Math / Calculus

QuestionGiven the function, $f(x)=x^{2}$ , using 6 rectangles of equal width, find an approximation to the area between the curve and the $x$ -axis over the interval $[2,6]$ if the heights of the rectangles are found by evaluating the function at the left endpoints, right endpoints, and midpoints of each subinterval created by a partition of the interval given.
Using left endpoints, the approximate area is: $\square$ Using right endpoints, the approximate area is: $\square$ Using midpoints, the approximate area is: $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the approximate area under the curve $f(x) = x^2$ from $x = 2$ to $x = 6$ using rectangles.
We'll do this three times: once using the left endpoints, once using the right endpoints, and once using the midpoints of each rectangle. Watch out! Don't forget to divide the interval into equal parts and use the correct points for calculating the height of each rectangle!

STEP 2

1. Divide the interval into subintervals
2. Calculate the area using left endpoints
3. Calculate the area using right endpoints
4. Calculate the area using midpoints

STEP 3

First, we need to divide the interval $[2, 6]$ into 6 equal parts.
The width of each subinterval, or rectangle, is found by subtracting the start of the interval from the end and then dividing by the number of rectangles.
So, we calculate:
$\text{Width of each rectangle} = \frac{6 - 2}{6} = \frac{4}{6} = \frac{2}{3}$
This means each rectangle will have a width of $\frac{2}{3}$ .

STEP 4

To find the area using the left endpoints, we evaluate the function at the left side of each subinterval.
The left endpoints are $2, 2 + \frac{2}{3}, 2 + 2\cdot\frac{2}{3}, \ldots, 2 + 5\cdot\frac{2}{3}$ .

STEP 5

Calculate the height of each rectangle using these left endpoints:
\begin{align} f(2) &= 2^2 = 4, \\ f\left(2 + \frac{2}{3}\right) &= \left(2 + \frac{2}{3}\right)^2, \\ f\left(2 + 2\cdot\frac{2}{3}\right) &= \left(2 + \frac{4}{3}\right)^2, \\ &\vdots \\ f\left(2 + 5\cdot\frac{2}{3}\right) &= \left(2 + \frac{10}{3}\right)^2. \end{align}

STEP 6

Now, calculate the area of each rectangle and sum them up:
$\text{Area} = \frac{2}{3} \left( f(2) + f\left(2 + \frac{2}{3}\right) + f\left(2 + 2\cdot\frac{2}{3}\right) + \ldots + f\left(2 + 5\cdot\frac{2}{3}\right) \right)$

STEP 7

For the right endpoints, we start from the first right endpoint of the first subinterval, which is $2 + \frac{2}{3}$ , and continue to the last point $6$ .

STEP 8

Calculate the height of each rectangle using these right endpoints:
\begin{align} f\left(2 + \frac{2}{3}\right) &= \left(2 + \frac{2}{3}\right)^2, \\ f\left(2 + 2\cdot\frac{2}{3}\right) &= \left(2 + \frac{4}{3}\right)^2, \\ &\vdots \\ f(6) &= 6^2 = 36. \end{align}

STEP 9

Now, calculate the area of each rectangle and sum them up:
$\text{Area} = \frac{2}{3} \left( f\left(2 + \frac{2}{3}\right) + f\left(2 + 2\cdot\frac{2}{3}\right) + \ldots + f(6) \right)$

STEP 10

For the midpoints, we find the midpoint of each subinterval.
The midpoints are $2 + \frac{1}{3}, 2 + \frac{3}{3}, 2 + \frac{5}{3}, \ldots, 2 + \frac{11}{3}$ .

STEP 11

Calculate the height of each rectangle using these midpoints:
\begin{align} f\left(2 + \frac{1}{3}\right) &= \left(2 + \frac{1}{3}\right)^2, \\ f\left(2 + \frac{3}{3}\right) &= \left(2 + 1\right)^2, \\ &\vdots \\ f\left(2 + \frac{11}{3}\right) &= \left(2 + \frac{11}{3}\right)^2. \end{align}

STEP 12

Now, calculate the area of each rectangle and sum them up:
$\text{Area} = \frac{2}{3} \left( f\left(2 + \frac{1}{3}\right) + f\left(2 + \frac{3}{3}\right) + \ldots + f\left(2 + \frac{11}{3}\right) \right)$

STEP 13

Using left endpoints, the approximate area is: $\frac{2}{3} \left( 4 + \left(2 + \frac{2}{3}\right)^2 + \left(2 + \frac{4}{3}\right)^2 + \ldots + \left(2 + \frac{10}{3}\right)^2 \right)$ .
Using right endpoints, the approximate area is: $\frac{2}{3} \left( \left(2 + \frac{2}{3}\right)^2 + \left(2 + \frac{4}{3}\right)^2 + \ldots + 36 \right)$ .
Using midpoints, the approximate area is: $\frac{2}{3} \left( \left(2 + \frac{1}{3}\right)^2 + \left(2 + 1\right)^2 + \ldots + \left(2 + \frac{11}{3}\right)^2 \right)$ .

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Studdy Solution

STEP 1

STEP 2

1. Divide the interval into subintervals
2. Calculate the area using left endpoints
3. Calculate the area using right endpoints
4. Calculate the area using midpoints

STEP 3

STEP 4

To find the area using the left endpoints, we evaluate the function at the left side of each subinterval.
The left endpoints are $2, 2 + \frac{2}{3}, 2 + 2\cdot\frac{2}{3}, \ldots, 2 + 5\cdot\frac{2}{3}$ .

STEP 5

STEP 6

Now, calculate the area of each rectangle and sum them up:
$\text{Area} = \frac{2}{3} \left( f(2) + f\left(2 + \frac{2}{3}\right) + f\left(2 + 2\cdot\frac{2}{3}\right) + \ldots + f\left(2 + 5\cdot\frac{2}{3}\right) \right)$

STEP 7

For the right endpoints, we start from the first right endpoint of the first subinterval, which is $2 + \frac{2}{3}$ , and continue to the last point $6$ .

STEP 8

Calculate the height of each rectangle using these right endpoints:
\begin{align} f\left(2 + \frac{2}{3}\right) &= \left(2 + \frac{2}{3}\right)^2, \\ f\left(2 + 2\cdot\frac{2}{3}\right) &= \left(2 + \frac{4}{3}\right)^2, \\ &\vdots \\ f(6) &= 6^2 = 36. \end{align}

STEP 9

Now, calculate the area of each rectangle and sum them up:
$\text{Area} = \frac{2}{3} \left( f\left(2 + \frac{2}{3}\right) + f\left(2 + 2\cdot\frac{2}{3}\right) + \ldots + f(6) \right)$

STEP 10

For the midpoints, we find the midpoint of each subinterval.
The midpoints are $2 + \frac{1}{3}, 2 + \frac{3}{3}, 2 + \frac{5}{3}, \ldots, 2 + \frac{11}{3}$ .

STEP 11

Calculate the height of each rectangle using these midpoints:
\begin{align} f\left(2 + \frac{1}{3}\right) &= \left(2 + \frac{1}{3}\right)^2, \\ f\left(2 + \frac{3}{3}\right) &= \left(2 + 1\right)^2, \\ &\vdots \\ f\left(2 + \frac{11}{3}\right) &= \left(2 + \frac{11}{3}\right)^2. \end{align}

STEP 12

Now, calculate the area of each rectangle and sum them up:
$\text{Area} = \frac{2}{3} \left( f\left(2 + \frac{1}{3}\right) + f\left(2 + \frac{3}{3}\right) + \ldots + f\left(2 + \frac{11}{3}\right) \right)$

STEP 13

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Studdy Solution

STEP 1

STEP 2

1. Divide the interval into subintervals2. Calculate the area using left endpoints3. Calculate the area using right endpoints4. Calculate the area using midpoints

STEP 3

STEP 4

To find the area using the **left endpoints**, we evaluate the function at the left side of each subinterval.The left endpoints are 2,2+23,2+2⋅23,…,2+5⋅232, 2 + \frac{2}{3}, 2 + 2\cdot\frac{2}{3}, \ldots, 2 + 5\cdot\frac{2}{3}2,2+32​,2+2⋅32​,…,2+5⋅32​.

STEP 5

STEP 6

STEP 7

For the **right endpoints**, we start from the first right endpoint of the first subinterval, which is 2+232 + \frac{2}{3}2+32​, and continue to the last point 666.

STEP 8

Calculate the height of each rectangle using these right endpoints:\begin{align*} f\left(2 + \frac{2}{3}\right) &= \left(2 + \frac{2}{3}\right)^2, \\ f\left(2 + 2\cdot\frac{2}{3}\right) &= \left(2 + \frac{4}{3}\right)^2, \\ &\vdots \\ f(6) &= 6^2 = 36. \end{align*}

STEP 9

Now, **calculate the area** of each rectangle and sum them up:Area=23(f(2+23)+f(2+2⋅23)+…+f(6))\text{Area} = \frac{2}{3} \left( f\left(2 + \frac{2}{3}\right) + f\left(2 + 2\cdot\frac{2}{3}\right) + \ldots + f(6) \right)Area=32​(f(2+32​)+f(2+2⋅32​)+…+f(6))

STEP 10

For the **midpoints**, we find the midpoint of each subinterval.The midpoints are 2+13,2+33,2+53,…,2+1132 + \frac{1}{3}, 2 + \frac{3}{3}, 2 + \frac{5}{3}, \ldots, 2 + \frac{11}{3}2+31​,2+33​,2+35​,…,2+311​.

STEP 11

Calculate the height of each rectangle using these midpoints:\begin{align*} f\left(2 + \frac{1}{3}\right) &= \left(2 + \frac{1}{3}\right)^2, \\ f\left(2 + \frac{3}{3}\right) &= \left(2 + 1\right)^2, \\ &\vdots \\ f\left(2 + \frac{11}{3}\right) &= \left(2 + \frac{11}{3}\right)^2. \end{align*}

STEP 12

Now, **calculate the area** of each rectangle and sum them up:Area=23(f(2+13)+f(2+33)+…+f(2+113))\text{Area} = \frac{2}{3} \left( f\left(2 + \frac{1}{3}\right) + f\left(2 + \frac{3}{3}\right) + \ldots + f\left(2 + \frac{11}{3}\right) \right)Area=32​(f(2+31​)+f(2+33​)+…+f(2+311​))

STEP 13

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1. Divide the interval into subintervals
2. Calculate the area using left endpoints
3. Calculate the area using right endpoints
4. Calculate the area using midpoints

To find the area using the left endpoints, we evaluate the function at the left side of each subinterval.
The left endpoints are $2, 2 + \frac{2}{3}, 2 + 2\cdot\frac{2}{3}, \ldots, 2 + 5\cdot\frac{2}{3}$ .

For the right endpoints, we start from the first right endpoint of the first subinterval, which is $2 + \frac{2}{3}$ , and continue to the last point $6$ .

Calculate the height of each rectangle using these right endpoints:
\begin{align} f\left(2 + \frac{2}{3}\right) &= \left(2 + \frac{2}{3}\right)^2, \\ f\left(2 + 2\cdot\frac{2}{3}\right) &= \left(2 + \frac{4}{3}\right)^2, \\ &\vdots \\ f(6) &= 6^2 = 36. \end{align}

Now, calculate the area of each rectangle and sum them up:
$\text{Area} = \frac{2}{3} \left( f\left(2 + \frac{2}{3}\right) + f\left(2 + 2\cdot\frac{2}{3}\right) + \ldots + f(6) \right)$

For the midpoints, we find the midpoint of each subinterval.
The midpoints are $2 + \frac{1}{3}, 2 + \frac{3}{3}, 2 + \frac{5}{3}, \ldots, 2 + \frac{11}{3}$ .

Calculate the height of each rectangle using these midpoints:
\begin{align} f\left(2 + \frac{1}{3}\right) &= \left(2 + \frac{1}{3}\right)^2, \\ f\left(2 + \frac{3}{3}\right) &= \left(2 + 1\right)^2, \\ &\vdots \\ f\left(2 + \frac{11}{3}\right) &= \left(2 + \frac{11}{3}\right)^2. \end{align}

Now, calculate the area of each rectangle and sum them up:
$\text{Area} = \frac{2}{3} \left( f\left(2 + \frac{1}{3}\right) + f\left(2 + \frac{3}{3}\right) + \ldots + f\left(2 + \frac{11}{3}\right) \right)$