Session 09-03: Tasks

Integral Calculus - Exam Review

Integral Calculus - Exam Review

Problem 1: Antiderivatives (x)

Find the antiderivative:

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

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

  3. \(\int \frac{7}{x} \, dx\)

  4. \(\int 6\sqrt{x} \, dx\)

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

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

a) We integrate term by term using the power rule \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\):

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

\[= x^5 - x^3 + 2x + C\]

b) Using \(\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C\) with \(a = 4\):

\[\int e^{4x} \, dx = \frac{1}{4} e^{4x} + C\]

c) Using \(\int \frac{1}{x} \, dx = \ln|x| + C\):

\[\int \frac{7}{x} \, dx = 7 \ln|x| + C\]

d) First rewrite \(6\sqrt{x} = 6x^{1/2}\), then apply the power rule:

\[\int 6x^{1/2} \, dx = 6 \cdot \frac{x^{3/2}}{3/2} + C = 6 \cdot \frac{2}{3} x^{3/2} + C = 4x^{3/2} + C\]

e) First rewrite \(\frac{3}{x^4} = 3x^{-4}\), then apply the power rule:

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

f) Integrate term by term:

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

Common error: Forgetting the factor \(\frac{1}{a}\) when integrating \(e^{ax}\). Always remember: \(\int e^{ax} \, dx = \frac{1}{a}e^{ax} + C\), not \(e^{ax} + C\).

Problem 2: Definite Integrals (x)

Evaluate:

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

  2. \(\int_0^2 (x^2 - 2x) \, dx\)

  3. \(\int_1^e \frac{2}{x} \, dx\)

  4. \(\int_0^{\ln 2} e^{3x} \, dx\)

a) Find the antiderivative, then evaluate at the bounds:

\[\int_1^3 (2x+1) \, dx = \left[ x^2 + x \right]_1^3\]

\[= (3^2 + 3) - (1^2 + 1) = 12 - 2 = 10\]

b) Find the antiderivative, then evaluate:

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

\[= \left(\frac{8}{3} - 4\right) - (0 - 0) = \frac{8}{3} - \frac{12}{3} = -\frac{4}{3}\]

Note: The negative value indicates that the function lies below the \(x\)-axis on part of the interval.

c) Using \(\int \frac{1}{x} \, dx = \ln|x|\):

\[\int_1^e \frac{2}{x} \, dx = \left[ 2\ln|x| \right]_1^e = 2\ln(e) - 2\ln(1) = 2 \cdot 1 - 2 \cdot 0 = 2\]

d) Using \(\int e^{3x} \, dx = \frac{1}{3}e^{3x}\):

\[\int_0^{\ln 2} e^{3x} \, dx = \left[ \frac{1}{3}e^{3x} \right]_0^{\ln 2}\]

\[= \frac{1}{3}e^{3\ln 2} - \frac{1}{3}e^0 = \frac{1}{3} \cdot e^{\ln 8} - \frac{1}{3} = \frac{1}{3} \cdot 8 - \frac{1}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}\]

Key step: \(e^{3\ln 2} = e^{\ln(2^3)} = e^{\ln 8} = 8\). This uses the exponent rule \(a \cdot \ln b = \ln(b^a)\) (Section 01 connection).

Problem 3: Basic Area Under a Curve (x)

  1. Find the area under \(f(x) = x^2 + 1\) from \(x=0\) to \(x=3\).

  2. Find the area under \(f(x) = e^x\) from \(x=0\) to \(x=2\).

a) Since \(f(x) = x^2 + 1 > 0\) for all \(x\), the area equals the definite integral:

\[A = \int_0^3 (x^2 + 1) \, dx = \left[ \frac{x^3}{3} + x \right]_0^3\]

\[= \left(\frac{27}{3} + 3\right) - (0 + 0) = 9 + 3 = 12\]

The area under the curve is \(12\) square units.

b) Since \(e^x > 0\) for all \(x\), the area equals:

\[A = \int_0^2 e^x \, dx = \left[ e^x \right]_0^2 = e^2 - e^0 = e^2 - 1 \approx 6.389\]

The area under the curve is \(e^2 - 1 \approx 6.389\) square units.

Problem 4: Area Between Two Curves (xx)

Find the area enclosed between:

  1. \(f(x) = 6-x^2\) and \(g(x) = x\) (find intersections first!)

  2. \(f(x) = x^2\) and \(g(x) = \sqrt{x}\) on \([0,1]\)

a) Step 1: Find intersections. Set \(6 - x^2 = x\):

\[x^2 + x - 6 = 0\]

Using the quadratic formula or factoring:

\[(x+3)(x-2) = 0 \implies x = -3 \text{ or } x = 2\]

Step 2: Determine which function is on top. Test \(x = 0\): \(f(0) = 6\) and \(g(0) = 0\), so \(f(x) > g(x)\) on \([-3, 2]\).

Step 3: Compute the area.

\[A = \int_{-3}^{2} \left[(6-x^2) - x\right] \, dx = \int_{-3}^{2} (6 - x - x^2) \, dx\]

Find the antiderivative:

\[= \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{-3}^{2}\]

Evaluate at \(x = 2\):

\[6(2) - \frac{4}{2} - \frac{8}{3} = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3}\]

Evaluate at \(x = -3\):

\[6(-3) - \frac{9}{2} - \frac{-27}{3} = -18 - \frac{9}{2} + 9 = -9 - \frac{9}{2} = -\frac{27}{2}\]

\[A = \frac{22}{3} - \left(-\frac{27}{2}\right) = \frac{22}{3} + \frac{27}{2} = \frac{44}{6} + \frac{81}{6} = \frac{125}{6} \approx 20.833\]

The enclosed area is \(\frac{125}{6}\) square units.

b) On \([0,1]\), compare: at \(x = 0.25\), \(f(0.25) = 0.0625\) and \(g(0.25) = 0.5\), so \(\sqrt{x} \geq x^2\) on \([0,1]\).

\[A = \int_0^1 \left(\sqrt{x} - x^2\right) \, dx = \int_0^1 \left(x^{1/2} - x^2\right) \, dx\]

\[= \left[ \frac{x^{3/2}}{3/2} - \frac{x^3}{3} \right]_0^1 = \left[ \frac{2}{3}x^{3/2} - \frac{x^3}{3} \right]_0^1\]

\[= \left(\frac{2}{3} - \frac{1}{3}\right) - 0 = \frac{1}{3}\]

The enclosed area is \(\frac{1}{3}\) square units.

Problem 5: Integration by Parts (xx)

Evaluate using integration by parts (\(\int u \, dv = uv - \int v \, du\)):

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

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

  3. \(\int \ln(x) \, dx\) (hint: let \(u = \ln(x)\), \(dv = dx\))

a) Choose \(u = x\) and \(dv = e^{3x} \, dx\).

Then \(du = dx\) and \(v = \frac{1}{3}e^{3x}\).

Apply IBP:

\[\int x \cdot e^{3x} \, dx = x \cdot \frac{1}{3}e^{3x} - \int \frac{1}{3}e^{3x} \, dx\]

\[= \frac{x}{3}e^{3x} - \frac{1}{3} \cdot \frac{1}{3}e^{3x} + C\]

\[= \frac{x}{3}e^{3x} - \frac{1}{9}e^{3x} + C = e^{3x}\left(\frac{x}{3} - \frac{1}{9}\right) + C\]

b) Choose \(u = x\) and \(dv = e^{-x} \, dx\).

Then \(du = dx\) and \(v = -e^{-x}\).

Apply IBP:

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

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

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

c) Choose \(u = \ln(x)\) and \(dv = dx\).

Then \(du = \frac{1}{x} \, dx\) and \(v = x\).

Apply IBP:

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

\[= x\ln(x) - \int 1 \, dx = x\ln(x) - x + C\]

Common error: In part (c), students often struggle because there is no obvious \(dv\). The trick is to set \(dv = dx\) so that \(v = x\), and the remaining integral simplifies nicely.

Problem 6: Consumer and Producer Surplus (xx)

A market has demand \(D(q) = 80 - 4q\) and supply \(S(q) = 20 + 2q\).

  1. Find the equilibrium quantity and price.

  2. Calculate consumer surplus.

  3. Calculate producer surplus.

  4. What is the total economic surplus?

a) Set \(D(q) = S(q)\):

\[80 - 4q = 20 + 2q\]

\[60 = 6q \implies q^* = 10\]

\[p^* = D(10) = 80 - 40 = 40\]

The equilibrium is at \(q^* = 10\) units and \(p^* = 40\).

b) Consumer surplus is the area between the demand curve and the equilibrium price:

\[CS = \int_0^{10} D(q) \, dq - p^* \cdot q^*\]

\[= \int_0^{10} (80 - 4q) \, dq - 40 \cdot 10\]

Find the antiderivative and evaluate:

\[= \left[ 80q - 2q^2 \right]_0^{10} - 400 = (800 - 200) - 400 = 600 - 400 = 200\]

The consumer surplus is \(CS = 200\).

c) Producer surplus is the area between the equilibrium price and the supply curve:

\[PS = p^* \cdot q^* - \int_0^{10} S(q) \, dq\]

\[= 400 - \int_0^{10} (20 + 2q) \, dq\]

\[= 400 - \left[ 20q + q^2 \right]_0^{10} = 400 - (200 + 100) = 400 - 300 = 100\]

The producer surplus is \(PS = 100\).

d) Total economic surplus:

\[TS = CS + PS = 200 + 100 = 300\]

The total economic surplus in the market is \(300\). This represents the total net benefit to all market participants from trade.

Problem 7: Substitution Integrals (xx)

Evaluate using substitution:

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

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

  3. \(\int x(x^2+1)^4 \, dx\)

a) Let \(u = x^2\), then \(du = 2x \, dx\).

\[\int 2x \cdot e^{x^2} \, dx = \int e^u \, du = e^u + C = e^{x^2} + C\]

b) Let \(u = x^2 + 4\), then \(du = 2x \, dx\), so \(x \, dx = \frac{1}{2} du\).

\[\int \frac{3x}{x^2+4} \, dx = 3 \int \frac{x \, dx}{x^2+4} = 3 \int \frac{1}{u} \cdot \frac{1}{2} \, du = \frac{3}{2} \int \frac{1}{u} \, du\]

\[= \frac{3}{2} \ln|u| + C = \frac{3}{2} \ln(x^2 + 4) + C\]

Note: We drop the absolute value since \(x^2 + 4 > 0\) always.

c) Let \(u = x^2 + 1\), then \(du = 2x \, dx\), so \(x \, dx = \frac{1}{2} du\).

\[\int x(x^2+1)^4 \, dx = \int u^4 \cdot \frac{1}{2} \, du = \frac{1}{2} \cdot \frac{u^5}{5} + C = \frac{(x^2+1)^5}{10} + C\]

Problem 8: Continuous Income Present Value (xx)

An investment generates a continuous income stream of \(f(t) = 2000 \cdot e^{0.03t}\) euros per year.

  1. Find the total income over 5 years.

  2. If the discount rate is \(r = 0.06\), find the present value: \(PV = \int_0^5 2000 \cdot e^{0.03t} \cdot e^{-0.06t} \, dt\)

  3. Is the investment worth more than 8,000 euros today?

a) Total income over 5 years:

\[I = \int_0^5 2000 \cdot e^{0.03t} \, dt = 2000 \cdot \left[ \frac{e^{0.03t}}{0.03} \right]_0^5\]

\[= \frac{2000}{0.03} \left( e^{0.15} - e^0 \right) = \frac{2000}{0.03} (e^{0.15} - 1)\]

\[= 66{,}666.67 \cdot (1.16183 - 1) = 66{,}666.67 \cdot 0.16183 \approx 10{,}788.84 \text{ euros}\]

The total (undiscounted) income over 5 years is approximately 10,788.84 euros.

b) Combine the exponents using the rule \(e^a \cdot e^b = e^{a+b}\) (Section 01 connection):

\[PV = \int_0^5 2000 \cdot e^{0.03t} \cdot e^{-0.06t} \, dt = \int_0^5 2000 \cdot e^{-0.03t} \, dt\]

\[= 2000 \cdot \left[ \frac{e^{-0.03t}}{-0.03} \right]_0^5 = \frac{2000}{-0.03} \left( e^{-0.15} - e^0 \right)\]

\[= -66{,}666.67 \cdot (e^{-0.15} - 1) = -66{,}666.67 \cdot (0.86071 - 1)\]

\[= -66{,}666.67 \cdot (-0.13929) \approx 9{,}286.01 \text{ euros}\]

The present value of the income stream is approximately 9,286.01 euros.

c) Since \(PV \approx 9{,}286.01 > 8{,}000\), yes, the investment is worth more than 8,000 euros today. It exceeds the threshold by approximately 1,286 euros.

Problem 9: Cost Function from Marginal Cost (xx)

A company’s marginal cost is \(MC(x) = 0.03x^2 - 2x + 50\) and fixed costs are 1,200 euros.

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

  2. Find the cost of producing the first 20 units.

  3. Find the average cost function \(\overline{C}(x) = \frac{C(x)}{x}\).

  4. Find the production level that minimizes average cost (uses derivatives – Section 05 connection).

a) The total cost function is the antiderivative of marginal cost plus fixed costs:

\[C(x) = \int MC(x) \, dx = \int (0.03x^2 - 2x + 50) \, dx\]

\[= 0.03 \cdot \frac{x^3}{3} - 2 \cdot \frac{x^2}{2} + 50x + K = 0.01x^3 - x^2 + 50x + K\]

Since fixed costs are \(C(0) = 1{,}200\), we get \(K = 1{,}200\).

\[C(x) = 0.01x^3 - x^2 + 50x + 1{,}200\]

b) The cost of producing the first 20 units:

\[C(20) = 0.01(8{,}000) - (400) + 50(20) + 1{,}200 = 80 - 400 + 1{,}000 + 1{,}200 = 1{,}880 \text{ euros}\]

Alternatively, using the definite integral of marginal cost:

\[\int_0^{20} MC(x) \, dx = C(20) - C(0) = 1{,}880 - 1{,}200 = 680 \text{ euros (variable cost only)}\]

The total cost including fixed costs is 1,880 euros.

c) Average cost function:

\[\overline{C}(x) = \frac{C(x)}{x} = \frac{0.01x^3 - x^2 + 50x + 1{,}200}{x} = 0.01x^2 - x + 50 + \frac{1{,}200}{x}\]

d) To minimize \(\overline{C}(x)\), take the derivative and set it to zero:

\[\overline{C}'(x) = 0.02x - 1 - \frac{1{,}200}{x^2}\]

Set \(\overline{C}'(x) = 0\):

\[0.02x - 1 - \frac{1{,}200}{x^2} = 0\]

Multiply by \(x^2\):

\[0.02x^3 - x^2 - 1{,}200 = 0\]

This is equivalent to \(0.02x^3 - x^2 = 1{,}200\). Testing \(x = 50\):

\[0.02(125{,}000) - 2{,}500 = 2{,}500 - 2{,}500 = 0 \neq 1{,}200\]

Testing \(x = 60\):

\[0.02(216{,}000) - 3{,}600 = 4{,}320 - 3{,}600 = 720 \neq 1{,}200\]

Testing \(x = 70\):

\[0.02(343{,}000) - 4{,}900 = 6{,}860 - 4{,}900 = 1{,}960\]

So the root lies between 60 and 70. Testing \(x = 65\):

\[0.02(274{,}625) - 4{,}225 = 5{,}492.5 - 4{,}225 = 1{,}267.5\]

Testing \(x = 64\):

\[0.02(262{,}144) - 4{,}096 = 5{,}242.88 - 4{,}096 = 1{,}146.88\]

The minimum average cost occurs at approximately \(x \approx 65\) units.

To verify this is a minimum, check \(\overline{C}''(x) = 0.02 + \frac{2{,}400}{x^3}\). Since this is always positive for \(x > 0\), the critical point is indeed a minimum.

The average cost at this production level is approximately:

\[\overline{C}(65) = 0.01(4{,}225) - 65 + 50 + \frac{1{,}200}{65} \approx 42.25 - 65 + 50 + 18.46 \approx 45.71 \text{ euros per unit}\]

Problem 10: Average Value of a Function (xx)

The average value of a function \(f\) on \([a,b]\) is \(\bar{f} = \frac{1}{b-a} \int_a^b f(x) \, dx\).

  1. Find the average value of \(f(x) = x^2\) on \([0, 3]\).

  2. Find the average value of \(g(x) = e^{-x}\) on \([0, 4]\).

  3. A company’s production rate over 8 hours is \(R(t) = 100 - 5t\) units/hour. Find the average production rate over the 8-hour period.

a) Average value:

\[\bar{f} = \frac{1}{3-0} \int_0^3 x^2 \, dx = \frac{1}{3} \left[ \frac{x^3}{3} \right]_0^3 = \frac{1}{3} \cdot \frac{27}{3} = \frac{1}{3} \cdot 9 = 3\]

The average value of \(f(x) = x^2\) on \([0,3]\) is \(3\).

b) Average value:

\[\bar{g} = \frac{1}{4-0} \int_0^4 e^{-x} \, dx = \frac{1}{4} \left[ -e^{-x} \right]_0^4\]

\[= \frac{1}{4} \left( -e^{-4} - (-e^0) \right) = \frac{1}{4} (1 - e^{-4}) = \frac{1}{4} (1 - 0.01832) \approx \frac{0.98168}{4} \approx 0.2454\]

The average value of \(g(x) = e^{-x}\) on \([0,4]\) is \(\frac{1-e^{-4}}{4} \approx 0.2454\).

c) Average production rate:

\[\bar{R} = \frac{1}{8-0} \int_0^8 (100 - 5t) \, dt = \frac{1}{8} \left[ 100t - \frac{5t^2}{2} \right]_0^8\]

Evaluate at \(t = 8\):

\[100(8) - \frac{5(64)}{2} = 800 - 160 = 640\]

Evaluate at \(t = 0\): \(0\).

\[\bar{R} = \frac{1}{8}(640 - 0) = \frac{640}{8} = 80 \text{ units/hour}\]

The average production rate over the 8-hour period is \(80\) units/hour.

Problem 11: Area Between Exponential and Linear (xxx)

Find the area between \(f(x) = e^x\) and \(g(x) = x + 1\) on the interval \([0, 1]\).

  1. Show that the curves intersect at \(x = 0\).

  2. Determine which function is on top on \((0, 1]\).

  3. Set up and evaluate the integral for the enclosed area.

a) At \(x = 0\):

\[f(0) = e^0 = 1 \quad \text{and} \quad g(0) = 0 + 1 = 1\]

Since \(f(0) = g(0) = 1\), the curves intersect at \(x = 0\). \(\checkmark\)

b) Test at \(x = 0.5\):

\[f(0.5) = e^{0.5} \approx 1.649 \quad \text{and} \quad g(0.5) = 0.5 + 1 = 1.5\]

Since \(f(0.5) > g(0.5)\), the exponential function \(f(x) = e^x\) is on top on \((0, 1]\).

This can also be seen from the fact that \(e^x > x + 1\) for all \(x > 0\) (since \(x + 1\) is the tangent line to \(e^x\) at \(x = 0\), and \(e^x\) is strictly convex).

c) Since \(e^x \geq x + 1\) on \([0, 1]\) with equality only at \(x = 0\):

\[A = \int_0^1 \left[ e^x - (x + 1) \right] \, dx\]

Find the antiderivative:

\[= \left[ e^x - \frac{x^2}{2} - x \right]_0^1\]

Evaluate at \(x = 1\):

\[e^1 - \frac{1}{2} - 1 = e - \frac{3}{2}\]

Evaluate at \(x = 0\):

\[e^0 - 0 - 0 = 1\]

\[A = \left(e - \frac{3}{2}\right) - 1 = e - \frac{5}{2} \approx 2.718 - 2.5 = 0.218\]

The area between the two curves on \([0, 1]\) is \(e - \frac{5}{2} \approx 0.218\) square units.

Problem 12: Waste Accumulation (xxx)

A chemical plant releases waste at a rate \(R(t) = 80 \cdot e^{-0.2t}\) tons/year.

  1. Find the total waste released in the first \(T\) years: \(Q(T) = \int_0^T 80 \cdot e^{-0.2t} \, dt\).

  2. Compute \(Q(5)\) and \(Q(20)\).

  3. What value does \(Q(T)\) approach as \(T\) becomes very large?

  4. After how many years has 90% of the long-run total been released? (Use logarithms.)

  5. Environmental regulations cap total waste at 350 tons. Will the plant comply in the long run?

a) Integrate the release rate over \([0, T]\):

\[Q(T) = \int_0^T 80 \cdot e^{-0.2t} \, dt = 80 \cdot \left[ \frac{e^{-0.2t}}{-0.2} \right]_0^T = -400 \left[ e^{-0.2t} \right]_0^T\]

\[= -400(e^{-0.2T} - e^0) = -400(e^{-0.2T} - 1) = 400(1 - e^{-0.2T})\]

\[Q(T) = 400(1 - e^{-0.2T}) \text{ tons}\]

b) Evaluate at \(T = 5\):

\[Q(5) = 400(1 - e^{-0.2 \cdot 5}) = 400(1 - e^{-1}) = 400(1 - 0.3679) = 400 \cdot 0.6321 \approx 252.85 \text{ tons}\]

Evaluate at \(T = 20\):

\[Q(20) = 400(1 - e^{-0.2 \cdot 20}) = 400(1 - e^{-4}) = 400(1 - 0.0183) = 400 \cdot 0.9817 \approx 392.67 \text{ tons}\]

After 5 years approximately 252.85 tons have been released, and after 20 years approximately 392.67 tons.

c) As \(T\) becomes very large, \(e^{-0.2T} \to 0\):

\[\lim_{T \to \infty} Q(T) = \lim_{T \to \infty} 400(1 - e^{-0.2T}) = 400(1 - 0) = 400 \text{ tons}\]

The total waste approaches \(400\) tons in the long run.

d) We need \(Q(T) = 0.9 \cdot 400 = 360\) tons:

\[400(1 - e^{-0.2T}) = 360\]

\[1 - e^{-0.2T} = 0.9\]

\[e^{-0.2T} = 0.1\]

Take the natural logarithm (Section 01 connection):

\[-0.2T = \ln(0.1) = -\ln(10)\]

\[T = \frac{\ln(10)}{0.2} = 5\ln(10) \approx 5 \cdot 2.3026 = 11.51 \text{ years}\]

After approximately 11.51 years, 90% of the long-run total waste has been released.

e) The long-run total waste is \(400\) tons, which exceeds the 350-ton cap. The plant will not comply with the regulation in the long run. To verify: \(Q(T) = 400(1 - e^{-0.2T})\) reaches 350 tons when \(e^{-0.2T} = 0.125\), i.e., at \(T = 5\ln(8) \approx 10.40\) years. After this point, cumulative waste exceeds the cap and continues growing toward 400 tons.

Problem 13: Combined Curve Sketching + Area (xxx)

For \(f(x) = x \cdot e^{-x}\):

  1. Find zeros.

  2. Find \(\lim_{x \to \infty} f(x)\).

  3. Find \(f'(x)\), critical points, and classify them.

  4. Find \(f''(x)\) and inflection points.

  5. Sketch the graph.

  6. Find the area under \(f(x)\) from \(x=0\) to \(x=3\): \(\int_0^3 x \cdot e^{-x} \, dx\) (use integration by parts).

a) Set \(f(x) = 0\):

\[x \cdot e^{-x} = 0\]

Since \(e^{-x} > 0\) for all \(x\), the only solution is \(x = 0\).

The function has a single zero at \(x = 0\).

b) As \(x \to \infty\): the exponential decay \(e^{-x}\) dominates the linear growth \(x\).

\[\lim_{x \to \infty} x \cdot e^{-x} = \lim_{x \to \infty} \frac{x}{e^x} = 0\]

This can be verified using L’Hopital’s rule: \(\lim_{x \to \infty} \frac{1}{e^x} = 0\).

c) Using the product rule:

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

Set \(f'(x) = 0\): Since \(e^{-x} \neq 0\), we need \(1 - x = 0\), so \(x = 1\).

Classification: \(f'(x) > 0\) for \(x < 1\) (increasing) and \(f'(x) < 0\) for \(x > 1\) (decreasing).

Therefore \(x = 1\) is a local maximum with \(f(1) = 1 \cdot e^{-1} = \frac{1}{e} \approx 0.368\).

d) Using the product rule on \(f'(x) = e^{-x}(1-x)\):

\[f''(x) = -e^{-x}(1-x) + e^{-x}(-1) = e^{-x}(-1+x-1) = e^{-x}(x-2)\]

Set \(f''(x) = 0\): Since \(e^{-x} \neq 0\), we need \(x - 2 = 0\), so \(x = 2\).

Check sign change: \(f''(1) = e^{-1}(-1) < 0\) and \(f''(3) = e^{-3}(1) > 0\).

The concavity changes at \(x = 2\), so there is an inflection point at \((2, 2e^{-2}) \approx (2, 0.271)\).

e) Graph sketch summary:

  • Zero at \(x = 0\); the function is negative for \(x < 0\)
  • Increasing on \((-\infty, 1)\), decreasing on \((1, \infty)\)
  • Local maximum at \((1, 1/e) \approx (1, 0.368)\)
  • Inflection point at \((2, 2/e^2) \approx (2, 0.271)\)
  • Concave down on \((-\infty, 2)\), concave up on \((2, \infty)\)
  • Approaches 0 from above as \(x \to \infty\)

The graph rises from the origin, peaks at \(x = 1\), then gradually decays toward zero.

f) Use integration by parts with \(u = x\), \(dv = e^{-x} \, dx\):

Then \(du = dx\) and \(v = -e^{-x}\).

\[\int_0^3 x \cdot e^{-x} \, dx = \left[ -x \cdot e^{-x} \right]_0^3 + \int_0^3 e^{-x} \, dx\]

Evaluate the boundary term:

\[\left[ -x \cdot e^{-x} \right]_0^3 = -3e^{-3} - 0 = -3e^{-3}\]

Evaluate the remaining integral:

\[\int_0^3 e^{-x} \, dx = \left[ -e^{-x} \right]_0^3 = -e^{-3} + 1 = 1 - e^{-3}\]

Combine:

\[\int_0^3 x \cdot e^{-x} \, dx = -3e^{-3} + 1 - e^{-3} = 1 - 4e^{-3}\]

\[= 1 - 4 \cdot 0.04979 = 1 - 0.19915 \approx 0.801\]

The area under \(f(x) = x \cdot e^{-x}\) from \(x = 0\) to \(x = 3\) is \(1 - 4e^{-3} \approx 0.801\) square units.

Problem 14: Revenue Optimization via Integration (xxx)

A firm’s marginal revenue is \(MR(x) = 200 - 4x\) and marginal cost is \(MC(x) = 40 + 2x\).

  1. Find the revenue function \(R(x)\) if \(R(0) = 0\).

  2. Find the cost function \(C(x)\) if fixed costs are 500 euros.

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

  4. Find the profit-maximizing quantity (using derivatives).

  5. Calculate the total profit at the optimal quantity.

  6. Calculate the consumer surplus at the optimal quantity, given \(D(x) = 200 - 4x\) is the demand curve.

a) Integrate marginal revenue:

\[R(x) = \int MR(x) \, dx = \int (200 - 4x) \, dx = 200x - 2x^2 + C_1\]

Since \(R(0) = 0\): \(0 = 0 - 0 + C_1\), so \(C_1 = 0\).

\[R(x) = 200x - 2x^2\]

b) Integrate marginal cost:

\[C(x) = \int MC(x) \, dx = \int (40 + 2x) \, dx = 40x + x^2 + C_2\]

Since fixed costs are 500 euros, \(C(0) = 500\): \(C_2 = 500\).

\[C(x) = x^2 + 40x + 500\]

c) Profit function:

\[P(x) = R(x) - C(x) = (200x - 2x^2) - (x^2 + 40x + 500)\]

\[= 200x - 2x^2 - x^2 - 40x - 500 = -3x^2 + 160x - 500\]

d) Set \(P'(x) = 0\):

\[P'(x) = -6x + 160 = 0 \implies x = \frac{160}{6} = \frac{80}{3} \approx 26.67\]

Check: \(P''(x) = -6 < 0\), confirming this is a maximum.

The profit-maximizing quantity is \(x^* = \frac{80}{3} \approx 26.67\) units.

e) Profit at the optimal quantity:

\[P\!\left(\frac{80}{3}\right) = -3\left(\frac{80}{3}\right)^2 + 160 \cdot \frac{80}{3} - 500\]

\[= -3 \cdot \frac{6{,}400}{9} + \frac{12{,}800}{3} - 500 = -\frac{6{,}400}{3} + \frac{12{,}800}{3} - 500\]

\[= \frac{6{,}400}{3} - 500 = \frac{6{,}400 - 1{,}500}{3} = \frac{4{,}900}{3} \approx 1{,}633.33 \text{ euros}\]

The maximum profit is approximately 1,633.33 euros.

f) At \(x^* = \frac{80}{3}\), the price is:

\[p^* = D\!\left(\frac{80}{3}\right) = 200 - 4 \cdot \frac{80}{3} = 200 - \frac{320}{3} = \frac{280}{3} \approx 93.33\]

Consumer surplus:

\[CS = \int_0^{80/3} D(x) \, dx - p^* \cdot x^*\]

\[= \int_0^{80/3} (200 - 4x) \, dx - \frac{280}{3} \cdot \frac{80}{3}\]

Evaluate the integral:

\[\int_0^{80/3} (200 - 4x) \, dx = \left[ 200x - 2x^2 \right]_0^{80/3}\]

\[= 200 \cdot \frac{80}{3} - 2 \cdot \frac{6{,}400}{9} = \frac{16{,}000}{3} - \frac{12{,}800}{9} = \frac{48{,}000}{9} - \frac{12{,}800}{9} = \frac{35{,}200}{9}\]

Now subtract \(p^* \cdot x^*\):

\[p^* \cdot x^* = \frac{280}{3} \cdot \frac{80}{3} = \frac{22{,}400}{9}\]

\[CS = \frac{35{,}200}{9} - \frac{22{,}400}{9} = \frac{12{,}800}{9} \approx 1{,}422.22 \text{ euros}\]

The consumer surplus at the optimal quantity is \(\frac{12{,}800}{9} \approx 1{,}422.22\) euros. This represents the total extra benefit consumers receive by paying less than their maximum willingness to pay.

Problem 15: Investment Comparison (xxx)

Compare two investment strategies over 10 years at discount rate \(r = 0.05\):

  • Strategy A: Lump sum of 50,000 euros invested today.
  • Strategy B: Continuous deposit of 6,000 euros per year.

  1. What is the present value of Strategy A? (Trivially 50,000 euros.)

  2. What is the present value of Strategy B? Compute \(PV_B = \int_0^{10} 6{,}000 \cdot e^{-0.05t} \, dt\).

  3. Which strategy is more attractive?

  4. At what annual deposit rate \(d\) (euros per year) would Strategy B have the same present value as Strategy A? Set up and solve the equation.

a) The present value of Strategy A is simply:

\[PV_A = 50{,}000 \text{ euros}\]

Since the investment is made today, no discounting is needed.

b) Present value of Strategy B:

\[PV_B = \int_0^{10} 6{,}000 \cdot e^{-0.05t} \, dt = 6{,}000 \cdot \left[ \frac{e^{-0.05t}}{-0.05} \right]_0^{10}\]

\[= -120{,}000 \left[ e^{-0.05t} \right]_0^{10} = -120{,}000 (e^{-0.5} - 1)\]

\[= 120{,}000 (1 - e^{-0.5}) = 120{,}000 (1 - 0.60653)\]

\[= 120{,}000 \cdot 0.39347 = 47{,}216.32 \text{ euros}\]

The present value of Strategy B is approximately 47,216.32 euros.

c) Since \(PV_A = 50{,}000 > PV_B \approx 47{,}216.32\), Strategy A is more attractive. The lump sum investment has a higher present value by approximately 2,783.68 euros.

d) Set \(PV_B = PV_A\):

\[\int_0^{10} d \cdot e^{-0.05t} \, dt = 50{,}000\]

\[d \cdot 120{,}000 \cdot (1 - e^{-0.5}) \cdot \frac{1}{6{,}000} \cdot 6{,}000 = 50{,}000\]

More directly:

\[d \cdot \left[ \frac{e^{-0.05t}}{-0.05} \right]_0^{10} = 50{,}000\]

\[d \cdot \frac{1 - e^{-0.5}}{0.05} = 50{,}000\]

\[d \cdot \frac{0.39347}{0.05} = 50{,}000\]

\[d \cdot 7.8694 = 50{,}000\]

\[d = \frac{50{,}000}{7.8694} \approx 6{,}353.68 \text{ euros per year}\]

Strategy B would need a continuous deposit rate of approximately 6,353.68 euros per year to match the present value of Strategy A. This is about 353.68 euros per year more than the current 6,000.

Problem 16: Surplus with Price Floor (xxxx)

A market has demand \(D(q) = 120 - q^2\) and supply \(S(q) = q^2 + 20\).

  1. Find the free-market equilibrium \((q^*, p^*)\).

  2. Calculate consumer surplus and producer surplus at equilibrium.

  3. The government imposes a price floor of \(p_f = 80\). Find the new quantities demanded and supplied.

  4. Calculate the deadweight loss caused by the price floor.

  5. Interpret the redistribution of surplus.

a) Set \(D(q) = S(q)\):

\[120 - q^2 = q^2 + 20\]

\[100 = 2q^2 \implies q^2 = 50 \implies q^* = \sqrt{50} = 5\sqrt{2} \approx 7.071\]

\[p^* = D(5\sqrt{2}) = 120 - 50 = 70\]

The free-market equilibrium is \((q^*, p^*) = (5\sqrt{2}, 70)\).

b) Consumer surplus:

\[CS = \int_0^{q^*} D(q) \, dq - p^* \cdot q^*\]

\[= \int_0^{5\sqrt{2}} (120 - q^2) \, dq - 70 \cdot 5\sqrt{2}\]

\[= \left[ 120q - \frac{q^3}{3} \right]_0^{5\sqrt{2}} - 350\sqrt{2}\]

\[= 120 \cdot 5\sqrt{2} - \frac{(5\sqrt{2})^3}{3} - 350\sqrt{2}\]

\[= 600\sqrt{2} - \frac{250\sqrt{2}}{3} - 350\sqrt{2}\]

\[= \left(600 - \frac{250}{3} - 350\right)\sqrt{2} = \left(250 - \frac{250}{3}\right)\sqrt{2} = \frac{500}{3}\sqrt{2} \approx 235.70\]

Producer surplus:

\[PS = p^* \cdot q^* - \int_0^{q^*} S(q) \, dq\]

\[= 350\sqrt{2} - \int_0^{5\sqrt{2}} (q^2 + 20) \, dq\]

\[= 350\sqrt{2} - \left[ \frac{q^3}{3} + 20q \right]_0^{5\sqrt{2}}\]

\[= 350\sqrt{2} - \frac{250\sqrt{2}}{3} - 100\sqrt{2}\]

\[= \left(350 - \frac{250}{3} - 100\right)\sqrt{2} = \left(250 - \frac{250}{3}\right)\sqrt{2} = \frac{500}{3}\sqrt{2} \approx 235.70\]

Note: \(CS = PS\) due to the symmetric structure of the demand and supply curves around the equilibrium.

Total surplus: \(TS = CS + PS = \frac{1000}{3}\sqrt{2} \approx 471.40\).

c) With price floor \(p_f = 80 > p^* = 70\), the price is artificially held above equilibrium.

Quantity demanded at \(p_f = 80\): Set \(D(q_d) = 80\):

\[120 - q_d^2 = 80 \implies q_d^2 = 40 \implies q_d = \sqrt{40} = 2\sqrt{10} \approx 6.325\]

Quantity supplied at \(p_f = 80\): Set \(S(q_s) = 80\):

\[q_s^2 + 20 = 80 \implies q_s^2 = 60 \implies q_s = \sqrt{60} = 2\sqrt{15} \approx 7.746\]

There is a surplus of goods: \(q_s - q_d = 2\sqrt{15} - 2\sqrt{10} \approx 1.42\) units. The actual quantity traded is \(q_d = 2\sqrt{10}\) (limited by demand).

d) The deadweight loss (DWL) is the loss in total surplus compared to the free market. It equals the area between the demand and supply curves from \(q_d\) to \(q^*\):

\[DWL = \int_{q_d}^{q^*} \left[ D(q) - S(q) \right] \, dq = \int_{2\sqrt{10}}^{5\sqrt{2}} \left[ (120 - q^2) - (q^2 + 20) \right] \, dq\]

\[= \int_{2\sqrt{10}}^{5\sqrt{2}} (100 - 2q^2) \, dq\]

\[= \left[ 100q - \frac{2q^3}{3} \right]_{2\sqrt{10}}^{5\sqrt{2}}\]

At \(q = 5\sqrt{2}\):

\[100 \cdot 5\sqrt{2} - \frac{2(5\sqrt{2})^3}{3} = 500\sqrt{2} - \frac{500\sqrt{2}}{3} = \frac{1000\sqrt{2}}{3}\]

At \(q = 2\sqrt{10}\):

\[100 \cdot 2\sqrt{10} - \frac{2(2\sqrt{10})^3}{3} = 200\sqrt{10} - \frac{2 \cdot 80\sqrt{10}}{3} = 200\sqrt{10} - \frac{160\sqrt{10}}{3} = \frac{440\sqrt{10}}{3}\]

\[DWL = \frac{1000\sqrt{2}}{3} - \frac{440\sqrt{10}}{3} = \frac{1000\sqrt{2} - 440\sqrt{10}}{3}\]

Numerically:

\[= \frac{1000(1.4142) - 440(3.1623)}{3} = \frac{1414.2 - 1391.4}{3} = \frac{22.8}{3} \approx 7.60\]

The deadweight loss from the price floor is approximately 7.60.

e) Interpretation of surplus redistribution:

  • Consumers lose because they pay a higher price (\(80\) instead of \(70\)) and buy fewer units (\(2\sqrt{10} \approx 6.32\) instead of \(5\sqrt{2} \approx 7.07\)). Consumer surplus decreases.
  • Producers who sell benefit from the higher price. However, fewer units are sold overall. Some of the former consumer surplus is transferred to producers.
  • The deadweight loss (\(\approx 7.60\)) represents transactions that would have been mutually beneficial but no longer occur. This is a net loss to society.
  • There is also a surplus of unsold goods (\(q_s - q_d \approx 1.42\) units), representing wasted production if suppliers produce at \(q_s\).

Problem 17: Full Business Scenario (xxxx)

A monopolist faces demand \(P(q) = 100 \cdot e^{-0.1q}\) and has cost function \(C(q) = 200 + 10q\).

  1. Find the revenue function \(R(q) = q \cdot P(q)\).

  2. Find the marginal revenue \(MR(q)\) (product rule!).

  3. Set \(MR = MC\) and solve for the profit-maximizing quantity.

  4. Find the monopoly price.

  5. Calculate consumer surplus under monopoly pricing: \(CS = \int_0^{q^*} P(q) \, dq - p^* \cdot q^*\).

  6. If the competitive price were \(P = MC = 10\), find the competitive quantity and compare total surplus.

a) Revenue function:

\[R(q) = q \cdot P(q) = q \cdot 100 \cdot e^{-0.1q} = 100q \cdot e^{-0.1q}\]

b) Using the product rule on \(R(q) = 100q \cdot e^{-0.1q}\):

\[MR(q) = R'(q) = 100 \cdot e^{-0.1q} + 100q \cdot (-0.1) \cdot e^{-0.1q}\]

\[= 100 \cdot e^{-0.1q}(1 - 0.1q)\]

c) The marginal cost is \(MC = C'(q) = 10\). Set \(MR = MC\):

\[100 \cdot e^{-0.1q}(1 - 0.1q) = 10\]

\[e^{-0.1q}(1 - 0.1q) = 0.1\]

This is a transcendental equation. We solve numerically.

Let \(h(q) = e^{-0.1q}(1 - 0.1q) - 0.1\).

  • \(h(0) = 1 \cdot 1 - 0.1 = 0.9 > 0\)
  • \(h(5) = e^{-0.5}(0.5) - 0.1 = 0.6065 \cdot 0.5 - 0.1 = 0.2033 > 0\)
  • \(h(8) = e^{-0.8}(0.2) - 0.1 = 0.4493 \cdot 0.2 - 0.1 = -0.0101 < 0\)
  • \(h(7) = e^{-0.7}(0.3) - 0.1 = 0.4966 \cdot 0.3 - 0.1 = 0.0490 > 0\)
  • \(h(7.5) = e^{-0.75}(0.25) - 0.1 = 0.4724 \cdot 0.25 - 0.1 = 0.0181 > 0\)
  • \(h(7.8) = e^{-0.78}(0.22) - 0.1 = 0.4584 \cdot 0.22 - 0.1 = 0.0008 > 0\)
  • \(h(7.9) = e^{-0.79}(0.21) - 0.1 = 0.4538 \cdot 0.21 - 0.1 = -0.0047 < 0\)

So \(q^* \approx 7.8\) units.

More precisely, \(q^* \approx 7.81\) units.

d) The monopoly price:

\[p^* = P(q^*) = 100 \cdot e^{-0.1 \cdot 7.81} = 100 \cdot e^{-0.781} = 100 \cdot 0.4578 \approx 45.78 \text{ euros}\]

e) Consumer surplus under monopoly:

\[CS = \int_0^{7.81} 100 \cdot e^{-0.1q} \, dq - p^* \cdot q^*\]

Evaluate the integral:

\[\int_0^{7.81} 100 \cdot e^{-0.1q} \, dq = 100 \cdot \left[ \frac{e^{-0.1q}}{-0.1} \right]_0^{7.81} = -1{,}000 \left[ e^{-0.1q} \right]_0^{7.81}\]

\[= -1{,}000 (e^{-0.781} - 1) = -1{,}000(0.4578 - 1) = 1{,}000 \cdot 0.5422 = 542.20\]

Revenue at monopoly output:

\[p^* \cdot q^* = 45.78 \cdot 7.81 \approx 357.54\]

\[CS = 542.20 - 357.54 \approx 184.66 \text{ euros}\]

The consumer surplus under monopoly pricing is approximately 184.66 euros.

f) Under competition, \(P = MC = 10\):

\[100 \cdot e^{-0.1q} = 10 \implies e^{-0.1q} = 0.1\]

Take the natural logarithm (Section 01 connection – using \(\ln(e^a) = a\)):

\[-0.1q = \ln(0.1) = -\ln(10) \approx -2.3026\]

\[q_c = \frac{2.3026}{0.1} = 23.026 \text{ units}\]

Competitive consumer surplus:

\[CS_c = \int_0^{23.026} 100 \cdot e^{-0.1q} \, dq - 10 \cdot 23.026\]

\[= -1{,}000(e^{-2.3026} - 1) - 230.26 = 1{,}000(1 - 0.1) - 230.26 = 900 - 230.26 = 669.74\]

Competitive producer surplus: Under perfect competition with constant \(MC = 10\), producer surplus is zero (price equals marginal cost, ignoring fixed costs).

Competitive total surplus (excluding fixed costs): \(TS_c = 669.74\)

Monopoly total surplus:

  • \(CS_m \approx 184.66\)
  • Monopoly profit: \(\pi = R(q^*) - C(q^*) = 357.54 - (200 + 10 \cdot 7.81) = 357.54 - 278.10 = 79.44\)
  • \(TS_m = CS_m + \pi = 184.66 + 79.44 = 264.10\)

Comparison: The monopoly causes a significant loss in total surplus:

\[DWL = TS_c - TS_m = 669.74 - 264.10 \approx 405.64 \text{ euros}\]

The monopolist restricts output from 23.03 to 7.81 units and raises the price from 10 to 45.78 euros, creating a substantial deadweight loss of approximately 405.64 euros. This illustrates the inefficiency of monopoly power – consumers face higher prices and reduced availability, and the monopolist’s profit gain does not compensate for the lost consumer surplus.