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5.2 Definite Integrals

Last time: if \( f(x) \ge 0 \), then the area underneath \( f(x) \) from \( x = a \) to \( x = b \) could be approximated by a Riemann Sum.

A graph showing a positive curve f(x) with vertical rectangles approximating the area under the curve from a to b.

What if \( f(x) < 0 \)?

For example, \( f(x) = \sin x \) on \( [\pi, \frac{3\pi}{2}] \).

Graph of sin(x) on the interval [pi, 3pi/2] where the curve is below the x-axis, with the region shaded.

Can't talk about area under the curve (it goes to \( -\infty \)).

But we can still calculate the area between the curve and x-axis (see shaded region on left).

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Example

\( f(x) = \sin x \) on \( [\pi, \frac{3\pi}{2}] \)

Right end point, \( n = 2 \) rectangles

Graph of sin(x) on [pi, 3pi/2] with two rectangles using right endpoints to approximate the area below the x-axis.

Right end of first subinterval: \( x = \frac{5\pi}{4} \)

Right end of second subinterval: \( x = \frac{3\pi}{2} \)

Width of each rectangle is \( \frac{b-a}{n} = \frac{\frac{3\pi}{2} - \pi}{2} = \frac{\pi}{4} \)

\[ R_2 = \left( \frac{\pi}{4} \right) \sin\left( \frac{5\pi}{4} \right) + \left( \frac{\pi}{4} \right) \sin\left( \frac{3\pi}{2} \right) \]

Note: width of (1) × height of (1) + (2)

\[ = \left( \frac{\pi}{4} \right) \left( -\frac{\sqrt{2}}{2} \right) + \left( \frac{\pi}{4} \right) (-1) \]

\[ = -1.341 \]
Note the "height" is negative for region below x-axis.
Notice if \( f(x) < 0 \) the area between the curve and x-axis is negative.
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Approximation of the area improves as more rectangles are used.

Example: Area under \( y = x \) on \( [0, 2] \)

Graph of y=x from x=0 to 2, forming a right triangle with base 2 and height 2.

Note: this is a right triangle so we know the true area \( = \frac{1}{2}(2)(2) = 2 \)

Now we'll see how more rectangles can improve approximation. Start with \( n = 2 \), right end point.

Graph of y=x with two right-endpoint rectangles of width 1, heights 1 and 2.
\[ R_2 = (1)(1) + (1)(2) = 3 \]

The terms correspond to rectangle 1 and rectangle 2 respectively.

Try \( R_4 \)

Graph of y=x with four right-endpoint rectangles of width 1/2, heights 1/2, 1, 3/2, and 2.
\[ R_4 = (\frac{1}{2})(\frac{1}{2}) + (\frac{1}{2})(1) + (\frac{1}{2})(\frac{3}{2}) + (\frac{1}{2})(2) \]\[ = 2.5 \]

The terms correspond to rectangles 1, 2, 3, and 4 respectively.

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\[ R_8 = (\frac{1}{4})(\frac{1}{4}) + (\frac{1}{4})(\frac{1}{2}) + (\frac{1}{4})(\frac{3}{4}) + (\frac{1}{4})(1) + (\frac{1}{4})(\frac{5}{4}) + (\frac{1}{4})(\frac{3}{2}) + (\frac{1}{4})(\frac{7}{4}) + (\frac{1}{4})(2) = 2.25 \]

So, as we can see, as \( n \to \infty \), the approx \( \to \) true value.

And, as \( n \to \infty \), the sample point we choose to use (right/left/mid point) becomes irrelevant.

\( \to \) they all lead to the same true value as \( n \to \infty \)

The approximation can be expressed as:

\[ \sum_{k=1}^{n} f(x_k) \Delta_k \]
  • \( f(x_k) \): Sample point (L/R/M)
  • \( \Delta_k = \frac{b-a}{n} \): width of rectangle

It becomes exact as \( n \to \infty \)

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So, \[ \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k) \Delta x = \text{true value of the area} \]

this cumbersome expression can be neatly expressed as a Definite Integral

\[ \lim_{n \to \infty} \sum_{k=1}^{n} f(x_k) \Delta x = \int_{a}^{b} f(x) \, dx \]

this represents the area between the curve \( f(x) \) and the x-axis from \( x = a \) to \( x = b \)

for example, \( \int_{0}^{2} x \, dx \) means the area between \( y = x \) from \( x = 0 \) to \( x = 2 \)

Graph of y=x from x=0 to 2, showing a shaded triangle with area equal to the integral from 0 to 2 of x dx.
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Example

Interpret \( \int_{-1}^{1} x \, dx \) in terms of areas and use them to evaluate

this definite integral is the area between \( y = x \) and x-axis from \( x = -1 \) to \( x = 1 \)

Graph of y=x from x=-1 to 1, showing two triangles labeled 1 and 2, one below and one above the x-axis.

notice \[ \int_{-1}^{1} x \, dx = \text{area of (1)} + \text{area of (2)} \]

both (1) and (2) are triangles

\( (1) : -\frac{1}{2}(1)(1) = -\frac{1}{2} \)

below x-axis

\( (2) : \frac{1}{2}(1)(1) = \frac{1}{2} \)

so, we see \[ \int_{-1}^{1} x \, dx = \underbrace{-\frac{1}{2}}_{(1)} + \underbrace{\frac{1}{2}}_{(2)} = 0 \]

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Integral Properties and Generalization

also notice

\[ \int_{-1}^{1} x \, dx = \underbrace{\int_{-1}^{0} x \, dx}_{\text{①}} + \underbrace{\int_{0}^{1} x \, dx}_{\text{②}} \]

this can be generalized into a useful property

\[ \int_{a}^{b} f(x) \, dx = \underbrace{\int_{a}^{c} f(x) \, dx}_{\text{①}} + \underbrace{\int_{c}^{b} f(x) \, dx}_{\text{②}} \]
Coordinate graph showing a curve f(x) over interval [a, b] divided at x=c into regions 1 and 2.

the whole region: \[ \int_{a}^{b} f(x) \, dx \]

other useful properties:

\[ \int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \]
\[ \int_{a}^{b} [f(x) - g(x)] \, dx = \int_{a}^{b} f(x) \, dx - \int_{a}^{b} g(x) \, dx \]
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Example: Area Under a Linear Function

\[ \text{example } \int_{0}^{3} (x+1) \, dx \]

this is the area between \( y = x + 1 \) and x-axis from \( x = 0 \) to \( x = 3 \)

Graph of y=x+1 from x=0 to x=3 showing a trapezoidal area.

but notice we can chop it up into two pieces

Trapezoid area split into a top triangle labeled 1 and a bottom rectangle labeled 2.

① is a triangle base 3 height 3 which is the same as

\[ \int_{0}^{3} x \, dx \]
Small diagram of a triangle with base 3 and height 3.

② is a rectangle length 3 width 1 same as

\[ \int_{0}^{3} 1 \, dx \]
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The property says

\[ \int_{0}^{3} (x+1) \, dx = \underbrace{\int_{0}^{3} x \, dx}_{\text{triangle part}} + \underbrace{\int_{0}^{3} 1 \, dx}_{\text{rectangle part}} \]

This demonstrates why

\[ \int_{a}^{b} [f(x) + g(x)] \, dx = \int_{a}^{b} f(x) \, dx + \int_{a}^{b} g(x) \, dx \]

is true.

A couple other useful properties

\[ \int_{a}^{b} k \cdot f(x) \, dx = k \int_{a}^{b} f(x) \, dx \]

k is a constant

\[ \int_{a}^{a} f(x) \, dx = 0 \]
\[ \int_{a}^{b} -f(x) \, dx = -\int_{a}^{b} f(x) \, dx \]
\[ \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \]

Can switch places of a and b by making the integral negative.