20.3 Fundamental Theorem of Calculus, Net Signed Area & Geometric Applications

20.3 Fundamental Theorem of Calculus, Net Signed Area & Geometric Applications

Riemann Sums, Partitions & The Definite Integral

The definite integral formalizes the process of continuous accumulation. Let $f$ be a function defined on a closed bounded interval $[a, b]$. A partition $P$ divides $[a, b]$ into $n$ subintervals by choosing points $a = x_0 < x_1 < x_2 < \dots < x_n = b$. The length of the $k$-th subinterval is $\Delta x_k = x_k - x_{k-1}$, and the norm of the partition, denoted $\|P\|$, is the maximum subinterval width: $\|P\| = \max_{1 \le k \le n} \Delta x_k$.

Choosing an arbitrary sample point $x_k^* \in [x_{k-1}, x_k]$ within each subinterval, the Riemann sum of $f$ associated with partition $P$ is:

$$R_n = \sum_{k=1}^n f(x_k^*) \Delta x_k$$

If $f$ is continuous on $[a, b]$ (or piecewise continuous with finitely many jump discontinuities), the limit of Riemann sums as $\|P\| \to 0$ ($n \to \infty$) exists and is independent of the choice of partition or sample points. This limit defines the definite integral:

$$\int_a^b f(x)\,dx = \lim_{\|P\| \to 0} \sum_{k=1}^n f(x_k^*) \Delta x_k$$

where $a$ is the lower limit of integration and $b$ is the upper limit of integration.

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Net Signed Area vs. Total Geometric Area

Geometrically, the definite integral computes the net signed area bounded between the curve $y = f(x)$ and the $x$-axis over $[a, b]$:

• Regions where $f(x) \ge 0$ lie above the $x$-axis and contribute positively to the integral: $A_{\text{above}} > 0$.
• Regions where $f(x) \le 0$ lie below the $x$-axis and contribute negatively to the integral: $A_{\text{below}} < 0$.

$$\int_a^b f(x)\,dx = A_{\text{above}} - A_{\text{below}}$$

If the objective is to find the total geometric area enclosed between the curve and the $x$-axis, all area contributions must be taken as positive values. This requires integrating the absolute value of the function:

$$\text{Total Area} = \int_a^b |f(x)|\,dx$$

To evaluate this manually, one must find all real roots where $f(x) = 0$ on $[a, b]$, partition the integral across these roots, and negate subintegrals where $f(x) < 0$.

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The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) creates the profound bridge between differential calculus and integral calculus, demonstrating that differentiation and integration are inverse processes.

Part 1: Accumulation Functions & Differentiation

If $f$ is continuous on $[a, b]$, the accumulation function defined by:

$$g(x) = \int_a^x f(t)\,dt \quad \text{for } a \le x \le b$$

is continuous on $[a, b]$, differentiable on the open interval $(a, b)$, and its derivative is simply the integrand evaluated at $x$:

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

When the limits of integration are functions of $x$, applying the Chain Rule produces the Extended Leibniz Rule:

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

Part 2: The Evaluation Theorem

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

$$\int_a^b f(x)\,dx = F(b) - F(a) = \left[ F(x) \right]_a^b$$

Part 2 eliminates the need to compute complex limits of Riemann sums, allowing exact calculation of definite integrals via algebraic antiderivatives.

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Algebraic & Structural Properties of Definite Integrals

Definite integrals satisfy fundamental operational rules:

1. Reversal of Bounds: $\int_b^a f(x)\,dx = -\int_a^b f(x)\,dx$.
2. Zero Width Interval: $\int_a^a f(x)\,dx = 0$.
3. Linearity: $\int_a^b [c_1 f(x) + c_2 g(x)]\,dx = c_1 \int_a^b f(x)\,dx + c_2 \int_a^b g(x)\,dx$.
4. Interval Additivity: For any three numbers $a, b, c$: $\int_a^b f(x)\,dx + \int_b^c f(x)\,dx = \int_a^c f(x)\,dx$.
5. Comparison Properties: If $f(x) \ge g(x)$ for all $x \in [a, b]$, then $\int_a^b f(x)\,dx \ge \int_a^b g(x)\,dx$.
6. Symmetry Properties:
   • If $f$ is an even function ($f(-x) = f(x)$): $\int_{-a}^a f(x)\,dx = 2 \int_0^a f(x)\,dx$.
   • If $f$ is an odd function ($f(-x) = -f(x)$): $\int_{-a}^a f(x)\,dx = 0$.

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Definite Integral Properties & Geometric Volume Formulas

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Kinematics: Net Displacement vs. Total Distance

In kinematics, the distinction between signed accumulation and absolute accumulation is critical:

• Net Displacement: The net change in position over time interval $[t_1, t_2]$ is given by $\int_{t_1}^{t_2} v(t)\,dt = s(t_2) - s(t_1)$. Positive velocity (forward motion) and negative velocity (backward motion) cancel out algebraically.
• Total Distance Traveled: The actual length of the path traversed is given by $\int_{t_1}^{t_2} |v(t)|\,dt$. Because distance is cumulative, intervals of negative velocity must be integrated with negated signs so all contributions are positive.

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Planar Area Between Curves

To calculate the area of a bounded region enclosed between two curves $y = f(x)$ and $y = g(x)$:

1. Find intersection points by setting $f(x) = g(x)$ and solving for $x$. These roots establish the integration bounds $[a, b]$.
2. Determine which curve is the upper boundary ($f_{\text{top}}$) and which is the lower boundary ($g_{\text{bottom}}$) on $[a, b]$ by testing an interior test value.
3. Integrate the difference: $$A = \int_a^b [f_{\text{top}}(x) - g_{\text{bottom}}(x)]\,dx$$

When boundaries are expressed more naturally as functions of $y$, integrate horizontally from the left boundary to the right boundary:

$$A = \int_c^d [f_{\text{right}}(y) - g_{\text{left}}(y)]\,dy$$

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Volumes of Solids of Revolution: Disk & Washer Methods

When a two-dimensional planar region is revolved $360^\circ$ around an axis of rotation, it sweeps out a three-dimensional solid of revolution.

The Disk Method

If the revolving region is flush against the axis of rotation, perpendicular cross-sections are solid circular disks with area $A = \pi R^2$. The volume is obtained by integrating cross-sectional circular disks:

• Revolved about the horizontal axis $y = k$: $V = \pi \int_a^b [R(x)]^2\,dx$, where $R(x) = |f(x) - k|$.
• Revolved about the vertical axis $x = h$: $V = \pi \int_c^d [R(y)]^2\,dy$, where $R(y) = |g(y) - h|$.

The Washer Method

If the revolving region has an open space or cavity between itself and the axis of rotation, perpendicular cross-sections are circular washers (annuli). The area of a washer cross-section is $A = \pi R_{\text{outer}}^2 - \pi r_{\text{inner}}^2 = \pi (R^2 - r^2)$:

• Revolved about the $x$-axis: $V = \pi \int_a^b \left([R(x)]^2 - [r(x)]^2\right)\,dx$
• Revolved about the $y$-axis: $V = \pi \int_c^d \left([R(y)]^2 - [r(y)]^2\right)\,dy$

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Worked Exemplar: Area Between Curves & Solid of Revolution

Problem:

1. Determine the exact area of the planar region bounded between the parabola $y = x^2$ and the line $y = 2x + 3$.
2. Compute the volume of the solid generated by revolving the region bounded by $y = \sqrt{x}$, the horizontal line $y = 0$ ($x$-axis), and the vertical line $x = 4$ around the $x$-axis.

Step 1: Find Intersection Points for Area. Set the two equations equal to determine intersection points:

$$x^2 = 2x + 3 \implies x^2 - 2x - 3 = 0 \implies (x - 3)(x + 1) = 0$$

The integration bounds are $x = -1$ and $x = 3$.

Step 2: Establish Boundary Orientation. Test $x = 0 \in (-1, 3)$: line value is $2(0) + 3 = 3$, parabola value is $0^2 = 0$. Because $3 > 0$, the line $y = 2x + 3$ is the upper boundary and the parabola $y = x^2$ is the lower boundary across the entire interval $[-1, 3]$.

Step 3: Integrate to Evaluate Enclosed Area. Apply the area formula:

$$A = \int_{-1}^3 \left[(2x + 3) - x^2\right]\,dx = \int_{-1}^3 (-x^2 + 2x + 3)\,dx$$

Find the antiderivative and evaluate at bounds:

$$A = \left[ -\frac{x^3}{3} + x^2 + 3x \right]_{-1}^3$$

Evaluate at upper bound $x = 3$:

$$\left(-\frac{27}{3} + (3)^2 + 3(3)\right) = -9 + 9 + 9 = 9$$

Evaluate at lower bound $x = -1$:

$$\left(-\frac{(-1)^3}{3} + (-1)^2 + 3(-1)\right) = \frac{1}{3} + 1 - 3 = -\frac{5}{3}$$

Subtract the evaluated bounds:

$$A = 9 - \left(-\frac{5}{3}\right) = \frac{27}{3} + \frac{5}{3} = \frac{32}{3}$$

Step 4: Formulate Volume of Solid of Revolution. For the region bounded by $y = \sqrt{x}$, $y = 0$, and $x = 4$ revolved about the $x$-axis:

• The axis of revolution is the $x$-axis ($y = 0$).
• The region is flush against the $x$-axis from $x = 0$ to $x = 4$.
• The radius of each cross-sectional disk is $R(x) = \sqrt{x} - 0 = \sqrt{x}$.
• Cross-sectional area is $A(x) = \pi [R(x)]^2 = \pi (\sqrt{x})^2 = \pi x$.

Step 5: Integrate for Volume via the Disk Method. Set up and evaluate the definite integral:

$$V = \pi \int_0^4 x\,dx = \pi \left[ \frac{x^2}{2} \right]_0^4 = \pi \left( \frac{4^2}{2} - 0 \right) = \pi \left(\frac{16}{2}\right) = 8\pi$$

The exact volume of the paraboloid solid is $8\pi\text{ cubic units}$.