Math / Calculus

QuestionConsider the integral $\int_{1}^{3} \frac{-x}{x^{2}+1} d x$ . Find the approximate value using the midpoint rule with 10 rectangles. Round your answer as a decimal correct to 5 decimal places.

Studdy Solution

STEP 1

What is this asking? We need to estimate the area under the curve $y = \frac{-x}{x^2 + 1}$ between $x = 1$ and $x = 3$ using 10 rectangles and the midpoint rule. Watch out! Don't forget to use the midpoint of each rectangle to determine its height!
Also, make sure to use the correct width for each rectangle based on the interval and the number of rectangles.

STEP 2

1. Calculate the width of each rectangle.
2. Determine the midpoints of each rectangle.
3. Calculate the height of each rectangle.
4. Calculate the area of each rectangle.
5. Sum the areas to find the approximate value of the integral.

STEP 3

We're integrating from $x = 1$ to $x = 3$ , so the total width of the interval is $3 - 1 = 2$ .
Since we're using 10 rectangles, the width of each rectangle, which we'll call $\Delta x$ , is $\frac{2}{10} = 0.2$ .
Boom!

STEP 4

The first rectangle starts at $x = 1$ .
Its midpoint is $1 + \frac{0.2}{2} = 1.1$ .
The next midpoint is $1.1 + 0.2 = 1.3$ .
We keep adding 0.2 until we get to the last midpoint.
The midpoints are 1.1, 1.3, 1.5, 1.7, 1.9, 2.1, 2.3, 2.5, 2.7, and 2.9.
Awesome!

STEP 5

The height of each rectangle is determined by the function $\frac{-x}{x^2 + 1}$ evaluated at the midpoint.
Let's calculate these heights!
For the first midpoint, $x = 1.1$ , the height is $\frac{-1.1}{1.1^2 + 1} = \frac{-1.1}{2.21} \approx -0.4977$ .
We do this for all the midpoints.

STEP 6

The area of each rectangle is its width times its height.
Since the width is always 0.2, we multiply each height from the previous step by 0.2.
For example, the area of the first rectangle is approximately $0.2 \cdot -0.4977 \approx -0.0995$ .

STEP 7

Now, we add up all the areas!
This gives us the approximate value of the integral. $\begin{array}{rcl} \text{Approximate Integral} & = & 0.2 \cdot \left(\frac{-1.1}{2.21} + \frac{-1.3}{2.69} + \frac{-1.5}{3.25} + \frac{-1.7}{3.89} + \frac{-1.9}{4.61} + \frac{-2.1}{5.41} + \frac{-2.3}{6.29} + \frac{-2.5}{7.25} + \frac{-2.7}{8.29} + \frac{-2.9}{9.41}\right) \\ & \approx & 0.2 \cdot (-0.4977 - 0.4833 - 0.4615 - 0.4370 - 0.4121 - 0.3881 - 0.3657 - 0.3456 - 0.3257 - 0.3084) \\ & \approx & 0.2 \cdot (-3.7251) \\ & \approx & -0.7450 \end{array}$

STEP 8

The approximate value of the integral using the midpoint rule with 10 rectangles is -0.74502.

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QuestionConsider the integral $\int_{1}^{3} \frac{-x}{x^{2}+1} d x$ . Find the approximate value using the midpoint rule with 10 rectangles. Round your answer as a decimal correct to 5 decimal places.

Studdy Solution

STEP 1

STEP 2

1. Calculate the width of each rectangle.
2. Determine the midpoints of each rectangle.
3. Calculate the height of each rectangle.
4. Calculate the area of each rectangle.
5. Sum the areas to find the approximate value of the integral.

STEP 3

We're integrating from $x = 1$ to $x = 3$ , so the total width of the interval is $3 - 1 = 2$ .
Since we're using 10 rectangles, the width of each rectangle, which we'll call $\Delta x$ , is $\frac{2}{10} = 0.2$ .
Boom!

STEP 4

The first rectangle starts at $x = 1$ .
Its midpoint is $1 + \frac{0.2}{2} = 1.1$ .
The next midpoint is $1.1 + 0.2 = 1.3$ .
We keep adding 0.2 until we get to the last midpoint.
The midpoints are 1.1, 1.3, 1.5, 1.7, 1.9, 2.1, 2.3, 2.5, 2.7, and 2.9.
Awesome!

STEP 5

The height of each rectangle is determined by the function $\frac{-x}{x^2 + 1}$ evaluated at the midpoint.
Let's calculate these heights!
For the first midpoint, $x = 1.1$ , the height is $\frac{-1.1}{1.1^2 + 1} = \frac{-1.1}{2.21} \approx -0.4977$ .
We do this for all the midpoints.

STEP 6

The area of each rectangle is its width times its height.
Since the width is always 0.2, we multiply each height from the previous step by 0.2.
For example, the area of the first rectangle is approximately $0.2 \cdot -0.4977 \approx -0.0995$ .

STEP 7

STEP 8

The approximate value of the integral using the midpoint rule with 10 rectangles is -0.74502.

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QuestionConsider the integral ∫13−xx2+1dx\int_{1}^{3} \frac{-x}{x^{2}+1} d x∫13​x2+1−x​dx. Find the approximate value using the midpoint rule with 10 rectangles. Round your answer as a decimal correct to 5 decimal places.

Studdy Solution

STEP 1

STEP 2

1. Calculate the width of each rectangle.2. Determine the midpoints of each rectangle.3. Calculate the height of each rectangle.4. Calculate the area of each rectangle.5. Sum the areas to find the approximate value of the integral.

STEP 3

We're integrating from x=1x = 1x=1 to x=3x = 3x=3, so the total width of the interval is 3−1=23 - 1 = 23−1=2.Since we're using **10** rectangles, the width of each rectangle, which we'll call Δx\Delta xΔx, is 210=0.2\frac{2}{10} = 0.2102​=0.2.Boom!

STEP 4

STEP 5

STEP 6

STEP 7

STEP 8

The approximate value of the integral using the midpoint rule with 10 rectangles is **-0.74502**.

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Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionConsider the integral $\int_{1}^{3} \frac{-x}{x^{2}+1} d x$ . Find the approximate value using the midpoint rule with 10 rectangles. Round your answer as a decimal correct to 5 decimal places.

1. Calculate the width of each rectangle.
2. Determine the midpoints of each rectangle.
3. Calculate the height of each rectangle.
4. Calculate the area of each rectangle.
5. Sum the areas to find the approximate value of the integral.

We're integrating from $x = 1$ to $x = 3$ , so the total width of the interval is $3 - 1 = 2$ .
Since we're using 10 rectangles, the width of each rectangle, which we'll call $\Delta x$ , is $\frac{2}{10} = 0.2$ .
Boom!

The approximate value of the integral using the midpoint rule with 10 rectangles is -0.74502.