Calculus: Definite Integrals and Integration by Parts

Definition and Calculus Fundamentals: A definite integral is an integral with specific bounds, denoted as $\int_a^b f(x) \, dx$ . The Fundamental Theorem of Calculus states that to evaluate this, one finds the antiderivative $F(x)$ and calculates the difference between the evaluation at the upper and lower bounds: $F(b) - F(a)$ .
Conceptual Representation: Unlike indefinite integrals, which result in a family of functions, a definite integral yields a single numerical value. This value represents the net signed area beneath the curve of the function from point $a$ to point $b$ .
Integration by Parts (IBP) Formula for Definite Integrals: When applying IBP to definite integrals, the formula is modified to account for the bounds at each step: $\int_a^b u \, dv = [uv]_a^b - \int_a^b v \, du$

Problem Statement: Evaluate $\int_0^{\pi/2} x \cos(x) \, dx$ .
Choosing Variables (LIATE/Heuristics):
- $u = x$ : This is chosen because its derivative ( $du = 1 \, dx$ ) is simpler than the original function.
- $dv = \cos(x) \, dx$ : This is chosen because its antiderivative ( $v = \sin(x)$ ) is easy to compute.
Applying the Formula: $\int_0^{\pi/2} x \cos(x) \, dx = [x \sin(x)]_0^{\pi/2} - \int_0^{\pi/2} \sin(x) \, dx$
Step-by-Step Evaluation:
- Evaluating the first part: $(\frac{\pi}{2} \sin(\frac{\pi}{2})) - (0 \sin(0)) = (\frac{\pi}{2} \times 1) - 0 = \frac{\pi}{2}$ .
- Finding the antiderivative of the second part: The antiderivative of $\sin(x)$ is $-\cos(x)$ .
- Applying the negative from the formula: $-(-\cos(x)) = \cos(x)$ .
- Evaluating $\cos(x)$ from $0$ to $\frac{\pi}{2}$ : $\cos(\frac{\pi}{2}) - \cos(0) = 0 - 1 = -1$ .
Final Result: $\frac{\pi}{2} - 1$ .
Observations: When sines and cosines are at their maximums/minimums at boundary points (like $0$ and $\frac{\pi}{2}$ ), one function typically evaluates to zero while the other evaluates to one (or vice-versa).

Problem Statement: Evaluate $\int_1^e \ln(x) \, dx$ .
Analysis: This function does not initially look like a product. However, because we do not have an elementary antiderivative for $\ln(x)$ , we treat it as a product of $\ln(x)$ and $1$ .
Selection:
- $u = \ln(x)$ , so $du = \frac{1}{x} \, dx$ .
- $dv = 1 \, dx$ , so $v = x$ .
Calculation:
- $[x \ln(x)]_1^e - \int_1^e x \left(\frac{1}{x}\right) \, dx$
- The term $x \left(\frac{1}{x}\right)$ simplifies to $1$ .
- Evaluating the first term: $(e \ln(e)) - (1 \ln(1))$ . Since $\ln(e) = 1$ and $\ln(1) = 0$ , this simplifies to $e - 0 = e$ .
- Integrating the second term: $\int_1^e 1 \, dx = [x]_1^e = e - 1$ .
Final Subtraction: $e - (e - 1) = 1$ .

Problem Statement: An integral involving $\ln(x^2)$ .
Algebraic Trick: Before applying IBP, use logarithmic properties to simplify the integrand: $\ln(x^r) = r \ln(x)$ . Thus, $\ln(x^2)$ can be rewritten as $2 \ln(x)$ .
Outcome: Moving the constant to the outside results in $2 \int \ln(x) \, dx$ , which is the same as the previous example multiplied by two. This is significantly more efficient than treating the squared term with the chain rule during differentiation.

Case Study: $\int_{-\pi}^{\pi} x \sin(x) \, dx$ .
Mathematical Principles:
- Even Functions: Satisfy $f(x) = f(-x)$ . They are symmetrical across the y-axis. The integral of an even function over boundary $[-a, a]$ is $2 \int_0^a f(x) \, dx$ .
- Odd Functions: Satisfy $f(-x) = -f(x)$ . They have rotational symmetry about the origin. The integral over boundary $[-a, a]$ for an odd function is always $0$ .
Evaluation of the Example:
- $x$ is an odd function.
- $\sin(x)$ is an odd function.
- The product of two odd functions is an even function. Therefore, the area should be twice the area from $0$ to $\pi$ , not zero.
Integration Results:
- Using $u = x$ and $dv = \sin(x) \, dx$
- $[-x \cos(x)]_{-\pi}^{\pi} + [\sin(x)]_{-\pi}^{\pi}$
- $(-\pi \cos(\pi) - (-(-\pi) \cos(-\pi))) + (0 - 0)$
- $(-\pi(-1) - (\pi(-1))) = \pi + \pi = 2\pi$ .

Power Reduction: For integrals like $\int x^n \sin(x) \, dx$ or $\int x^n e^x \, dx$ , the process of IBP must be repeated $n$ times (the power of the polynomial) to eventually eliminate the polynomial term.
Evaluation Strategy:
- Method A (Step-by-Step): Evaluate limits at every intermediate step. This ofen saves space and identifies terms that cancel out to zero early on.
- Method B (Wait-until-the-end): Keep everything as a single large expression and plug in the bounds once at the very end. This carries a higher risk of algebraic errors due to complex distribution of signs and constants.

Methodology:
- Choose $u = \ln(x^2 + 1)$ , so $du = \frac{2x}{x^2 + 1} \, dx$ .
- Choose $dv = dx$ , so $v = x$ .
Intermediate Result: $x \ln(x^2 + 1) - \int \frac{2x^2}{x^2 + 1} \, dx$ .
Solving the Rational Integral:
- Perform a "clever trick" by rewriting the numerator $\frac{x^2}{x^2 + 1}$ as $\frac{x^2 + 1 - 1}{x^2 + 1}$ .
- This splits into $\int (1 - \frac{1}{x^2 + 1}) \, dx$ .
- The antiderivative of $1$ is $x$ , and the antiderivative of $\frac{1}{x^2 + 1}$ is $\arctan(x)$ .
Finalizing Solutions: In many cases (like bound $3$ ), $\arctan(3)$ does not have a