Definite Integrals: Definition and Fundamental Theorems

Definite Integrals as the Limit of a Sum

While indefinite integrals represent a family of functions (antiderivatives), definite integrals represent a specific numerical value. Historically, the definite integral was conceived to solve the problem of finding the area of regions bounded by curves. This concept is formalized through the idea of approximating the area using sums of rectangles and taking a limit.

Concept: The Area Problem

Consider a non-negative function $y = f(x)$ that is continuous over a closed interval $[a, b]$. We want to find the area of the region bounded by the graph of $f(x)$, the x-axis, and the vertical lines $x=a$ and $x=b$. While simple geometric formulas exist for areas of polygons, this region has a curved boundary $y=f(x)$.

The fundamental approach to finding this area is to approximate it using shapes whose areas we know, namely rectangles.

Approximation using Rectangles (Riemann Sums)

The strategy to approximate the area under the curve is to divide the interval $[a, b]$ into many small subintervals and construct rectangles over these subintervals. The sum of the areas of these rectangles can approximate the area under the curve.

1. Partition the Interval: Divide the closed interval $[a, b]$ into $n$ smaller subintervals of equal width. The width of each subinterval, denoted by $\Delta x$ (or $h$), is given by $\Delta x = \frac{b-a}{n}$. Let the endpoints of these subintervals be $x_0, x_1, x_2, \dots, x_n$, where $x_0 = a$, $x_n = b$, and $x_i = a + i \Delta x$ for $i = 1, 2, \dots, n$. The $i$-th subinterval is $[x_{i-1}, x_i]$.
2. Construct Rectangles: For each subinterval $[x_{i-1}, x_i]$, choose a sample point $x_i^*$. This point can be any value within the subinterval, such as the left endpoint ($x_i^* = x_{i-1}$), the right endpoint ($x_i^* = x_i$), the midpoint ($x_i^* = (x_{i-1} + x_i)/2$), or any other point in the interval. Construct a rectangle over the $i$-th subinterval with width $\Delta x$ and height determined by the function value at the sample point, $f(x_i^*)$.
3. Sum the Areas: The area of the $i$-th rectangle is $A_i = \text{width} \times \text{height} = \Delta x \cdot f(x_i^*) = f(x_i^*) \Delta x$. The sum of the areas of all $n$ rectangles gives an approximation of the total area under the curve:

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

   This sum is called a Riemann Sum, named after mathematician Bernhard Riemann. Different choices for the sample points $x_i^*$ lead to different types of Riemann sums (e.g., Left Riemann Sum, Right Riemann Sum, Midpoint Riemann Sum).

The accuracy of this approximation increases as the number of rectangles $n$ increases (and consequently, as the width of the rectangles $\Delta x$ decreases).

Definition of the Definite Integral

To obtain the exact area under the curve, we take the limit of the Riemann sum as the number of subintervals $n$ approaches infinity. If this limit exists, it defines the definite integral of $f(x)$ from $a$ to $b$.

Definition: The definite integral of a function $f(x)$ from $a$ to $b$ is defined as the limit of the Riemann Sum as the number of subintervals $n$ approaches infinity (or equivalently, as the width of the subintervals $\Delta x$ approaches zero), provided this limit exists and is independent of the choice of sample points $x_i^*$.

$\int_{a}^{b} f(x) dx = \lim\limits_{n \to \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x$

where $\Delta x = \frac{b-a}{n}$ and $x_i^*$ is any point in the $i$-th subinterval $[x_{i-1}, x_i]$.

Terminology associated with the definite integral $\int_{a}^{b} f(x) dx$:

• $\int$ is the integral sign (same as for the indefinite integral).
• $a$ is the lower limit of integration (the starting point of the interval).
• $b$ is the upper limit of integration (the ending point of the interval).
• $f(x)$ is the integrand (the function being integrated).
• $dx$ indicates the variable of integration.

The value of a definite integral $\int_{a}^{b} f(x) dx$ is a single real number. This is a key difference from the indefinite integral, which is a family of functions.

Interpretation as Net Signed Area: If the function $f(x)$ takes on negative values over the interval $[a, b]$, the definite integral represents the "net signed area". Areas of regions above the x-axis are counted as positive, and areas of regions below the x-axis are counted as negative.

For the definite integral to exist, the function $f$ needs to be "well-behaved" over the interval $[a, b]$. A sufficient condition is that $f$ is continuous on $[a, b]$. If $f$ is continuous, the limit of the Riemann sum is guaranteed to exist.

Fundamental Theorems of Integral Calculus (Evaluation of Definite Integrals)

Calculating definite integrals directly from the definition as the limit of a Riemann sum is a rigorous process but is extremely difficult and time-consuming for most functions. The remarkable connection between differentiation and integration, revealed by the Fundamental Theorems of Calculus, provides a much more practical method for evaluating definite integrals.

First Fundamental Theorem of Calculus (FTC1)

This theorem establishes that differentiation and integration are essentially inverse processes. It connects the definite integral to the concept of an antiderivative.

Statement: If $f$ is a continuous function on a closed interval $[a, b]$, then the function $G$ defined as the definite integral from $a$ to a variable upper limit $x$,

is continuous on $[a, b]$, differentiable on the open interval $(a, b)$, and its derivative is given by:

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

Interpretation: The function $G(x)$ represents the accumulated area under the curve $f(t)$ from a fixed lower limit $a$ up to a variable upper limit $x$. The First Fundamental Theorem states that the rate at which this accumulated area changes as the upper limit $x$ changes is equal to the value of the function $f(x)$ at that upper limit. In simpler terms, differentiating the definite integral with a variable upper limit "recovers" the original integrand function evaluated at that upper limit. This shows that finding the area accumulation function is the inverse of differentiation.

Second Fundamental Theorem of Calculus (FTC2) - The Evaluation Theorem

This is the theorem that is most commonly used to evaluate definite integrals in practice. It provides a direct method using antiderivatives.

Statement: If $f$ is a continuous function on the closed interval $[a, b]$, and $F$ is **any** antiderivative of $f$ on $[a, b]$ (meaning $F'(x) = f(x)$ for all $x$ in $[a, b]$), then the definite integral of $f$ from $a$ to $b$ can be evaluated as the difference between the antiderivative evaluated at the upper limit and the antiderivative evaluated at the lower limit.

$\int_{a}^{b} f(x) dx = F(b) - F(a)$

Notation: The difference $F(b) - F(a)$ is often written using the notation $[F(x)]_{a}^{b}$ or $F(x) \bigg|_{a}^{b}$.

So, the theorem is frequently stated as:

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

Significance and How to Use FTC2: This theorem is incredibly significant because it connects the definite integral (defined as a limit of sums related to area) directly to the concept of an antiderivative. It tells us that to find the value of a definite integral, we do not need to compute the limit of a Riemann sum. Instead, we can use the process of antidifferentiation.

The steps for evaluating a definite integral $\int_{a}^{b} f(x) dx$ using FTC2 are:

1. Find any antiderivative of the integrand $f(x)$. Let this be $F(x)$. (When finding an antiderivative for a definite integral, you can choose any one, typically by setting the constant of integration $C=0$, as the constant will cancel out in $F(b)-F(a)$).
2. Evaluate this antiderivative $F(x)$ at the upper limit of integration, $b$. This gives the value $F(b)$.
3. Evaluate the same antiderivative $F(x)$ at the lower limit of integration, $a$. This gives the value $F(a)$.
4. Subtract the value at the lower limit from the value at the upper limit: $F(b) - F(a)$. This difference is the value of the definite integral.

This theorem is the cornerstone of practical definite integral evaluation.

Answer:

Definite Integrals as Area under the Curve (Geometric Interpretation)

The definite integral, defined as the limit of a Riemann sum, has a powerful and intuitive geometric interpretation, particularly when the integrand is a non-negative function. It represents the area of the region under the curve. When the function takes on negative values, the interpretation extends to the concept of net signed area.

Area Interpretation (for Non-Negative Functions)

If $f(x)$ is a continuous and non-negative function (i.e., $f(x) \ge 0$) on the closed interval $[a, b]$, then the definite integral $\int_{a}^{b} f(x) dx$ has a direct geometric meaning:

The value of the definite integral $\int_{a}^{b} f(x) dx$ is exactly equal to the area of the region bounded by:

• The graph of the function $y = f(x)$,
• The x-axis (the line $y=0$),
• The vertical line $x = a$ (the left boundary), and
• The vertical line $x = b$ (the right boundary).

This interpretation is a direct consequence of the definition of the definite integral as the limit of Riemann sums, where each $f(x_i^*) \Delta x$ term represents the area of a rectangle approximating a part of the region, and the limit sums these areas exactly.

Net Signed Area (for Functions with Positive and Negative Values)

If the function $f(x)$ takes on both positive and negative values over the interval $[a, b]$, the definite integral $\int_{a}^{b} f(x) dx$ no longer represents the total geometric area. Instead, it represents the net signed area.

The definite integral calculates the sum of the areas of the regions between the curve and the x-axis, but areas of regions that lie above the x-axis are counted as positive contributions, and areas of regions that lie below the x-axis are counted as negative contributions.

$\int_{a}^{b} f(x) dx = (\text{Area above x-axis}) - (\text{Area below x-axis})$

Let's say the interval $[a, b]$ is divided into subintervals where $f(x)$ is positive and subintervals where $f(x)$ is negative. Let $A_{above}$ be the total area of the regions where $f(x) > 0$, and $A_{below}$ be the total area of the regions where $f(x) < 0$ (where $A_{below}$ is taken as a positive value representing the magnitude of the area below the axis). Then:

To find the total geometric area (where all areas are counted as positive) between the curve $y=f(x)$ and the x-axis from $a$ to $b$ when $f(x)$ changes sign, you need to integrate the absolute value of the function:

Total Geometric Area $= \int_{a}^{b} |f(x)| dx$

Calculating $\int_{a}^{b} |f(x)| dx$ typically involves finding the roots of $f(x)$ within the interval $(a, b)$, splitting the integral into a sum of integrals over sub-intervals where $f(x)$ has a constant sign, and taking the absolute value of each resulting integral before summing them up.

Answer:

Geometric Interpretation:

The integrand is $f(x) = \sin x$. We are integrating over the interval $[0, \pi]$. Let's consider the graph of $y = \sin x$ on this interval.

On the interval $[0, \pi]$, the values of $\sin x$ are non-negative ($\sin x \ge 0$). Since the function is non-negative and continuous on the closed interval $[0, \pi]$, the definite integral $\int_{0}^{\pi} \sin x dx$ represents the exact geometric area of the region bounded by the curve $y = \sin x$, the x-axis, the vertical line $x=0$ (the y-axis), and the vertical line $x=\pi$. This region is the familiar arch of the sine curve above the x-axis from 0 to $\pi$.

Evaluation using FTC2:

We evaluate the definite integral using the Second Fundamental Theorem of Calculus, $\int_{a}^{b} f(x) dx = F(b) - F(a)$, where $F(x)$ is any antiderivative of $f(x) = \sin x$.

An antiderivative of $\sin x$ is $F(x) = -\cos x$ (since $\frac{d}{dx}(-\cos x) = \sin x$).

The lower limit is $a=0$ and the upper limit is $b=\pi$.

"$\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi}$"

[Apply FTC2]

Evaluate the antiderivative at the upper limit ($\pi$) and the lower limit (0) and subtract:

"$= (-\cos \pi) - (-\cos 0)$"

[Evaluate at limits and subtract]

We know that $\cos \pi = -1$ and $\cos 0 = 1$.

$= (-(-1)) - (-1)$"

[Substitute values]

$= 1 - (-1) = 1 + 1 = 2$"

The value of the definite integral is 2.

Conclusion: The definite integral $\int_{0}^{\pi} \sin x dx$ represents the area under the curve $y=\sin x$ from $x=0$ to $x=\pi$. The value of this area is 2 square units.