Math Problem Statement

Let f be the function defined by f(x) = 5√x. If four subintervals of equal length are used, what is the value of the midpoint Riemann sum approximation for ∫₁⁹ 5√x dx? Round to the nearest thousandth if necessary.

Solution

We are asked to approximate the integral 195xdx\int_1^9 5\sqrt{x} \, dx using the midpoint Riemann sum with 4 subintervals of equal length.

Step 1: Define the function and interval

The function given is: f(x)=5xf(x) = 5\sqrt{x} The interval is [1,9][1, 9].

Step 2: Determine the width of each subinterval

The total interval length is 91=89 - 1 = 8. Since we are dividing the interval into 4 subintervals, the width of each subinterval, Δx\Delta x, is: Δx=84=2\Delta x = \frac{8}{4} = 2

Step 3: Identify the midpoints of each subinterval

The midpoints of the subintervals are found by taking the average of the endpoints of each subinterval. The subintervals are:

  • [1,3][1, 3]
  • [3,5][3, 5]
  • [5,7][5, 7]
  • [7,9][7, 9]

The midpoints are:

  • x1=1+32=2x_1 = \frac{1 + 3}{2} = 2
  • x2=3+52=4x_2 = \frac{3 + 5}{2} = 4
  • x3=5+72=6x_3 = \frac{5 + 7}{2} = 6
  • x4=7+92=8x_4 = \frac{7 + 9}{2} = 8

Step 4: Calculate the function values at the midpoints

We now evaluate f(x)f(x) at each midpoint: f(2)=52,f(4)=54,f(6)=56,f(8)=58f(2) = 5\sqrt{2}, \quad f(4) = 5\sqrt{4}, \quad f(6) = 5\sqrt{6}, \quad f(8) = 5\sqrt{8}

These simplify to: f(2)=52,f(4)=10,f(6)=56,f(8)=102f(2) = 5\sqrt{2}, \quad f(4) = 10, \quad f(6) = 5\sqrt{6}, \quad f(8) = 10\sqrt{2}

Step 5: Calculate the Riemann sum

The midpoint Riemann sum is given by: Riemann sum=Δx(f(x1)+f(x2)+f(x3)+f(x4))\text{Riemann sum} = \Delta x \left( f(x_1) + f(x_2) + f(x_3) + f(x_4) \right)

Substituting the values: Riemann sum=2(52+10+56+102)\text{Riemann sum} = 2 \left( 5\sqrt{2} + 10 + 5\sqrt{6} + 10\sqrt{2} \right) Let's calculate the approximate numerical value of this expression.

I will compute this now.The value of the midpoint Riemann sum approximation for the integral 195xdx\int_1^9 5\sqrt{x} \, dx, using four subintervals, is approximately 86.921 when rounded to the nearest thousandth.

Would you like further details or clarifications? Here are some related questions for deeper understanding:

  1. How does the midpoint Riemann sum differ from left or right Riemann sums?
  2. What are the steps to calculate the definite integral exactly instead of approximating?
  3. How does the accuracy of the approximation change with more subintervals?
  4. Can the midpoint rule be applied to functions that are not continuous?
  5. What other methods, besides Riemann sums, are commonly used for approximating integrals?

Tip: Increasing the number of subintervals generally improves the accuracy of Riemann sum approximations!

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Math Problem Analysis

Mathematical Concepts

Integral Approximation
Midpoint Riemann Sum
Square Roots

Formulas

f(x) = 5√x
Midpoint Riemann Sum Formula: Δx * (f(x₁) + f(x₂) + ... + f(xₙ))

Theorems

Riemann Sum Theorem

Suitable Grade Level

High School, Grades 10-12