Math / Calculus

Question9 AM Sun Nov 17 AA webassign.net HSCI130 Fall 2024 BURNABY - CI... Cengage Learning OA10 - MATH 154, [-/13 Points] DETAILS MY NOTES SBIOCALC1 5.1.005. (a) Estimate the area under the graph of $f(x)=1+4 x^{2}$ from $x=-1$ to $x=2$ using three rectangles and right endpoints. $R_{3}=$ $\square$ Then improve your estimate by using six rectangles. $R_{6}=$ $\square$ Sketch the curve and the approximating rectangles for $R_{3}$ . $\begin{array}{r} y \\ 15 \end{array}$

Studdy Solution

STEP 1

What is this asking? We need to estimate the area under a curve using rectangles, first with three and then with six, and then visually show what the three-rectangle estimate looks like. Watch out! Make sure to use the right endpoints of each rectangle to determine its height.
Also, remember that the width of the rectangles changes when we go from three to six rectangles!

STEP 2

1. Calculate the width of the rectangles for $R_3$ .
2. Calculate $R_3$ .
3. Calculate the width of the rectangles for $R_6$ .
4. Calculate $R_6$ .
5. Sketch the curve and the approximating rectangles for $R_3$ .

STEP 3

The total width of the area we're looking at goes from $x = -1$ to $x = 2$ , which is a width of $2 - (-1) = 2 + 1 = \mathbf{3}$ .

STEP 4

We're using three rectangles, so the width of each rectangle is the total width divided by the number of rectangles: $\frac{3}{3} = \mathbf{1}$ .

STEP 5

Since we're using right endpoints, the $x$ -values we'll plug into our function are $x = 0$ , $x = 1$ , and $x = 2$ .
The width of each rectangle is $\mathbf{1}$ , as we found earlier.

STEP 6

Let's calculate the height of each rectangle.
For $x = 0$ , $f(0) = 1 + 4 \cdot (0)^2 = 1$ .
For $x = 1$ , $f(1) = 1 + 4 \cdot (1)^2 = 5$ .
For $x = 2$ , $f(2) = 1 + 4 \cdot (2)^2 = 17$ .

STEP 7

Now, let's find the area of each rectangle by multiplying its width by its height, and then add them up!
The first rectangle has area $1 \cdot 1 = 1$ .
The second rectangle has area $1 \cdot 5 = 5$ .
The third rectangle has area $1 \cdot 17 = 17$ .
So, $R_3 = 1 + 5 + 17 = \mathbf{23}$ .

STEP 8

The total width is still $\mathbf{3}$ .

STEP 9

Now we have six rectangles, so the width of each is $\frac{3}{6} = \mathbf{0.5}$ .

STEP 10

With right endpoints and a width of $\mathbf{0.5}$ , our $x$ -values are $-0.5$ , $0$ , $0.5$ , $1$ , $1.5$ , and $2$ .

STEP 11

Calculate the heights: $f(-0.5) = 2$ , $f(0) = 1$ , $f(0.5) = 2$ , $f(1) = 5$ , $f(1.5) = 10$ , and $f(2) = 17$ .

STEP 12

Calculate the area of each rectangle and add them up: $0.5 \cdot 2 + 0.5 \cdot 1 + 0.5 \cdot 2 + 0.5 \cdot 5 + 0.5 \cdot 10 + 0.5 \cdot 17 = 1 + 0.5 + 1 + 2.5 + 5 + 8.5 = \mathbf{18.5}$ .

STEP 13

Sketch the curve $f(x) = 1 + 4x^2$ from $x = -1$ to $x = 2$ .

STEP 14

Draw three rectangles under the curve.
The first rectangle spans $x = -1$ to $x = 0$ with height $f(0) = 1$ .
The second rectangle spans $x = 0$ to $x = 1$ with height $f(1) = 5$ .
The third rectangle spans $x = 1$ to $x = 2$ with height $f(2) = 17$ .

STEP 15

$R_3 = 23$ $R_6 = 18.5$ The sketch should show the curve and the three rectangles as described above.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the width of the rectangles for $R_3$ .
2. Calculate $R_3$ .
3. Calculate the width of the rectangles for $R_6$ .
4. Calculate $R_6$ .
5. Sketch the curve and the approximating rectangles for $R_3$ .

STEP 3

The total width of the area we're looking at goes from $x = -1$ to $x = 2$ , which is a width of $2 - (-1) = 2 + 1 = \mathbf{3}$ .

STEP 4

We're using three rectangles, so the width of each rectangle is the total width divided by the number of rectangles: $\frac{3}{3} = \mathbf{1}$ .

STEP 5

Since we're using right endpoints, the $x$ -values we'll plug into our function are $x = 0$ , $x = 1$ , and $x = 2$ .
The width of each rectangle is $\mathbf{1}$ , as we found earlier.

STEP 6

Let's calculate the height of each rectangle.
For $x = 0$ , $f(0) = 1 + 4 \cdot (0)^2 = 1$ .
For $x = 1$ , $f(1) = 1 + 4 \cdot (1)^2 = 5$ .
For $x = 2$ , $f(2) = 1 + 4 \cdot (2)^2 = 17$ .

STEP 7

Now, let's find the area of each rectangle by multiplying its width by its height, and then add them up!
The first rectangle has area $1 \cdot 1 = 1$ .
The second rectangle has area $1 \cdot 5 = 5$ .
The third rectangle has area $1 \cdot 17 = 17$ .
So, $R_3 = 1 + 5 + 17 = \mathbf{23}$ .

STEP 8

The total width is still $\mathbf{3}$ .

STEP 9

Now we have six rectangles, so the width of each is $\frac{3}{6} = \mathbf{0.5}$ .

STEP 10

With right endpoints and a width of $\mathbf{0.5}$ , our $x$ -values are $-0.5$ , $0$ , $0.5$ , $1$ , $1.5$ , and $2$ .

STEP 11

Calculate the heights: $f(-0.5) = 2$ , $f(0) = 1$ , $f(0.5) = 2$ , $f(1) = 5$ , $f(1.5) = 10$ , and $f(2) = 17$ .

STEP 12

Calculate the area of each rectangle and add them up: $0.5 \cdot 2 + 0.5 \cdot 1 + 0.5 \cdot 2 + 0.5 \cdot 5 + 0.5 \cdot 10 + 0.5 \cdot 17 = 1 + 0.5 + 1 + 2.5 + 5 + 8.5 = \mathbf{18.5}$ .

STEP 13

Sketch the curve $f(x) = 1 + 4x^2$ from $x = -1$ to $x = 2$ .

STEP 14

Draw three rectangles under the curve.
The first rectangle spans $x = -1$ to $x = 0$ with height $f(0) = 1$ .
The second rectangle spans $x = 0$ to $x = 1$ with height $f(1) = 5$ .
The third rectangle spans $x = 1$ to $x = 2$ with height $f(2) = 17$ .

STEP 15

$R_3 = 23$ $R_6 = 18.5$ The sketch should show the curve and the three rectangles as described above.

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Studdy Solution

STEP 1

STEP 2

1. Calculate the width of the rectangles for R3R_3R3​.2. Calculate R3R_3R3​.3. Calculate the width of the rectangles for R6R_6R6​.4. Calculate R6R_6R6​.5. Sketch the curve and the approximating rectangles for R3R_3R3​.

STEP 3

The **total width** of the area we're looking at goes from x=−1x = -1x=−1 to x=2x = 2x=2, which is a width of 2−(−1)=2+1=32 - (-1) = 2 + 1 = \mathbf{3}2−(−1)=2+1=3.

STEP 4

We're using **three** rectangles, so the width of each rectangle is the **total width** divided by the **number of rectangles**: 33=1\frac{3}{3} = \mathbf{1}33​=1.

STEP 5

Since we're using *right endpoints*, the xxx-values we'll plug into our function are x=0x = 0x=0, x=1x = 1x=1, and x=2x = 2x=2.The width of each rectangle is 1\mathbf{1}1, as we found earlier.

STEP 6

Let's calculate the height of each rectangle.For x=0x = 0x=0, f(0)=1+4⋅(0)2=1f(0) = 1 + 4 \cdot (0)^2 = 1f(0)=1+4⋅(0)2=1.For x=1x = 1x=1, f(1)=1+4⋅(1)2=5f(1) = 1 + 4 \cdot (1)^2 = 5f(1)=1+4⋅(1)2=5.For x=2x = 2x=2, f(2)=1+4⋅(2)2=17f(2) = 1 + 4 \cdot (2)^2 = 17f(2)=1+4⋅(2)2=17.

STEP 7

STEP 8

The **total width** is still 3\mathbf{3}3.

STEP 9

Now we have **six** rectangles, so the width of each is 36=0.5\frac{3}{6} = \mathbf{0.5}63​=0.5.

STEP 10

With *right endpoints* and a width of 0.5\mathbf{0.5}0.5, our xxx-values are −0.5-0.5−0.5, 000, 0.50.50.5, 111, 1.51.51.5, and 222.

STEP 11

Calculate the heights: f(−0.5)=2f(-0.5) = 2f(−0.5)=2, f(0)=1f(0) = 1f(0)=1, f(0.5)=2f(0.5) = 2f(0.5)=2, f(1)=5f(1) = 5f(1)=5, f(1.5)=10f(1.5) = 10f(1.5)=10, and f(2)=17f(2) = 17f(2)=17.

STEP 12

STEP 13

Sketch the curve f(x)=1+4x2f(x) = 1 + 4x^2f(x)=1+4x2 from x=−1x = -1x=−1 to x=2x = 2x=2.

STEP 14

STEP 15

R3=23R_3 = 23R3​=23 R6=18.5R_6 = 18.5R6​=18.5 The sketch should show the curve and the three rectangles as described above.

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1. Calculate the width of the rectangles for $R_3$ .
2. Calculate $R_3$ .
3. Calculate the width of the rectangles for $R_6$ .
4. Calculate $R_6$ .
5. Sketch the curve and the approximating rectangles for $R_3$ .

The total width of the area we're looking at goes from $x = -1$ to $x = 2$ , which is a width of $2 - (-1) = 2 + 1 = \mathbf{3}$ .

We're using three rectangles, so the width of each rectangle is the total width divided by the number of rectangles: $\frac{3}{3} = \mathbf{1}$ .

Since we're using right endpoints, the $x$ -values we'll plug into our function are $x = 0$ , $x = 1$ , and $x = 2$ .
The width of each rectangle is $\mathbf{1}$ , as we found earlier.

Let's calculate the height of each rectangle.
For $x = 0$ , $f(0) = 1 + 4 \cdot (0)^2 = 1$ .
For $x = 1$ , $f(1) = 1 + 4 \cdot (1)^2 = 5$ .
For $x = 2$ , $f(2) = 1 + 4 \cdot (2)^2 = 17$ .

The total width is still $\mathbf{3}$ .

Now we have six rectangles, so the width of each is $\frac{3}{6} = \mathbf{0.5}$ .

With right endpoints and a width of $\mathbf{0.5}$ , our $x$ -values are $-0.5$ , $0$ , $0.5$ , $1$ , $1.5$ , and $2$ .

Calculate the heights: $f(-0.5) = 2$ , $f(0) = 1$ , $f(0.5) = 2$ , $f(1) = 5$ , $f(1.5) = 10$ , and $f(2) = 17$ .

Sketch the curve $f(x) = 1 + 4x^2$ from $x = -1$ to $x = 2$ .

$R_3 = 23$ $R_6 = 18.5$ The sketch should show the curve and the three rectangles as described above.