Integration (Riemann sums, Fundamental Theorem of Calculus, u-substitution) Knowledge-Based Study

Study notes

1. The Area Problem and Partitions

   A fundamental goal of integral calculus is finding the exact area under a continuous curve $f(x)$ over a closed interval $[a, b]$, where $f(x) \ge 0$. We approximate this region by dividing $[a, b]$ into $n$ subintervals of equal width:

   $$\Delta x = \frac{b - a}{n}$$

   The endpoints of these subintervals are given by $x_i = a + i\Delta x$ for $i = 0, 1, 2, \dots, n$.

   Riemann Sum Definition

   By selecting a sample point $x_i^*$ in each subinterval $[x_{i-1}, x_i]$, we construct rectangles with base $\Delta x$ and height $f(x_i^*)$. The total approximate area is given by the Riemann sum:

   $$S_n = \sum_{i=1}^{n} f(x_i^*) \Delta x$$

   Common choices for $x_i^*$ include:

   • Left Riemann Sum: $x_i^* = x_{i-1}$
   • Right Riemann Sum: $x_i^* = x_i$
   • Midpoint Rule: $x_i^* = \frac{x_{i-1} + x_i}{2}$

   From Sums to Definite Integrals

   As the number of subdivisions $n$ approaches infinity ($\Delta x o 0$), the approximation approaches the exact net signed area. If $f$ is continuous on $[a, b]$, this limit exists and defines the definite integral:

   $$\int_{a}^{b} f(x) \, dx = \lim_{n o \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x$$

   Connecting Differentiation and Integration

   The Fundamental Theorem of Calculus (FTC) bridges the two central branches of calculus: differential calculus (slopes) and integral calculus (accumulated area).

   FTC Part 1: The Accumulation Function

   Let $f$ be a continuous function on $[a, b]$. The accumulation function defined by:

   $$g(x) = \int_{a}^{x} f(t) \, dt$$

   is continuous on $[a, b]$, differentiable on $(a, b)$, and its derivative is:

   $$g'(x) = \frac{d}{dx} \left[ \int_{a}^{x} f(t) \, dt \right] = f(x)$$

   Chain Rule Extension: If the upper limit is a differentiable function $u(x)$, then:

   $$\frac{d}{dx} \left[ \int_{a}^{u(x)} f(t) \, dt \right] = f(u(x)) \cdot u'(x)$$

   FTC Part 2: The Evaluation Theorem

   If $f$ is continuous on $[a, b]$ and $F$ is any antiderivative of $f$ such that $F'(x) = f(x)$, then:

   $$\int_{a}^{b} f(x) \, dx = F(b) - F(a) = \Big[ F(x) \Big]_{a}^{b}$$

   This allows us to evaluate definite integrals algebraically without computing limits of Riemann sums.

   The Chain Rule in Reverse

   Many integrands are composite functions multiplied by the derivative of the inner function. Integration by substitution reverses the differentiation chain rule:

   $$\frac{d}{dx} \Big[ F(g(x)) \Big] = F'(g(x)) \cdot g'(x) = f(g(x)) \cdot g'(x)$$

   Integrating both sides yields the substitution formula:

   $$\int f(g(x)) g'(x) \, dx = \int f(u) \, du = F(u) + C$$

   Step-by-Step Procedure

   1. Identify the inner function: Choose $u = g(x)$ such that its differential $du = g'(x) \, dx$ appears in the integrand (up to a constant multiple).
   2. Compute the differential: Find $du = g'(x) \, dx$ and solve for $dx$ or directly rewrite $g'(x) \, dx$ as $du$.
   3. Rewrite entirely in terms of $u$: Transform the entire integral into $\int f(u) \, du$. Ensure no terms with $x$ remain.
   4. Evaluate the integral: Find the antiderivative with respect to $u$.
   5. Back-substitute: Replace $u$ with $g(x)$ to express the final answer in terms of the original variable $x$.

   Handling Limits of Integration

   When applying $u$-substitution to a definite integral, there are two valid approaches:

   • Method 1 (Changing Limits): Transform the bounds $x = a$ and $x = b$ into $u = g(a)$ and $u = g(b)$:
     $$\int_{a}^{b} f(g(x)) g'(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, du$$
     This avoids back-substituting to $x$ at the end.
   • Method 2 (Back-Substitution): Evaluate the indefinite integral in terms of $x$ first, then apply the original bounds $a$ and $b$.

   Worked Example 1: FTC Part 1 with Chain Rule

   Find $\frac{d}{dx} \left[ \int_{1}^{x^3} \cos(t) \, dt \right]$.

   Using FTC Part 1 and the Chain Rule with $u = x^3$ and $\frac{du}{dx} = 3x^2$:

   $$\frac{d}{dx} \left[ \int_{1}^{x^3} \cos(t) \, dt \right] = \cos(x^3) \cdot \frac{d}{dx}(x^3) = 3x^2 \cos(x^3)$$

   Worked Example 2: Definite Integral via Substitution

   Evaluate $\int_{0}^{2} 2x \sqrt{x^2 + 1} \, dx$.

   Let $u = x^2 + 1$. Then $du = 2x \, dx$.
   Update the limits of integration:

   • When $x = 0$, $u = 0^2 + 1 = 1$.
   • When $x = 2$, $u = 2^2 + 1 = 5$.

   Transform and evaluate:

   $$\int_{1}^{5} \sqrt{u} \, du = \int_{1}^{5} u^{1/2} \, du = \left[ \frac{2}{3} u^{3/2} \right]_{1}^{5} = \frac{2}{3} \left( 5^{3/2} - 1^{3/2} \right) = \frac{2}{3}(5\sqrt{5} - 1)$$