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Complete the following steps for the given function, interval, and value of \( n \). a. Sketch the graph of the function on the given interval. b. Calculate \( \Delta x \) and the grid points \( x_{0}, x_{1}, \ldots, x_{n} \). c. Illustrate the midpoint Riemann sum by sketching the appropriate rectangles. d. Calculate the midpoint Riemann sum. \( f(x)=2 x^{2}+2 \) on \( [1,5] ; n=4 \)

Ask by Malone Sullivan. in the United States
Mar 09,2025

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Answer

**a. Sketching the Graph** The function \( f(x) = 2x^2 + 2 \) is a parabola opening upward. On the interval \([1,5]\), it starts at \((1,4)\), passes through \((3,20)\), and ends at \((5,52)\). **b. Calculating \( \Delta x \) and Grid Points** Divide the interval into 4 equal parts: \[ \Delta x = 1 \] Grid points: \[ x_0 = 1,\quad x_1 = 2,\quad x_2 = 3,\quad x_3 = 4,\quad x_4 = 5 \] **c. Midpoint Riemann Sum Rectangles** Midpoints: \[ m_1 = 1.5,\quad m_2 = 2.5,\quad m_3 = 3.5,\quad m_4 = 4.5 \] **d. Calculating the Midpoint Riemann Sum** Compute \( f(m_i) \) for each midpoint: \[ f(1.5) = 6.5,\quad f(2.5) = 14.5,\quad f(3.5) = 26.5,\quad f(4.5) = 42.5 \] Midpoint Sum: \[ \text{Midpoint Sum} = 1 \times (6.5 + 14.5 + 26.5 + 42.5) = 90 \] **Final Answer:** The midpoint Riemann sum is **90**.

Solution

**a. Sketching the Graph** The function is \[ f(x)=2x^2+2. \] This is a parabola opening upward with vertex at the smallest \( x \)‐value on the interval. On the interval \([1,5]\), some key points are: - At \( x=1 \): \[ f(1)=2(1)^2+2=2+2=4. \] - At \( x=3 \): \[ f(3)=2(3)^2+2=2(9)+2=18+2=20. \] - At \( x=5 \): \[ f(5)=2(5)^2+2=2(25)+2=50+2=52. \] The graph will be a steeply rising upward-opening parabola starting at the point \((1,4)\), passing through \((3,20)\), and ending at \((5,52)\). **b. \(\Delta x\) and Grid Points** Since we are dividing the interval from \(x=1\) to \(x=5\) into \(n=4\) subintervals, the width of each subinterval is \[ \Delta x = \frac{5-1}{4} = 1. \] Thus, the grid points are: \[ x_0 = 1,\quad x_1 = 2,\quad x_2 = 3,\quad x_3 = 4,\quad x_4 = 5. \] **c. Midpoint Riemann Sum Rectangles** For the midpoint Riemann sum, we use the midpoint of each subinterval \([x_{i-1}, x_i]\) to determine the height of the rectangle. The midpoints are calculated as follows: - Midpoint of \([1,2]\): \[ m_1 = \frac{1+2}{2} = 1.5. \] - Midpoint of \([2,3]\): \[ m_2 = \frac{2+3}{2} = 2.5. \] - Midpoint of \([3,4]\): \[ m_3 = \frac{3+4}{2} = 3.5. \] - Midpoint of \([4,5]\): \[ m_4 = \frac{4+5}{2} = 4.5. \] Each rectangle has: - Base (width) \( \Delta x = 1 \) - Height \( f(m_i) \) where \( m_i \) is the midpoint of the \(i\)th subinterval. **d. Calculating the Midpoint Riemann Sum** We compute the value of \( f(x) \) at each midpoint: 1. For \( m_1 = 1.5 \): \[ f(1.5)=2(1.5)^2+2=2(2.25)+2=4.5+2=6.5. \] 2. For \( m_2 = 2.5 \): \[ f(2.5)=2(2.5)^2+2=2(6.25)+2=12.5+2=14.5. \] 3. For \( m_3 = 3.5 \): \[ f(3.5)=2(3.5)^2+2=2(12.25)+2=24.5+2=26.5. \] 4. For \( m_4 = 4.5 \): \[ f(4.5)=2(4.5)^2+2=2(20.25)+2=40.5+2=42.5. \] The midpoint Riemann sum is then: \[ \text{Midpoint Sum} = \Delta x \left[f(1.5)+f(2.5)+f(3.5)+f(4.5)\right]. \] Substitute the values: \[ \text{Midpoint Sum} = 1 \times \left[6.5+14.5+26.5+42.5\right]. \] Adding the function values: \[ 6.5+14.5=21,\quad 26.5+42.5=69,\quad \text{and}\quad 21+69=90. \] Thus, the midpoint Riemann sum is: \[ \boxed{90}. \]

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Let’s dive into this mathematical journey step by step! To start off, let's sketch the function \( f(x) = 2x^2 + 2 \) on the interval \([1, 5]\): The graph resembles a parabola that opens upwards, starting at \( (1,4) \) and reaching \( (5,52) \). You can visualize this curve as a joyful arc that rapidly climbs as \( x \) increases, showcasing the quadratic nature of the function. For the values of \( \Delta x \) and the grid points \( x_0, x_1, \ldots, x_n \): First, we find \( \Delta x \) by taking the formula \(\Delta x = \frac{b-a}{n} = \frac{5-1}{4} = 1\). The grid points will be \( x_0 = 1 \), \( x_1 = 2 \), \( x_2 = 3 \), \( x_3 = 4 \), and \( x_4 = 5 \), which nicely split the interval into equal sections. Now, imagine placing rectangles under the curve for the midpoint Riemann sum where you find the midpoints of each interval: these spots are \( x_{0.5} = 1.5 \), \( x_{1.5} = 2.5 \), \( x_{2.5} = 3.5 \), \( x_{3.5} = 4.5 \). Sketching the rectangles will be visually satisfying, as you’ll show how each rectangle rises to the height of \( f \) evaluated at these midpoints. Lastly, let's calculate the midpoint Riemann sum. 1. Evaluate \( f(1.5) = 2(1.5)^2 + 2 = 6.5 \), 2. Evaluate \( f(2.5) = 2(2.5)^2 + 2 = 15.5 \), 3. Evaluate \( f(3.5) = 2(3.5)^2 + 2 = 26.5 \), 4. Evaluate \( f(4.5) = 2(4.5)^2 + 2 = 38.5 \). Now, putting it all together: Midpoint Riemann Sum \( = \Delta x \times (f(1.5) + f(2.5) + f(3.5) + f(4.5)) = 1 \times (6.5 + 15.5 + 26.5 + 38.5) = 87.0 \). Voilà! You’ve beautifully computed this sum while sketching a vibrant representation of the function!

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