Math Problem Statement

Arvutage integraali ∫₀⁴ (8x² + √x) dx ligikaudne väärtus ristkülikvalemiga võttes lõigus [0, 4] kolm osalõiku. Vastus esitage ühe komakohaga täpsusega.

Solution

The task involves calculating the integral

04(8x2+x)dx\int_{0}^{4} \left( 8x^2 + \sqrt{x} \right) dx

using an approximation method (Riemann sums) with the interval [0,4][0, 4] divided into three subintervals.

Step 1: Divide the interval [0,4][0, 4] into three subintervals

The interval [0,4][0, 4] is divided into three subintervals of equal width: Δx=403=43.\Delta x = \frac{4 - 0}{3} = \frac{4}{3}.

The subinterval endpoints are: x0=0,x1=43,x2=83,x3=4.x_0 = 0, \quad x_1 = \frac{4}{3}, \quad x_2 = \frac{8}{3}, \quad x_3 = 4.

Step 2: Use the midpoint of each subinterval for the approximation

The midpoints for the subintervals are: m1=x0+x12=0+432=23,m_1 = \frac{x_0 + x_1}{2} = \frac{0 + \frac{4}{3}}{2} = \frac{2}{3}, m2=x1+x22=43+832=2,m_2 = \frac{x_1 + x_2}{2} = \frac{\frac{4}{3} + \frac{8}{3}}{2} = 2, m3=x2+x32=83+42=103.m_3 = \frac{x_2 + x_3}{2} = \frac{\frac{8}{3} + 4}{2} = \frac{10}{3}.

Step 3: Evaluate the function at each midpoint

The function is: f(x)=8x2+x.f(x) = 8x^2 + \sqrt{x}.

Evaluate f(x)f(x) at each midpoint:

  1. At m1=23m_1 = \frac{2}{3}: f(23)=8(23)2+23=849+23=329+23.f\left(\frac{2}{3}\right) = 8\left(\frac{2}{3}\right)^2 + \sqrt{\frac{2}{3}} = 8 \cdot \frac{4}{9} + \sqrt{\frac{2}{3}} = \frac{32}{9} + \sqrt{\frac{2}{3}}.

  2. At m2=2m_2 = 2: f(2)=8(2)2+2=84+2=32+2.f(2) = 8(2)^2 + \sqrt{2} = 8 \cdot 4 + \sqrt{2} = 32 + \sqrt{2}.

  3. At m3=103m_3 = \frac{10}{3}: f(103)=8(103)2+103=81009+103=8009+103.f\left(\frac{10}{3}\right) = 8\left(\frac{10}{3}\right)^2 + \sqrt{\frac{10}{3}} = 8 \cdot \frac{100}{9} + \sqrt{\frac{10}{3}} = \frac{800}{9} + \sqrt{\frac{10}{3}}.

Step 4: Apply the midpoint rule

The approximate integral is: 04(8x2+x)dxΔx[f(m1)+f(m2)+f(m3)].\int_{0}^{4} \left( 8x^2 + \sqrt{x} \right) dx \approx \Delta x \cdot \left[ f(m_1) + f(m_2) + f(m_3) \right].

Substitute Δx=43\Delta x = \frac{4}{3}: 04(8x2+x)dx43[f(23)+f(2)+f(103)].\int_{0}^{4} \left( 8x^2 + \sqrt{x} \right) dx \approx \frac{4}{3} \left[ f\left(\frac{2}{3}\right) + f(2) + f\left(\frac{10}{3}\right) \right].

Substitute the evaluated values: 04(8x2+x)dx43[(329+23)+(32+2)+(8009+103)].\int_{0}^{4} \left( 8x^2 + \sqrt{x} \right) dx \approx \frac{4}{3} \left[ \left(\frac{32}{9} + \sqrt{\frac{2}{3}}\right) + (32 + \sqrt{2}) + \left(\frac{800}{9} + \sqrt{\frac{10}{3}}\right) \right].

Step 5: Simplify and compute the result

  1. Combine terms: 329+8009=8329,\frac{32}{9} + \frac{800}{9} = \frac{832}{9}, and the constants involving square roots sum up: 23+2+103.\sqrt{\frac{2}{3}} + \sqrt{2} + \sqrt{\frac{10}{3}}.

  2. Factor the 43\frac{4}{3} and evaluate: This requires numerical approximation for the square roots and the fractions to provide a final answer rounded to one decimal place.

Let me compute the exact numerical result.import math

Midpoints

m1 = 2/3 m2 = 2 m3 = 10/3

Function values at midpoints

f_m1 = 8 * (m12) + math.sqrt(m1) f_m2 = 8 * (m22) + math.sqrt(m2) f_m3 = 8 * (m3**2) + math.sqrt(m3)

Width of subintervals

delta_x = 4 / 3

Approximation of the integral

integral_approx = delta_x * (f_m1 + f_m2 + f_m3) integral_approx171.33452859479442171.33452859479442successThe approximate value of the integral is 171.3171.3, rounded to one decimal place.

Would you like further details on the steps or have any questions? Here are five related questions you might explore:

  1. How does the choice of midpoints affect the accuracy of the approximation?
  2. Can we calculate this integral using a different method, like the trapezoidal rule?
  3. What is the exact value of the integral, and how does it compare to this approximation?
  4. How can increasing the number of subintervals improve the accuracy of this method?
  5. How is the Riemann sum applied to functions with more complex behaviors?

Tip: When approximating an integral, dividing the interval into more subintervals generally yields more accurate results.

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Math Problem Analysis

Mathematical Concepts

Integral Approximation
Riemann Sums
Midpoint Rule

Formulas

Midpoint rule: ∫ₐᵇ f(x) dx ≈ Δx * Σ f(mᵢ)
Δx = (b - a) / n, where n is the number of subintervals
f(x) = 8x² + √x

Theorems

Riemann Sum Approximation
Properties of Definite Integrals

Suitable Grade Level

High School (Grades 10-12)