx 0.718\)

  5. \(\int_0^2 (x + 1) \cdot e^x \, dx\)

    Using \(\int (x+1) e^x dx = xe^x\) (from Problem 1d):

    \[= [xe^x]_0^2 = 2e^2 - 0 = 2e^2\]

    Answer: \(2e^2 \approx 14.78\)

Problem 12: Exam-Style Problem - 2025 Format (xx)

Consider the function \(f(x) = (x + 1)e^x - 1\).

  1. Find \(\int f(x) \, dx\)

  2. Evaluate \(\int_0^2 f(x) \, dx\)

  3. Find where \(f(x) = 0\) and interpret what this means for the integral.

  4. Sketch the graph of \(f(x)\) for \(-3 \leq x \leq 2\) and shade the region whose area is computed in part (b).

  1. \(\int f(x) \, dx = \int [(x + 1)e^x - 1] \, dx\)

    \[= \int (x+1)e^x \, dx - \int 1 \, dx\]

    For \(\int (x+1)e^x dx\): Let \(u = x+1\), \(dv = e^x dx\), so \(du = dx\), \(v = e^x\)

    \[\int (x+1)e^x dx = (x+1)e^x - e^x = xe^x\]

    Therefore: \[\int f(x) \, dx = xe^x - x + C\]

    Verification: \(\frac{d}{dx}[xe^x - x] = e^x + xe^x - 1 = (x+1)e^x - 1\) ✓

  2. \(\int_0^2 f(x) \, dx = [xe^x - x]_0^2 = (2e^2 - 2) - (0 - 0) = 2e^2 - 2\)

    Answer: \(2e^2 - 2 \approx 12.78\)

  3. Solve \((x + 1)e^x - 1 = 0\):

    \((x + 1)e^x = 1\)

    This requires numerical methods. Testing values:

    • At \(x = 0\): \((0+1)e^0 - 1 = 1 - 1 = 0\) ✓

    So \(f(0) = 0\), meaning \(x = 0\) is a zero of \(f\).

    Interpretation: The integral from 0 to 2 starts exactly at a zero of the function, so we’re computing the area from where the function crosses the x-axis.

  4. Graph characteristics:

    • \(f(0) = 0\)
    • \(f(-3) = (-2)e^{-3} - 1 \approx -1.10\)
    • \(f(2) = 3e^2 - 1 \approx 21.17\)
    • \(f'(x) = e^x + (x+1)e^x = (x+2)e^x\)
    • \(f'(x) = 0\) when \(x = -2\)

The graph shows f(x) = (x+1)e^x - 1, which crosses zero at x = 0 (marked with a point) and has a local minimum near x = -2. The region between x = 0 and x = 2 is shaded under the curve, representing the integral area of 2e^2 - 2. The function grows rapidly for x > 0, reaching approximately 21 at x = 2. A dashed horizontal line marks y = 0.

Problem 13: Exam-Style Problem - 2023 Format (xxx)

Evaluate \(\int_0^1 x^2 \cdot e^{-x} \, dx\).

Show all steps clearly, including:

  1. Setting up integration by parts (identify \(u\) and \(dv\))
  2. The first application of integration by parts
  3. The second application of integration by parts
  4. Combining results and evaluating the definite integral
  5. Express your answer in exact form and as a decimal approximation

Setup: We need to evaluate \(\int_0^1 x^2 \cdot e^{-x} \, dx\)

This requires integration by parts twice since we have \(x^2\) multiplied by \(e^{-x}\).

a) First setup:

  • \(u = x^2\) (algebraic - decreases degree when differentiated)
  • \(dv = e^{-x} \, dx\)
  • \(du = 2x \, dx\)
  • \(v = -e^{-x}\)

b) First application:

\[\int x^2 \cdot e^{-x} \, dx = x^2 \cdot (-e^{-x}) - \int (-e^{-x}) \cdot 2x \, dx\]

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

c) Second application: For \(\int x \cdot e^{-x} \, dx\):

  • \(u = x\), \(dv = e^{-x} \, dx\)
  • \(du = dx\), \(v = -e^{-x}\)

\[\int x \cdot e^{-x} \, dx = x \cdot (-e^{-x}) - \int (-e^{-x}) \, dx\]

\[= -xe^{-x} + \int e^{-x} \, dx = -xe^{-x} - e^{-x}\]

\[= -e^{-x}(x + 1)\]

d) Combining and evaluating:

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

\[= -x^2 e^{-x} - 2xe^{-x} - 2e^{-x}\]

\[= -e^{-x}(x^2 + 2x + 2)\]

Now evaluate from 0 to 1:

\[\int_0^1 x^2 \cdot e^{-x} \, dx = [-e^{-x}(x^2 + 2x + 2)]_0^1\]

\[= [-e^{-1}(1 + 2 + 2)] - [-e^0(0 + 0 + 2)]\]

\[= -\frac{5}{e} - (-2)\]

\[= 2 - \frac{5}{e}\]

e) Final answers:

Exact form: \(2 - \frac{5}{e}\) or equivalently \(\frac{2e - 5}{e}\)

Decimal approximation: \(2 - \frac{5}{2.718...} \approx 2 - 1.839 \approx 0.161\)

Verification: The antiderivative is \(F(x) = -e^{-x}(x^2 + 2x + 2)\)

Check: \(F'(x) = e^{-x}(x^2 + 2x + 2) + (-e^{-x})(2x + 2) = e^{-x}(x^2 + 2x + 2 - 2x - 2) = x^2 e^{-x}\) ✓

Problem 14: Business Application - Accumulated Profit (xx)

A company’s marginal profit function (rate of profit in thousands of euros per month) is given by:

\[MP(t) = P'(t) = (20 - 2t) \cdot e^{-0.1t}\]

where \(t\) is months since product launch.

  1. Find the accumulated profit function \(P(t)\), given that at launch (\(t = 0\)), the initial investment creates a loss of €50,000, so \(P(0) = -50\).

  2. Calculate the total profit from month 0 to month 12.

  3. At what time does the marginal profit equal zero? What does this mean for the business?

  4. What is the maximum accumulated profit, and when does it occur?

  1. Finding \(P(t)\):

\[P(t) = \int (20 - 2t) e^{-0.1t} \, dt\]

Split into two parts: \[= \int 20e^{-0.1t} dt - \int 2t \cdot e^{-0.1t} dt\]

First integral: \[\int 20e^{-0.1t} dt = 20 \cdot \frac{e^{-0.1t}}{-0.1} = -200e^{-0.1t}\]

Second integral: (using integration by parts with \(u = 2t\), \(dv = e^{-0.1t}dt\))

\(v = -10e^{-0.1t}\), \(du = 2dt\)

\[\int 2t \cdot e^{-0.1t} dt = 2t(-10e^{-0.1t}) - \int (-10e^{-0.1t}) \cdot 2 \, dt\] \[= -20te^{-0.1t} + 20 \int e^{-0.1t} dt = -20te^{-0.1t} - 200e^{-0.1t}\]

Combining: \[P(t) = -200e^{-0.1t} - (-20te^{-0.1t} - 200e^{-0.1t}) + C\] \[= -200e^{-0.1t} + 20te^{-0.1t} + 200e^{-0.1t} + C = 20te^{-0.1t} + C\]

Using initial condition \(P(0) = -50\): \[P(0) = 0 + C = -50 \implies C = -50\]

Answer: \(P(t) = 20te^{-0.1t} - 50\) (thousands of euros)

  1. Total profit from month 0 to 12:

\[\int_0^{12} MP(t) \, dt = P(12) - P(0)\] \[= [20 \cdot 12 \cdot e^{-1.2} - 50] - [-50] = 240e^{-1.2} = 240 \cdot 0.301 \approx 72.3\]

Or using the accumulated profit directly: \[P(12) = 20(12)e^{-1.2} - 50 \approx 72.3 - 50 = 22.3\]

Answer: Total profit gain of approximately €72,300 over 12 months, with accumulated profit of €22,300 at month 12.

  1. When \(MP(t) = 0\):

\[(20 - 2t)e^{-0.1t} = 0\]

Since \(e^{-0.1t} > 0\) for all \(t\), we need: \[20 - 2t = 0 \implies t = 10 \text{ months}\]

Interpretation: At \(t = 10\) months, marginal profit becomes zero. Before this, each additional month adds positive profit (though decreasing). After this, each additional month would add negative profit (losses).

  1. Maximum accumulated profit:

The accumulated profit is maximized when \(P'(t) = MP(t) = 0\), which occurs at \(t = 10\).

\[P(10) = 20(10)e^{-1} - 50 = 200e^{-1} - 50 \approx 73.6 - 50 = 23.6\]

Answer: Maximum accumulated profit of approximately €23,600 occurs at month 10.

After month 10, the marginal profit becomes negative (the company starts losing money on each additional month of operation), so the accumulated profit decreases.