Calculus II – Integration Techniques and Business Applications

1. What is an Integral?

An indefinite integral (antiderivative) of a function $f(x)$ is a family of functions $F(x)$ such that

$$\frac{d}{dx}F(x)=f(x)\text{or}\int f(x)dx=F(x)+C,$$

where $C$ is the constant of integration.

A definite integral computes the net signed area under the curve $y=f(x)$ between limits $a$ and $b$:

$${\int }_a^{b}f(x)dx=F(b)-F(a)\text{(Fundamental Theorem of Calculus)}.$$

The definite integral has a clear geometric meaning:

The shaded region represents ${\int }_0^2{x}^{2}dx=\frac{8}{3}$.

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2. Basic Integration Rules

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3. Techniques of Integration

3.1 Substitution (u‑substitution)

When an integrand contains a function and its derivative, set $u=g(x)$. Then $du=g^{\prime }(x)dx$.

Worked Example 1 – Revenue Accumulation

A digital product sells $q(p)=200-5p$ units per day at price $p$ (in NPR). The daily revenue is $R(p)=pq(p)=p(200-5p)=200p-5p^2$. Find the total revenue when price rises from NPR 10 to NPR 30.

$$\text{Total revenue}={\int }_{10}^{30}(200p-5p^2)dp.$$

Integrate directly (no substitution needed), but we illustrate the steps:

$$\int (200p-5p^2)dp=100p^2-\frac{5}{3}{p}^3+C.$$

Evaluate:

$$\begin{array}{rl}{R}_{\text{total}}& =[100p^2-\frac{5}{3}{p}^3{]}_{10}^{30}\\ & =(100\cdot 30^2-\frac{5}{3}\cdot 30^3)-(100\cdot 10^2-\frac{5}{3}\cdot 10^3)\\ & =(100\cdot 900-\frac{5}{3}\cdot 27000)-(100\cdot 100-\frac{5}{3}\cdot 1000)\\ & =(90000-45000)-(10000-1667)\\ & =45000-8333=36667\text{NPR}.\end{array}$$

Thus, raising the price from NPR 10 to NPR 30 yields a total revenue of NPR 36,667.

3.2 Integration by Parts

Based on the product rule:

$$\int udv=uv-\int vdu.$$

Worked Example 2 – Consumer Surplus for a Linear Demand Curve

Demand: $p=100-2q$. Consumer surplus (CS) when 20 units are sold is

$$CS={\int }_0^{20}[(100-2q)-p_{\text{market}}]dq,$$

where market price $p_{\text{market}}=60$ NPR.

First, rewrite integrand: $100-2q-60=40-2q$.

$$CS={\int }_0^{20}(40-2q)dq.$$

Integrate directly (no parts needed), but to illustrate parts, suppose we had $\int q{e}^{q}dq$. The steps would be:

$$u=q,dv=e^{q}dq\Rightarrow du=dq,v=e^q.$$ $$\int q{e}^{q}dq=q{e}^q-\int e^{q}dq=q{e}^q-e^q+C.$$

Returning to CS:

$$\int (40-2q)dq=40q-q^2+C.$$

Evaluate:

$$CS=[40q-q^2{]}_0^{20}=(800-400)-0=400\text{NPR}.$$

Interpretation: Consumers gain NPR 400 in surplus over the market price.

3.3 Partial Fractions

Useful when the integrand is a rational function with factorable denominator.

Worked Example 3 – Inventory Holding Cost

Holding cost per unit per day: $h(t)=\frac{5}{t^2+t}$. Find total holding cost from day 1 to day 4.

Decompose:

$$\frac{5}{t^2+t}=\frac{5}{t(t+1)}=\frac{A}{t}+\frac{B}{t+1}.$$

Solve:

$$5=A(t+1)+Bt\Rightarrow \{\begin{array}{l}A+B=0\\ A=5\end{array}\Rightarrow A=5,B=-5.$$

Thus

$${\int }_1^4\frac{5}{t^2+t}dt={\int }_1^4(\frac{5}{t}-\frac{5}{t+1})dt=5[\ln |t|-\ln |t+1|{]}_1^4.$$

Compute:

$$5[(\ln 4-\ln 5)-(\ln 1-\ln 2)]=5[(\ln 4-\ln 5)-(-\ln 2)]=5[\ln 4-\ln 5+\ln 2]=5\ln \frac{8}{5}\approx 5\cdot 0.470=2.35\text{NPR}.$$

3.4 Improper Integrals

When limits are infinite or the integrand is unbounded.

Example – Discounted Cash Flow (DCF) with Perpetuity

A perpetual cash flow of $C=100$ NPR per year discounted at continuous rate $r=0.08$. Present value (PV):

$$PV={\int }_0^{\infty }100e^{-0.08t}dt.$$

Treat as improper integral:

$$PV=\underset{b\to \infty }{\lim }{\int }_0^{b}100e^{-0.08t}dt=\underset{b\to \infty }{\lim }[-\frac{100}{0.08}{e}^{-0.08t}{]}_0^b=\frac{100}{0.08}(1-0)=1250\text{NPR}.$$

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4. Applications in Business

4.1 Area Between Curves – Profit Region

Profit $\Pi (q)=R(q)-C(q)$. The region where $\Pi (q)>0$ is the area between revenue and cost curves.

The shaded region (0 ≤ q ≤ 6) gives ${\int }_0^6[5q-(2q+4)]dq={\int }_0^6(3q-4)dq=36$ NPR profit.

4.2 Volume of Revolution – Production Capacity

A cylindrical tank of radius $r=3$ m and height $h=5$ m stores liquid product. Volume $V=\pi r^{2}h$. Using integration:

$$V={\int }_0^5\pi (3{)}^{2}dx=9\pi {\int }_0^{5}dx=45\pi {\text{m}}^3.$$

4.3 Accumulation Functions – Inventory Build‑up

If daily inflow of raw material is $I(t)=20e^{0.02t}$ units, total inventory after 30 days:

$$I_{\text{total}}={\int }_0^{30}20e^{0.02t}dt=\frac{20}{0.02}(e^{0.6}-1)\approx 1000(0.822)=822\text{units}.$$

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5. Comparison of Integration Techniques

Advantages / Disadvantages

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6. Real‑World Connections

In the real world

1. eSewa calculates the total amount transferred each day by integrating the instantaneous transaction rate $r(t)$ (transactions per minute). If $r(t)=120+30\sin (\pi t/12)$, the daily total is ${\int }_0^{24}r(t)dt=2880$ transactions.

2. Daraz uses the volume‑of‑revolution concept to model warehouse space. A cylindrical storage silo of radius 4 m and height 10 m holds $\pi r^{2}h=502.65$ m³ of goods; integration confirms the capacity for seasonal sales spikes.

3. NEPSE analysts compute the area under the price‑time curve to gauge market momentum. For a price function $p(t)=1500+200\sin (\pi t/6)$ over a 6‑hour window, the integrated price change ${\int }_0^{6}p(t)dt$ yields the total value moved, informing trading strategies.

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7. Worked Example Integrated with a Real Situation

Scenario: A Nepali bank offers a loan with simple interest rate $i=12\%$ per annum, compounded continuously. The loan amount is NPR 500,000. Find the total interest accrued over 3 years using integration.

Continuous compounding formula derived from integration:

$$A(t)=P{e}^{it},\text{Interest }=A(t)-P=P(e^{it}-1).$$

Derivation:

$$\frac{dA}{dt}=iA\Rightarrow \frac{dA}{A}=idt\Rightarrow \int \frac{dA}{A}=i\int dt\Rightarrow \ln A=it+C.$$

Exponentiate and apply $A(0)=P$ to get $C=\ln P$.

Now compute:

$${\text{Interest}}_{3y}=500,000(e^{0.12\times 3}-1)=500,000(e^{0.36}-1)\approx 500,000(1.433-1)=216,500\text{NPR}.$$

Thus, the bank earns NPR 216,500 interest over three years.

The curve shows exponential growth; the area between the balance curve and the horizontal line at 500,000 represents accumulated interest.

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8. Summary Checklist

• Know the notation $\int f(x)dx$ and the meaning of $C$.
• Master the power, constant‑multiple, and sum rules.
• Identify which technique fits a given integrand.
• Apply the Fundamental Theorem of Calculus for definite integrals.
• Translate business problems (revenue, cost, inventory) into integrals.
• Verify convergence for improper integrals before evaluating.

Exam tip

Most university exams ask you to (1) choose the correct technique, (2) show every algebraic step, and (3) interpret the result in business terms.

• Quick scan: Look for a derivative inside the integrand → substitution; a product of algebraic and transcendental → parts; a rational function → partial fractions.
• Write the antiderivative first, then plug limits; never skip the “+ C” for indefinite integrals.
• For applications, always label what the integral represents (area, volume, accumulated cost, etc.) before computing.
• Time‑saving: Memorise the standard forms in the table above; they reduce algebraic manipulation during the exam.

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