Solved on Feb 11, 2024

Use trapezoidal rule with 4 intervals to approximate the area under $f(x)=\sqrt{x+1}$ on $[2,4.5]$ . Round answer to 2 decimal places.

STEP 1

Assumptions
1. The function to integrate is $f(x) = \sqrt{x+1}$ .
2. The interval of integration is $[2, 4.5]$ .
3. We are using the trapezoidal rule for approximation.
4. The number of intervals (trapezoids) is 4.
5. The trapezoidal rule formula is given by: $\text{Area} \approx \frac{\Delta x}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + \ldots + 2f(x_{n-1}) + f(x_n)]$ where $\Delta x$ is the width of each interval and $x_0, x_1, \ldots, x_n$ are the endpoints and midpoints of the intervals.

STEP 2

Calculate the width of each interval $\Delta x$ by subtracting the lower bound of the interval from the upper bound and dividing by the number of intervals.
$\Delta x = \frac{4.5 - 2}{4}$

STEP 3

Compute the value of $\Delta x$ .
$\Delta x = \frac{2.5}{4} = 0.625$

STEP 4

Identify the x-values that will be used in the trapezoidal rule. These are the endpoints of the intervals and are given by: $x_0 = 2, \quad x_1 = 2 + \Delta x, \quad x_2 = 2 + 2\Delta x, \quad x_3 = 2 + 3\Delta x, \quad x_4 = 4.5$

STEP 5

Calculate the x-values using the value of $\Delta x$ found in STEP_3.
$x_1 = 2 + 0.625 = 2.625, \quad x_2 = 2 + 2 \cdot 0.625 = 3.25, \quad x_3 = 2 + 3 \cdot 0.625 = 3.875$

STEP 6

Evaluate the function $f(x) = \sqrt{x+1}$ at the x-values $x_0, x_1, x_2, x_3, x_4$ .
$f(x_0) = \sqrt{2 + 1}, \quad f(x_1) = \sqrt{2.625 + 1}, \quad f(x_2) = \sqrt{3.25 + 1}, \quad f(x_3) = \sqrt{3.875 + 1}, \quad f(x_4) = \sqrt{4.5 + 1}$

STEP 7

Calculate the function values.
$f(x_0) = \sqrt{3}, \quad f(x_1) = \sqrt{3.625}, \quad f(x_2) = \sqrt{4.25}, \quad f(x_3) = \sqrt{4.875}, \quad f(x_4) = \sqrt{5.5}$

STEP 8

Compute the numerical values of the function at each x-value.
$f(x_0) = \sqrt{3} \approx 1.732, \quad f(x_1) = \sqrt{3.625} \approx 1.904, \quad f(x_2) = \sqrt{4.25} \approx 2.062, \quad f(x_3) = \sqrt{4.875} \approx 2.209, \quad f(x_4) = \sqrt{5.5} \approx 2.345$

STEP 9

Apply the trapezoidal rule formula using the function values and $\Delta x$ .
$\text{Area} \approx \frac{0.625}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4)]$

STEP 10

Substitute the function values into the trapezoidal rule formula.
$\text{Area} \approx \frac{0.625}{2} [1.732 + 2(1.904) + 2(2.062) + 2(2.209) + 2.345]$

STEP 11

Perform the multiplication and addition inside the brackets.
$\text{Area} \approx \frac{0.625}{2} [1.732 + 3.808 + 4.124 + 4.418 + 2.345]$

STEP 12

Sum the values inside the brackets.
$\text{Area} \approx \frac{0.625}{2} [16.427]$

STEP 13

Multiply by $\frac{0.625}{2}$ to find the approximate area.
$\text{Area} \approx \frac{0.625}{2} \cdot 16.427$

STEP 14

Calculate the area.
$\text{Area} \approx 0.3125 \cdot 16.427 \approx 5.133375$

STEP 15

Round the result to two decimal places as required by the problem.
$\text{Area} \approx 5.13$
The approximation for the area under the curve of $f(x) = \sqrt{x+1}$ in the interval $[2, 4.5]$ using the trapezoidal rule with 4 even intervals is approximately $5.13$ .

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Use trapezoidal rule with 4 intervals to approximate the area under $f(x)=\sqrt{x+1}$ on $[2,4.5]$ . Round answer to 2 decimal places.

STEP 1

STEP 2

Calculate the width of each interval $\Delta x$ by subtracting the lower bound of the interval from the upper bound and dividing by the number of intervals.
$\Delta x = \frac{4.5 - 2}{4}$

STEP 3

Compute the value of $\Delta x$ .
$\Delta x = \frac{2.5}{4} = 0.625$

STEP 4

Identify the x-values that will be used in the trapezoidal rule. These are the endpoints of the intervals and are given by: $x_0 = 2, \quad x_1 = 2 + \Delta x, \quad x_2 = 2 + 2\Delta x, \quad x_3 = 2 + 3\Delta x, \quad x_4 = 4.5$

STEP 5

Calculate the x-values using the value of $\Delta x$ found in STEP_3.
$x_1 = 2 + 0.625 = 2.625, \quad x_2 = 2 + 2 \cdot 0.625 = 3.25, \quad x_3 = 2 + 3 \cdot 0.625 = 3.875$

STEP 6

Evaluate the function $f(x) = \sqrt{x+1}$ at the x-values $x_0, x_1, x_2, x_3, x_4$ .
$f(x_0) = \sqrt{2 + 1}, \quad f(x_1) = \sqrt{2.625 + 1}, \quad f(x_2) = \sqrt{3.25 + 1}, \quad f(x_3) = \sqrt{3.875 + 1}, \quad f(x_4) = \sqrt{4.5 + 1}$

STEP 7

Calculate the function values.
$f(x_0) = \sqrt{3}, \quad f(x_1) = \sqrt{3.625}, \quad f(x_2) = \sqrt{4.25}, \quad f(x_3) = \sqrt{4.875}, \quad f(x_4) = \sqrt{5.5}$

STEP 8

Compute the numerical values of the function at each x-value.
$f(x_0) = \sqrt{3} \approx 1.732, \quad f(x_1) = \sqrt{3.625} \approx 1.904, \quad f(x_2) = \sqrt{4.25} \approx 2.062, \quad f(x_3) = \sqrt{4.875} \approx 2.209, \quad f(x_4) = \sqrt{5.5} \approx 2.345$

STEP 9

Apply the trapezoidal rule formula using the function values and $\Delta x$ .
$\text{Area} \approx \frac{0.625}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4)]$

STEP 10

Substitute the function values into the trapezoidal rule formula.
$\text{Area} \approx \frac{0.625}{2} [1.732 + 2(1.904) + 2(2.062) + 2(2.209) + 2.345]$

STEP 11

Perform the multiplication and addition inside the brackets.
$\text{Area} \approx \frac{0.625}{2} [1.732 + 3.808 + 4.124 + 4.418 + 2.345]$

STEP 12

Sum the values inside the brackets.
$\text{Area} \approx \frac{0.625}{2} [16.427]$

STEP 13

Multiply by $\frac{0.625}{2}$ to find the approximate area.
$\text{Area} \approx \frac{0.625}{2} \cdot 16.427$

STEP 14

Calculate the area.
$\text{Area} \approx 0.3125 \cdot 16.427 \approx 5.133375$

STEP 15

Round the result to two decimal places as required by the problem.
$\text{Area} \approx 5.13$
The approximation for the area under the curve of $f(x) = \sqrt{x+1}$ in the interval $[2, 4.5]$ using the trapezoidal rule with 4 even intervals is approximately $5.13$ .

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Use trapezoidal rule with 4 intervals to approximate the area under f(x)=x+1f(x)=\sqrt{x+1}f(x)=x+1​ on [2,4.5][2,4.5][2,4.5]. Round answer to 2 decimal places.

STEP 1

STEP 2

Calculate the width of each interval Δx\Delta xΔx by subtracting the lower bound of the interval from the upper bound and dividing by the number of intervals.Δx=4.5−24 \Delta x = \frac{4.5 - 2}{4} Δx=44.5−2​

STEP 3

Compute the value of Δx\Delta xΔx.Δx=2.54=0.625 \Delta x = \frac{2.5}{4} = 0.625 Δx=42.5​=0.625

STEP 4

STEP 5

Calculate the x-values using the value of Δx\Delta xΔx found in STEP_3.x1=2+0.625=2.625,x2=2+2⋅0.625=3.25,x3=2+3⋅0.625=3.875 x_1 = 2 + 0.625 = 2.625, \quad x_2 = 2 + 2 \cdot 0.625 = 3.25, \quad x_3 = 2 + 3 \cdot 0.625 = 3.875 x1​=2+0.625=2.625,x2​=2+2⋅0.625=3.25,x3​=2+3⋅0.625=3.875

STEP 6

STEP 7

Calculate the function values.f(x0)=3,f(x1)=3.625,f(x2)=4.25,f(x3)=4.875,f(x4)=5.5 f(x_0) = \sqrt{3}, \quad f(x_1) = \sqrt{3.625}, \quad f(x_2) = \sqrt{4.25}, \quad f(x_3) = \sqrt{4.875}, \quad f(x_4) = \sqrt{5.5} f(x0​)=3​,f(x1​)=3.625​,f(x2​)=4.25​,f(x3​)=4.875​,f(x4​)=5.5​

STEP 8

STEP 9

Apply the trapezoidal rule formula using the function values and Δx\Delta xΔx.Area≈0.6252[f(x0)+2f(x1)+2f(x2)+2f(x3)+f(x4)] \text{Area} \approx \frac{0.625}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4)] Area≈20.625​[f(x0​)+2f(x1​)+2f(x2​)+2f(x3​)+f(x4​)]

STEP 10

Substitute the function values into the trapezoidal rule formula.Area≈0.6252[1.732+2(1.904)+2(2.062)+2(2.209)+2.345] \text{Area} \approx \frac{0.625}{2} [1.732 + 2(1.904) + 2(2.062) + 2(2.209) + 2.345] Area≈20.625​[1.732+2(1.904)+2(2.062)+2(2.209)+2.345]

STEP 11

Perform the multiplication and addition inside the brackets.Area≈0.6252[1.732+3.808+4.124+4.418+2.345] \text{Area} \approx \frac{0.625}{2} [1.732 + 3.808 + 4.124 + 4.418 + 2.345] Area≈20.625​[1.732+3.808+4.124+4.418+2.345]

STEP 12

Sum the values inside the brackets.Area≈0.6252[16.427] \text{Area} \approx \frac{0.625}{2} [16.427] Area≈20.625​[16.427]

STEP 13

Multiply by 0.6252\frac{0.625}{2}20.625​ to find the approximate area.Area≈0.6252⋅16.427 \text{Area} \approx \frac{0.625}{2} \cdot 16.427 Area≈20.625​⋅16.427

STEP 14

Calculate the area.Area≈0.3125⋅16.427≈5.133375 \text{Area} \approx 0.3125 \cdot 16.427 \approx 5.133375 Area≈0.3125⋅16.427≈5.133375

STEP 15

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Use trapezoidal rule with 4 intervals to approximate the area under $f(x)=\sqrt{x+1}$ on $[2,4.5]$ . Round answer to 2 decimal places.

Calculate the width of each interval $\Delta x$ by subtracting the lower bound of the interval from the upper bound and dividing by the number of intervals.
$\Delta x = \frac{4.5 - 2}{4}$

Compute the value of $\Delta x$ .
$\Delta x = \frac{2.5}{4} = 0.625$

Calculate the x-values using the value of $\Delta x$ found in STEP_3.
$x_1 = 2 + 0.625 = 2.625, \quad x_2 = 2 + 2 \cdot 0.625 = 3.25, \quad x_3 = 2 + 3 \cdot 0.625 = 3.875$

Calculate the function values.
$f(x_0) = \sqrt{3}, \quad f(x_1) = \sqrt{3.625}, \quad f(x_2) = \sqrt{4.25}, \quad f(x_3) = \sqrt{4.875}, \quad f(x_4) = \sqrt{5.5}$

Apply the trapezoidal rule formula using the function values and $\Delta x$ .
$\text{Area} \approx \frac{0.625}{2} [f(x_0) + 2f(x_1) + 2f(x_2) + 2f(x_3) + f(x_4)]$

Substitute the function values into the trapezoidal rule formula.
$\text{Area} \approx \frac{0.625}{2} [1.732 + 2(1.904) + 2(2.062) + 2(2.209) + 2.345]$

Perform the multiplication and addition inside the brackets.
$\text{Area} \approx \frac{0.625}{2} [1.732 + 3.808 + 4.124 + 4.418 + 2.345]$

Sum the values inside the brackets.
$\text{Area} \approx \frac{0.625}{2} [16.427]$

Multiply by $\frac{0.625}{2}$ to find the approximate area.
$\text{Area} \approx \frac{0.625}{2} \cdot 16.427$

Calculate the area.
$\text{Area} \approx 0.3125 \cdot 16.427 \approx 5.133375$