Lesson 31 (11/11/26)

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Lesson 31 (11/11/26)

Approximating Area under Curves (5.1)

Announcements:

  • Exam 3 on Wednesday (11/18), 8pm (Lesson 21 - 31)
  • Study guide on Brightspace

Office Hours:

MWF 1:30pm - 2:30pm

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Warmup Example:

Suppose you are driving on a straight road for 6 hours at a constant speed of 50 mi/hr. How far are you from the start?

A velocity-time graph showing a horizontal line at velocity v = 50. The x-axis represents time (labeled 'time' at the far right), and the y-axis represents velocity (labeled 'Velocity' at the top). The horizontal line is drawn from time t=0 to t=6. The area under the curve is represented by a rectangular box with shaded diagonal lines, spanning from v=0 to v=50 on the vertical axis and t=0 to t=6 on the horizontal axis.
Visual Description: A velocity-time graph showing a horizontal line at velocity v = 50. The x-axis represents time (labeled 'time' at the far right), and the y-axis represents velocity (labeled 'Velocity' at the top). The horizontal line is drawn from time t=0 to t=6. The area under the curve is represented by a rectangular box with shaded diagonal lines, spanning from v=0 to v=50 on the vertical axis and t=0 to t=6 on the horizontal axis.

Area = 300

\( V(t) = 50 = \) derivative of \( S(t) \)

\( S(t) = \) Antiderivative of \( V(t) \)

\( = 50t + c \)

If \( S(0) = 0 \implies C = 0 \)

\( S(t) = 50t \)

\( S(6) = 50 \times 6 = 300 \)

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Example: Find Area under \(f(x) = -32x + 48\) on \([0, 3/2]\)

A Cartesian coordinate system plot showing the function f(x) = -32x + 48. The y-axis intercept is at 48, and the x-axis intercept is at 3/2. A triangle is formed bounded by the y-axis, the x-axis, and the line representing the function. The interior of this triangle is shaded with diagonal hatch marks.
Visual Description: A Cartesian coordinate system plot showing the function f(x) = -32x + 48. The y-axis intercept is at 48, and the x-axis intercept is at 3/2. A triangle is formed bounded by the y-axis, the x-axis, and the line representing the function. The interior of this triangle is shaded with diagonal hatch marks.

Area = Area of triangle

= \(\frac{1}{2} \times \text{base} \times \text{height}\)

= \(\frac{1}{2} \times \frac{3}{2} \times 48\)

= 36

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Example: Find Area under \( y = \sqrt{1 - x^2} \) on \( [-1, 1] \)

From \( y^2 = 1 - x^2 \), we get \( x^2 + y^2 = 1 \), but \( y = \sqrt{1 - x^2} \geq 0 \).

A Cartesian coordinate system showing a semi-circle centered at the origin (0,0) with radius 1. The curve represents the function y = sqrt(1-x^2), spanning from x = -1 to x = 1 along the x-axis, with its peak at (0,1) on the y-axis. The area under the curve is shaded with diagonal hatch marks. The points (0,1) and (1,0) are explicitly labeled on the graph.
Visual Description: A Cartesian coordinate system showing a semi-circle centered at the origin (0,0) with radius 1. The curve represents the function y = sqrt(1-x^2), spanning from x = -1 to x = 1 along the x-axis, with its peak at (0,1) on the y-axis. The area under the curve is shaded with diagonal hatch marks. The points (0,1) and (1,0) are explicitly labeled on the graph.

Area = Area of Semi-circle of Radius 1

\[ = \frac{1}{2} \times \pi \times 1^2 = \frac{\pi}{2} \]

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Example: Find Area under the curve \( f(x) = 4x - x^2 \) on \( [0, 4] \)

\( f'(x) = 4 - 2x = 0 \implies x = 2 \)

\( x < 2 \implies f'(x) > 0 \uparrow \)

\( x > 2 \implies f'(x) < 0 \downarrow \)

\( f''(x) = -2 < 0 \) (Concave down)

A hand-drawn graph on a grid showing the parabola f(x) = 4x - x^2. The x-axis ranges from 0 to 4, and the y-axis shows a peak at y=4 when x=2. The area under the curve is being approximated by vertical rectangles. The parabola is shown in blue, and the approximating rectangles are drawn in red. A dashed vertical line indicates the axis of symmetry at x=2.
Visual Description: A hand-drawn graph on a grid showing the parabola f(x) = 4x - x^2. The x-axis ranges from 0 to 4, and the y-axis shows a peak at y=4 when x=2. The area under the curve is being approximated by vertical rectangles. The parabola is shown in blue, and the approximating rectangles are drawn in red. A dashed vertical line indicates the axis of symmetry at x=2.

Area cannot be computed from formulas we know

But

\( 0 < \text{Area under } 4x - x^2 < 16 \)

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Idea: Divide into small rectangles and approximate Area

eg: 4 Rectangles

A graph on a Cartesian plane showing a continuous curve resembling a downward-opening parabola starting at the origin (0) and ending on the x-axis at x=4. The area under this curve is approximated using four vertical rectangles of equal width. The rectangles are drawn between x=0 and x=4. The regions inside the rectangles that fall under the curve are shaded with blue diagonal lines. Regions where the rectangles exceed the curve are highlighted in pink, and regions under the curve not covered by the rectangles are highlighted in yellow, illustrating the approximation error of the Riemann sum method.
Visual Description: A graph on a Cartesian plane showing a continuous curve resembling a downward-opening parabola starting at the origin (0) and ending on the x-axis at x=4. The area under this curve is approximated using four vertical rectangles of equal width. The rectangles are drawn between x=0 and x=4. The regions inside the rectangles that fall under the curve are shaded with blue diagonal lines. Regions where the rectangles exceed the curve are highlighted in pink, and regions under the curve not covered by the rectangles are highlighted in yellow, illustrating the approximation error of the Riemann sum method.

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Using 8-Rectangles

A Cartesian coordinate system graph on a grid background showing a downward-opening parabola starting at the origin (0,0) and intersecting the x-axis at x=4. The area under the curve is approximated using 8 vertical rectangles. The x-axis is marked with intervals at 0, 1/2, 1, 3/2, 2, 5/2, 3, 7/2, and 4. The rectangles illustrate a Riemann sum approximation. Some rectangles extend above the parabola curve (filled in magenta), and some fall below the curve (filled in yellow), representing an alternating over-approximation and under-approximation of the area under the curve.
Visual Description: A Cartesian coordinate system graph on a grid background showing a downward-opening parabola starting at the origin (0,0) and intersecting the x-axis at x=4. The area under the curve is approximated using 8 vertical rectangles. The x-axis is marked with intervals at 0, 1/2, 1, 3/2, 2, 5/2, 3, 7/2, and 4. The rectangles illustrate a Riemann sum approximation. Some rectangles extend above the parabola curve (filled in magenta), and some fall below the curve (filled in yellow), representing an alternating over-approximation and under-approximation of the area under the curve.

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Riemann Sum: Approximates area under \( y = f(x) \) on \( [a, b] \) using rectangles.

1. Divide [a, b] into n equal parts

Each has width \( \Delta x = \frac{b-a}{n} \).

2. In each small interval choose a sample point \( x_k^* \) & form rectangle with height \( f(x_k^*) \)

Area of that rectangle \( f(x_k^*) \Delta x \).

A coordinate plane showing a graph of a function y = f(x) curving upward. The region under the curve between points a and b on the x-axis is shaded with blue diagonal lines. Two specific rectangular bars are highlighted in red, representing the area under the curve in small intervals with width Δx. Tick marks are visible along the x-axis between a and b.
Visual Description: A coordinate plane showing a graph of a function y = f(x) curving upward. The region under the curve between points a and b on the x-axis is shaded with blue diagonal lines. Two specific rectangular bars are highlighted in red, representing the area under the curve in small intervals with width Δx. Tick marks are visible along the x-axis between a and b.

3. Approximate Area by Adding Areas of all the small rectangles

Approx Area = \( f(x_1^*) \Delta x + f(x_2^*) \Delta x + f(x_3^*) \Delta x + \dots + f(x_n^*) \Delta x \)

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Approximate Area under \( f(x) = 4x - x^2 \) on \( [0, 4] \) using 4 rectangles and left end points.

Notation: \( L_4 \)

  1. Divide \( [0, 4] \) into 4 equal size rectangles of width \( \Delta x = \frac{4-0}{4} = 1 \).

  2. Sample points

    \( x_1^* = 0 \)
    \( x_2^* = 1 \)
    \( x_3^* = 2 \)
    \( x_4^* = 3 \)

    Heights = \( f(x_1^*) = 0 \)
    \( f(x_2^*) = 3 \)
    \( f(x_3^*) = 4 \)
    \( f(x_4^*) = 3 \)

  3. Approx Area = \( f(x_1^*) \Delta x + f(x_2^*) \Delta x + f(x_3^*) \Delta x + f(x_4^*) \Delta x \)

    = \( 0(1) + 3(1) + 4(1) + 3(1) \)

    = 10

A coordinate plane graph displaying the downward-opening parabola f(x) = 4x - x^2. The x-axis is marked with points 0, 1, 2, 3, and 4. Four rectangles are drawn under the curve to approximate the area using left endpoints. The first rectangle has a height of 0 at x=0. The second rectangle has a width from x=1 to x=2 with height 3. The third rectangle has a width from x=2 to x=3 with height 4. The fourth rectangle has a width from x=3 to x=4 with height 3. The top left corner of each rectangle touches the parabola curve.
Visual Description: A coordinate plane graph displaying the downward-opening parabola f(x) = 4x - x^2. The x-axis is marked with points 0, 1, 2, 3, and 4. Four rectangles are drawn under the curve to approximate the area using left endpoints. The first rectangle has a height of 0 at x=0. The second rectangle has a width from x=1 to x=2 with height 3. The third rectangle has a width from x=2 to x=3 with height 4. The fourth rectangle has a width from x=3 to x=4 with height 3. The top left corner of each rectangle touches the parabola curve.

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Approximate Area under \( f(x) = 4x - x^2 \) on \( [0, 4] \)

using 4 Rectangles and mid points

Notation: \( M_4 \)

A coordinate plane showing the downward-opening parabola f(x) = 4x - x^2. Four red rectangles are drawn beneath the curve to approximate the area. Each rectangle has a width of 1 unit. The height of each rectangle is determined by the function value at the midpoint of the interval (0.5, 1.5, 2.5, 3.5). The x-axis is marked with values 1, 2, and 3. The curve passes through (0,0), (2,4), and (4,0).
Visual Description: A coordinate plane showing the downward-opening parabola f(x) = 4x - x^2. Four red rectangles are drawn beneath the curve to approximate the area. Each rectangle has a width of 1 unit. The height of each rectangle is determined by the function value at the midpoint of the interval (0.5, 1.5, 2.5, 3.5). The x-axis is marked with values 1, 2, and 3. The curve passes through (0,0), (2,4), and (4,0).

1. 4 equal rectangles of width

\[ \Delta x = \frac{4-0}{4} = 1 \]

2. Sample points

\[ x_1^* = 1/2 \]

\[ x_2^* = 3/2 \]

\[ x_3^* = 5/2 \]

\[ x_4^* = 7/2 \]

3. Approx Area:

\[ = f(x_1^*)\Delta x + f(x_2^*)\Delta x + f(x_3^*)\Delta x + f(x_4^*)\Delta x \]

\[ = f(1/2) + f(3/2) + f(5/2) + f(7/2) = 11 \]

Note: Actual = 10.667.

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Divide into smaller rectangles to get better approx.

Example: 16 rectangles, right end point, \( f(x) = 4x - x^2 \) on \( [0, 4] \).

Margin Note: \( R_{16} \)

  1. Width of each small rectangle \( = \frac{4-0}{16} = \frac{1}{4} = 0.25 \)

  2. Sample points:

    \( x_1^* = 0.25 \)

    \( x_2^* = 0.25 \times 2 \)

    \( x_3^* = 0.25 \times 3 \)

    \( \vdots \)

    \( x_{16}^* = 0.25 \times 16 = 4 \)

  3. Add areas to get approx area

    Approx Area \( = R_{16} = f(x_1^*) \Delta x + \dots + f(x_{16}^*) \Delta x = 10.625 \)

    Margin Note: Actual \( 10.667 \)

    \( = \sum_{k=1}^{16} f(x_k^*) \Delta x \)

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Summation Notation

\[ \sum_{k=1}^{n} a_k = a_1 + a_2 + \dots + a_n \]

Using this notation: \( n^{\text{th}} \) Riemann Sum for \( f(x) \) on \( [a, b] \)

  1. width \( \Delta x = \frac{b-a}{n} \)

  2. Sample points = \( x_1^*, x_2^*, \dots, x_n^* \)

Approx Area = \[ \sum_{k=1}^{n} f(x_k^*) \Delta x \]

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Some Important Summation Formulas

1. \(\sum_{k=1}^{n} k = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}\)

\(\sum_{k=1}^{6} k = 1 + 2 + 3 + 4 + 5 + 6 = 21 = \frac{6(6+1)}{2} = 21\)

\(\sum_{k=1}^{100} k = \frac{(100)(101)}{2} = 5050\)

2. \(\sum_{k=1}^{n} k^2 = 1^2 + 2^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6}\)

\(1^2 + 2^2 + 3^2 = \frac{3(3+1)(6+1)}{6} = 14\)

3. \(\sum_{k=1}^{n} k^3 = 1^3 + 2^3 + \dots + n^3 = \frac{n^2(n+1)^2}{4}\)

\(1^3 + 2^3 + 3^3 + 4^3 = \frac{4^2 \cdot 5^2}{4} = 100\)

\(1 + 8 + 27 + 64\)