14.6: Definite Integrals Notes

14.6: Definite Integrals

Definition of Definite Integral

The definite integral of a function $f(x)$ from $a$ to $b$, denoted as $\int_{a}^{b} f(x) dx$, represents the area under the curve $y = f(x)$ between the lines $x = a$ and $x = b$. Here,

• $a$ and $b$ are called the "bounds" or "limits of integration."

• $a$ is the "lower bound" or "lower limit."

• $b$ is the "upper bound" or "upper limit."

$\int_{a}^{b} f(x) dx$ is called a "definite integral."

Approximating Definite Integrals

Definite integrals are calculated by approximating the area under the curve using rectangles.

• Lower Sum: An underestimate of the area.

• Upper Sum: An overestimate of the area.

Approximating Sums

These approximating sums have the form:

$f(x<em>1) \Delta x + f(x</em>2) \Delta x + \ldots + f(x<em>n) \Delta x = \sum</em>{i=1}^{n} f(x_i) \Delta x$

The $x_i$ are called "sample points"; they are the points where the height of the rectangle is measured.

• Right Endpoints: $x_i$ is the right edge of the rectangle.

• Left Endpoints: $x_i$ is the left edge of the rectangle.

Example: Approximating $\int_{0}^{1} 2x dx$

Consider the function $f(x) = 2x$ with $a = 0$ and $b = 1$. Let's find $\int<em>{0}^{1} f(x) dx = \int</em>{0}^{1} 2x dx$.

Dividing the interval into $n$ subintervals gives:

$x<em>0 = 0, x</em>1 = \frac{1}{n}, x<em>2 = \frac{2}{n}, \ldots, x</em>n = \frac{n}{n} = 1$

The width of the rectangles is $\Delta x = \frac{1}{n}$.

The heights of the rectangles are:

$f(x<em>1) = 2 \cdot \frac{1}{n} = \frac{2}{n}, f(x</em>2) = 2 \cdot \frac{2}{n} = \frac{4}{n}, \ldots, f(x_n) = 2 \cdot \frac{n}{n} = 2$

So the sum using right endpoints is:

$<br>S<em>n = \sum</em>{i=1}^{n} f(x_i) \Delta x = \frac{2}{n} \cdot \frac{1}{n} + \frac{4}{n} \cdot \frac{1}{n} + \ldots + \frac{2n}{n} \cdot \frac{1}{n} = \frac{2}{n^2} (1 + 2 + \ldots + n)<br>$

Using the arithmetic progression formula:

$1 + 2 + \ldots + n = \frac{n(n+1)}{2}$

We have:

$<br>S_n = \frac{2}{n^2} \cdot \frac{n(n+1)}{2} = \frac{n(n+1)}{n^2} = \frac{n^2 + n}{n^2} = \frac{n+1}{n}<br>$

Thus,

$\int<em>{0}^{1} 2x dx = \lim</em>{n \to \infty} S<em>n = \lim</em>{n \to \infty} \frac{n+1}{n} = \lim_{n \to \infty} \frac{n}{n} = 1$

Key Facts

• $\int_{a}^{b} f(x) dx$ is the limit of approximating sums.

• Larger values of $n$ give better approximations.

• Changing the sample points can affect whether the sum is an upper or lower sum.

Examples

Example 1: Left and Right Sums

Suppose $f(1) = 3$, $f(2) = 4$, $f(3) = 3$, and $f(4) = 2$. Find left and right sums approximating $\int_{1}^{4} f(x) dx$.

• $\Delta x = 1$

• Left Sum: $f(1) \cdot \Delta x + f(2) \cdot \Delta x + f(3) \cdot \Delta x = 3 \cdot 1 + 4 \cdot 1 + 3 \cdot 1 = 10$

• Right Sum: $f(2) \cdot \Delta x + f(3) \cdot \Delta x + f(4) \cdot \Delta x = 4 \cdot 1 + 3 \cdot 1 + 2 \cdot 1 = 9$

Example 2: Another Left and Right Sum

Suppose $g(1) = 0$, $g(\frac{3}{2}) = 1$, $g(2) = 2$, $g(\frac{5}{2}) = 4$, and $g(3) = 8$. Find left and right sums approximating $\int_{1}^{3} g(x) dx$.

• $\Delta x = \frac{1}{2}$

• Left Sum: $g(1) \cdot \Delta x + g(\frac{3}{2}) \cdot \Delta x + g(2) \cdot \Delta x + g(\frac{5}{2}) \cdot \Delta x = 0 \cdot \frac{1}{2} + 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} + 4 \cdot \frac{1}{2} = \frac{7}{2}$

• Right Sum: $g(\frac{3}{2}) \cdot \Delta x + g(2) \cdot \Delta x + g(\frac{5}{2}) \cdot \Delta x + g(3) \cdot \Delta x = 1 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} + 4 \cdot \frac{1}{2} + 8 \cdot \frac{1}{2} = \frac{15}{2}$

Example 3: Car's Speedometer

A car's speedometer reads 20 mph at noon, 30 mph at 12:30 pm, 20 mph at 1 pm, and 40 mph at 1:30 pm. Estimate the distance traveled using left and right sums.

• $\Delta t = \frac{1}{2}$ (half an hour)

• Left Sum: $20 \cdot \frac{1}{2} + 30 \cdot \frac{1}{2} + 20 \cdot \frac{1}{2} = 10 + 15 + 10 = 35$

• Right Sum: $30 \cdot \frac{1}{2} + 20 \cdot \frac{1}{2} + 40 \cdot \frac{1}{2} = 15 + 10 + 20 = 45$

Note: $30 \cdot \frac{1}{2}$ means "half an hour at 30 mph."

Example 4: Approximating $\int_{2}^{4} x^2 dx$

Approximate $\int_{2}^{4} x^2 dx$ using left and right sums with $n = 3$.

First, divide $[2, 4]$ into 3 subintervals:

$x<em>0 = 2, x</em>1 = \frac{8}{3}, x<em>2 = \frac{10}{3}, x</em>3 = 4$ and $\Delta x = \frac{2}{3}$

Next, evaluate $f(x) = x^2$ at each point:

$f(x<em>0) = 4, f(x</em>1) = \frac{64}{9}, f(x<em>2) = \frac{100}{9}, f(x</em>3) = 16$

• Left Sum:

$<br>f(x<em>0) \cdot \Delta x + f(x</em>1) \cdot \Delta x + f(x_2) \cdot \Delta x = 4 \cdot \frac{2}{3} + \frac{64}{9} \cdot \frac{2}{3} + \frac{100}{9} \cdot \frac{2}{3} = \frac{8}{3} + \frac{128}{27} + \frac{200}{27} = \frac{72 + 128 + 200}{27} = \frac{400}{27}<br>$

• Right Sum:

$<br>f(x<em>1) \cdot \Delta x + f(x</em>2) \cdot \Delta x + f(x_3) \cdot \Delta x = \frac{64}{9} \cdot \frac{2}{3} + \frac{100}{9} \cdot \frac{2}{3} + 16 \cdot \frac{2}{3} = \frac{128}{27} + \frac{200}{27} + \frac{32}{3} = \frac{128 + 200 + 288}{27} = \frac{616}{27}<br>$