Math / Calculus

QuestionUse the Trapezoidal Rule to find $T_{n}$ using the indicated value of $n$ . Round to three decimal places. $\int_{0}^{1} \frac{1}{x^{2}+1} d x ; n=4$ A. $T_{4}=0.783$ B. $T_{4}=0.970$ C. $T_{4}=1.383$ D. $T_{4}=1.566$

Studdy Solution

STEP 1

What is this asking? We need to approximate the area under the curve $y = \frac{1}{x^2 + 1}$ from $x = 0$ to $x = 1$ using the Trapezoidal Rule with four trapezoids. Watch out! Don't forget to divide the interval into equal parts and be careful with the function values at each point!

STEP 2

1. Calculate Delta x
2. Calculate Function Values
3. Apply the Trapezoidal Rule

STEP 3

Alright, let's start by finding the width of each trapezoid, which we call $\Delta x$ .
This is like slicing our interval from $x=0$ to $x=1$ into four equal pieces.

STEP 4

The formula for $\Delta x$ is $\frac{b - a}{n}$ , where 'a' is the start of our interval ( $0$ ), 'b' is the end ( $1$ ), and 'n' is the number of trapezoids ( $4$ ).

STEP 5

So, $\Delta x = \frac{1 - 0}{4} = \frac{1}{4} = 0.25$ .
Each trapezoid will have a width of 0.25.

STEP 6

Now, we need the height of each trapezoid.
We get these heights by plugging the $x$ values at the edges of each trapezoid into our function, $\frac{1}{x^2 + 1}$ .

STEP 7

Our $x$ values are $0$ , $0.25$ , $0.5$ , $0.75$ , and $1$ .
Let's plug them in!

STEP 8

* $f(0) = \frac{1}{0^2 + 1} = 1$ * $f(0.25) = \frac{1}{(0.25)^2 + 1} = \frac{1}{0.0625 + 1} = \frac{1}{1.0625} \approx 0.941176$ * $f(0.5) = \frac{1}{(0.5)^2 + 1} = \frac{1}{0.25 + 1} = \frac{1}{1.25} = 0.8$ * $f(0.75) = \frac{1}{(0.75)^2 + 1} = \frac{1}{0.5625 + 1} = \frac{1}{1.5625} \approx 0.64$ * $f(1) = \frac{1}{1^2 + 1} = \frac{1}{2} = 0.5$

STEP 9

The Trapezoidal Rule is: $T_n = \frac{\Delta x}{2} \cdot [f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)]$

STEP 10

Let's plug in our values: $T_4 = \frac{0.25}{2} \cdot [1 + 2(0.941176) + 2(0.8) + 2(0.64) + 0.5]$ $T_4 = 0.125 \cdot [1 + 1.882352 + 1.6 + 1.28 + 0.5]$ $T_4 = 0.125 \cdot [6.262352]$ $T_4 \approx 0.782794$

STEP 11

Rounding to three decimal places, we get $T_4 \approx 0.783$ .

STEP 12

The answer is A, $T_4 = 0.783$ .

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QuestionUse the Trapezoidal Rule to find $T_{n}$ using the indicated value of $n$ . Round to three decimal places. $\int_{0}^{1} \frac{1}{x^{2}+1} d x ; n=4$ A. $T_{4}=0.783$ B. $T_{4}=0.970$ C. $T_{4}=1.383$ D. $T_{4}=1.566$

Studdy Solution

STEP 1

What is this asking? We need to approximate the area under the curve $y = \frac{1}{x^2 + 1}$ from $x = 0$ to $x = 1$ using the Trapezoidal Rule with four trapezoids. Watch out! Don't forget to divide the interval into equal parts and be careful with the function values at each point!

STEP 2

1. Calculate Delta x
2. Calculate Function Values
3. Apply the Trapezoidal Rule

STEP 3

Alright, let's start by finding the width of each trapezoid, which we call $\Delta x$ .
This is like slicing our interval from $x=0$ to $x=1$ into four equal pieces.

STEP 4

The formula for $\Delta x$ is $\frac{b - a}{n}$ , where 'a' is the start of our interval ( $0$ ), 'b' is the end ( $1$ ), and 'n' is the number of trapezoids ( $4$ ).

STEP 5

So, $\Delta x = \frac{1 - 0}{4} = \frac{1}{4} = 0.25$ .
Each trapezoid will have a width of 0.25.

STEP 6

Now, we need the height of each trapezoid.
We get these heights by plugging the $x$ values at the edges of each trapezoid into our function, $\frac{1}{x^2 + 1}$ .

STEP 7

Our $x$ values are $0$ , $0.25$ , $0.5$ , $0.75$ , and $1$ .
Let's plug them in!

STEP 8

STEP 9

The Trapezoidal Rule is: $T_n = \frac{\Delta x}{2} \cdot [f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)]$

STEP 10

Let's plug in our values: $T_4 = \frac{0.25}{2} \cdot [1 + 2(0.941176) + 2(0.8) + 2(0.64) + 0.5]$ $T_4 = 0.125 \cdot [1 + 1.882352 + 1.6 + 1.28 + 0.5]$ $T_4 = 0.125 \cdot [6.262352]$ $T_4 \approx 0.782794$

STEP 11

Rounding to three decimal places, we get $T_4 \approx 0.783$ .

STEP 12

The answer is A, $T_4 = 0.783$ .

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Studdy Solution

STEP 1

STEP 2

1. Calculate Delta x2. Calculate Function Values3. Apply the Trapezoidal Rule

STEP 3

Alright, let's **start** by finding the width of each trapezoid, which we call Δx\Delta xΔx.This is like slicing our interval from x=0x=0x=0 to x=1x=1x=1 into **four** equal pieces.

STEP 4

The formula for Δx\Delta xΔx is b−an\frac{b - a}{n}nb−a​, where **'a'** is the **start** of our interval (000), **'b'** is the **end** (111), and **'n'** is the number of trapezoids (444).

STEP 5

So, Δx=1−04=14=0.25\Delta x = \frac{1 - 0}{4} = \frac{1}{4} = 0.25Δx=41−0​=41​=0.25.Each trapezoid will have a width of **0.25**.

STEP 6

Now, we need the height of each trapezoid.We get these heights by plugging the xxx values at the edges of each trapezoid into our function, 1x2+1\frac{1}{x^2 + 1}x2+11​.

STEP 7

Our xxx values are 000, 0.250.250.25, 0.50.50.5, 0.750.750.75, and 111.Let's plug them in!

STEP 8

STEP 9

The Trapezoidal Rule is: Tn=Δx2⋅[f(x0)+2f(x1)+2f(x2)+...+2f(xn−1)+f(xn)]T_n = \frac{\Delta x}{2} \cdot [f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)]Tn​=2Δx​⋅[f(x0​)+2f(x1​)+2f(x2​)+...+2f(xn−1​)+f(xn​)]

STEP 10

STEP 11

Rounding to three decimal places, we get T4≈0.783T_4 \approx 0.783T4​≈0.783.

STEP 12

The answer is A, T4=0.783T_4 = 0.783T4​=0.783.

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1. Calculate Delta x
2. Calculate Function Values
3. Apply the Trapezoidal Rule

Alright, let's start by finding the width of each trapezoid, which we call $\Delta x$ .
This is like slicing our interval from $x=0$ to $x=1$ into four equal pieces.

The formula for $\Delta x$ is $\frac{b - a}{n}$ , where 'a' is the start of our interval ( $0$ ), 'b' is the end ( $1$ ), and 'n' is the number of trapezoids ( $4$ ).

So, $\Delta x = \frac{1 - 0}{4} = \frac{1}{4} = 0.25$ .
Each trapezoid will have a width of 0.25.

Now, we need the height of each trapezoid.
We get these heights by plugging the $x$ values at the edges of each trapezoid into our function, $\frac{1}{x^2 + 1}$ .

Our $x$ values are $0$ , $0.25$ , $0.5$ , $0.75$ , and $1$ .
Let's plug them in!

The Trapezoidal Rule is: $T_n = \frac{\Delta x}{2} \cdot [f(x_0) + 2f(x_1) + 2f(x_2) + ... + 2f(x_{n-1}) + f(x_n)]$

Rounding to three decimal places, we get $T_4 \approx 0.783$ .

The answer is A, $T_4 = 0.783$ .