Integration by Parts and Present Value

Integration by parts is derived from the Product Rule for differentiation: $\frac{d}{dx}[uv] = u \frac{dv}{dx} + v \frac{du}{dx}$ .
Fundamental Formula: $\int u \, dv = uv - \int v \, du$ .
This technique is specifically applied to integrands involving products of algebraic, exponential, or logarithmic functions.

Choosing $dv$ : Select the most complicated part of the integrand that fits a basic integration rule. Note that $dv$ must always include the $dx$ portion of the original integrand.
Choosing $u$ : Select the portion of the integrand whose derivative is simpler than the original function.

Algebraic and Exponential: To find $\int x e^x \, dx$ , let $u = x$ and $dv = e^x \, dx$ . The resulting antiderivative is $x e^x - e^x + C$ .
Logarithmic: To find $\int x^2 \ln(x) \, dx$ , let $u = \ln(x)$ and $dv = x^2 \, dx$ . The derivative of $\ln(x)$ is simpler than the function itself. The result is $\frac{x^3}{3} \ln(x) - \frac{x^3}{9} + C$ .
Repeated Application: For integrals like $\int x^2 e^x \, dx$ , the technique is used twice. The first application reduces the power of the algebraic term, and the second solves the remaining integral, resulting in $e^x (x^2 - 2x + 2) + C$ .

Present Value is the amount that must be deposited today, at an annual interest or inflation rate $r$ (expressed as a decimal and compounded continuously), to produce a future payment.
Actual Total Income over $t_{1}$ years is calculated as: $\int_{0}^{t_{1}} c(t) \, dt$ .
Present Value Formula: $\text{Present Value} = \int_{0}^{t_{1}} c(t) e^{-rt} \, dt$ .
Case Study (Example 7): A company with an income function $c(t) = 100,000t$ over $5$ years ( $0 \le t \le 5$ ) and an annual inflation rate of $4\%$ ( $r = 0.04$ ) uses integration by parts to determine if its present value exceeds $\$ 1 \text{ million}$ .