Solved on Feb 11, 2024

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

STEP 1

Assumptions
1. The function to integrate is f(x)=x+1f(x) = \sqrt{x+1}.
2. The interval of integration is [2,4.5][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: AreaΔx2[f(x0)+2f(x1)+2f(x2)++2f(xn1)+f(xn)] \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 Δx\Delta x is the width of each interval and x0,x1,,xnx_0, x_1, \ldots, x_n are the endpoints and midpoints of the intervals.

STEP 2

Calculate the width of each interval Δx\Delta x by subtracting the lower bound of the interval from the upper bound and dividing by the number of intervals.
Δx=4.524 \Delta x = \frac{4.5 - 2}{4}

STEP 3

Compute the value of Δx\Delta x.
Δx=2.54=0.625 \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: x0=2,x1=2+Δx,x2=2+2Δx,x3=2+3Δx,x4=4.5 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 Δx\Delta x found in STEP_3.
x1=2+0.625=2.625,x2=2+20.625=3.25,x3=2+30.625=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

STEP 6

Evaluate the function f(x)=x+1f(x) = \sqrt{x+1} at the x-values x0,x1,x2,x3,x4x_0, x_1, x_2, x_3, x_4.
f(x0)=2+1,f(x1)=2.625+1,f(x2)=3.25+1,f(x3)=3.875+1,f(x4)=4.5+1 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(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}

STEP 8

Compute the numerical values of the function at each x-value.
f(x0)=31.732,f(x1)=3.6251.904,f(x2)=4.252.062,f(x3)=4.8752.209,f(x4)=5.52.345 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 Δx\Delta x.
Area0.6252[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)]

STEP 10

Substitute the function values into the trapezoidal rule formula.
Area0.6252[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]

STEP 11

Perform the multiplication and addition inside the brackets.
Area0.6252[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]

STEP 12

Sum the values inside the brackets.
Area0.6252[16.427] \text{Area} \approx \frac{0.625}{2} [16.427]

STEP 13

Multiply by 0.6252\frac{0.625}{2} to find the approximate area.
Area0.625216.427 \text{Area} \approx \frac{0.625}{2} \cdot 16.427

STEP 14

Calculate the area.
Area0.312516.4275.133375 \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.
Area5.13 \text{Area} \approx 5.13
The approximation for the area under the curve of f(x)=x+1f(x) = \sqrt{x+1} in the interval [2,4.5][2, 4.5] using the trapezoidal rule with 4 even intervals is approximately 5.135.13.

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