Find The Indicated Midpoint Rule Approximation For The Following Integral: ∫ 1 4 5 Sin ⁡ ( Π 4 X ) D X \int_1^4 5 \sin \left(\frac{\pi}{4} X\right) \, Dx ∫ 1 4 ​ 5 Sin ( 4 Π ​ X ) D X Using N = 6 N = 6 N = 6 Subintervals.

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Introduction

The Midpoint Rule is a numerical integration technique used to approximate the value of a definite integral. It is based on the idea of approximating the area under a curve by dividing the area into small rectangles and summing the areas of these rectangles. In this article, we will discuss how to find the Midpoint Rule approximation for the given integral: ∫145sin⁡(π4x) dx\int_1^4 5 \sin \left(\frac{\pi}{4} x\right) \, dx using n=6n = 6 subintervals.

The Midpoint Rule Formula

The Midpoint Rule formula is given by:

∫abf(x) dx≈Δx[f(x0+Δx2)+f(x1+Δx2)+⋯+f(xn−1+Δx2)]\int_a^b f(x) \, dx \approx \Delta x \left[ f\left(x_0 + \frac{\Delta x}{2}\right) + f\left(x_1 + \frac{\Delta x}{2}\right) + \cdots + f\left(x_{n-1} + \frac{\Delta x}{2}\right) \right]

where Δx=b−an\Delta x = \frac{b-a}{n} is the width of each subinterval, and xix_i are the points of subdivision.

Calculating the Width of Each Subinterval

To calculate the width of each subinterval, we need to divide the total length of the interval [a,b][a,b] by the number of subintervals nn. In this case, the total length of the interval is 4−1=34-1=3, and the number of subintervals is n=6n=6. Therefore, the width of each subinterval is:

Δx=b−an=4−16=36=12\Delta x = \frac{b-a}{n} = \frac{4-1}{6} = \frac{3}{6} = \frac{1}{2}

Finding the Midpoints of Each Subinterval

To find the Midpoint Rule approximation, we need to find the midpoints of each subinterval. The midpoints are given by:

xi+Δx2=xi+12x_i + \frac{\Delta x}{2} = x_i + \frac{1}{2}

for i=0,1,2,3,4,5i=0,1,2,3,4,5.

Evaluating the Function at the Midpoints

We need to evaluate the function f(x)=5sin⁡(π4x)f(x) = 5 \sin \left(\frac{\pi}{4} x\right) at the midpoints of each subinterval. The function values are:

f(x0+Δx2)=f(1+12)=f(32)=5sin⁡(π4⋅32)=5sin⁡(3π8)f\left(x_0 + \frac{\Delta x}{2}\right) = f\left(1 + \frac{1}{2}\right) = f\left(\frac{3}{2}\right) = 5 \sin \left(\frac{\pi}{4} \cdot \frac{3}{2}\right) = 5 \sin \left(\frac{3\pi}{8}\right)

f(x1+Δx2)=f(32+12)=f(2)=5sin⁡(π4⋅2)=5sin⁡(π2)f\left(x_1 + \frac{\Delta x}{2}\right) = f\left(\frac{3}{2} + \frac{1}{2}\right) = f(2) = 5 \sin \left(\frac{\pi}{4} \cdot 2\right) = 5 \sin \left(\frac{\pi}{2}\right)

f(x2+Δx2)=f(2+12)=f(72)=5sin⁡(π4⋅72)=5sin⁡(7π8)f\left(x_2 + \frac{\Delta x}{2}\right) = f\left(2 + \frac{1}{2}\right) = f\left(\frac{7}{2}\right) = 5 \sin \left(\frac{\pi}{4} \cdot \frac{7}{2}\right) = 5 \sin \left(\frac{7\pi}{8}\right)

f(x3+Δx2)=f(72+12)=f(3)=5sin⁡(π4⋅3)=5sin⁡(3π4)f\left(x_3 + \frac{\Delta x}{2}\right) = f\left(\frac{7}{2} + \frac{1}{2}\right) = f(3) = 5 \sin \left(\frac{\pi}{4} \cdot 3\right) = 5 \sin \left(\frac{3\pi}{4}\right)

f(x4+Δx2)=f(3+12)=f(92)=5sin⁡(π4⋅92)=5sin⁡(9π8)f\left(x_4 + \frac{\Delta x}{2}\right) = f\left(3 + \frac{1}{2}\right) = f\left(\frac{9}{2}\right) = 5 \sin \left(\frac{\pi}{4} \cdot \frac{9}{2}\right) = 5 \sin \left(\frac{9\pi}{8}\right)

f(x5+Δx2)=f(92+12)=f(4)=5sin⁡(π4⋅4)=5sin⁡(π)f\left(x_5 + \frac{\Delta x}{2}\right) = f\left(\frac{9}{2} + \frac{1}{2}\right) = f(4) = 5 \sin \left(\frac{\pi}{4} \cdot 4\right) = 5 \sin \left(\pi\right)

Calculating the Midpoint Rule Approximation

Now that we have the function values at the midpoints of each subinterval, we can calculate the Midpoint Rule approximation:

∫145sin⁡(π4x) dx≈Δx[f(x0+Δx2)+f(x1+Δx2)+⋯+f(xn−1+Δx2)]\int_1^4 5 \sin \left(\frac{\pi}{4} x\right) \, dx \approx \Delta x \left[ f\left(x_0 + \frac{\Delta x}{2}\right) + f\left(x_1 + \frac{\Delta x}{2}\right) + \cdots + f\left(x_{n-1} + \frac{\Delta x}{2}\right) \right]

=12[5sin⁡(3π8)+5sin⁡(π2)+5sin⁡(7π8)+5sin⁡(3π4)+5sin⁡(9π8)+5sin⁡(π)]= \frac{1}{2} \left[ 5 \sin \left(\frac{3\pi}{8}\right) + 5 \sin \left(\frac{\pi}{2}\right) + 5 \sin \left(\frac{7\pi}{8}\right) + 5 \sin \left(\frac{3\pi}{4}\right) + 5 \sin \left(\frac{9\pi}{8}\right) + 5 \sin \left(\pi\right) \right]

=12[5sin⁡(3π8)+5⋅1+5sin⁡(7π8)+5⋅22+5sin⁡(9π8)+5⋅0]= \frac{1}{2} \left[ 5 \sin \left(\frac{3\pi}{8}\right) + 5 \cdot 1 + 5 \sin \left(\frac{7\pi}{8}\right) + 5 \cdot \frac{\sqrt{2}}{2} + 5 \sin \left(\frac{9\pi}{8}\right) + 5 \cdot 0 \right]

=12[5sin⁡(3π8)+5+5sin⁡(7π8)+522+5sin⁡(9π8)]= \frac{1}{2} \left[ 5 \sin \left(\frac{3\pi}{8}\right) + 5 + 5 \sin \left(\frac{7\pi}{8}\right) + \frac{5\sqrt{2}}{2} + 5 \sin \left(\frac{9\pi}{8}\right) \right]

Simplifying the Expression

To simplify the expression, we can use the following trigonometric identities:

sin⁡(3π8)=sin⁡(π2−π8)=cos⁡(π8)\sin \left(\frac{3\pi}{8}\right) = \sin \left(\frac{\pi}{2} - \frac{\pi}{8}\right) = \cos \left(\frac{\pi}{8}\right)

sin⁡(7π8)=sin⁡(π2+π8)=cos⁡(π8)\sin \left(\frac{7\pi}{8}\right) = \sin \left(\frac{\pi}{2} + \frac{\pi}{8}\right) = \cos \left(\frac{\pi}{8}\right)

sin⁡(9π8)=sin⁡(π2+π8)=cos⁡(π8)\sin \left(\frac{9\pi}{8}\right) = \sin \left(\frac{\pi}{2} + \frac{\pi}{8}\right) = \cos \left(\frac{\pi}{8}\right)

Substituting these identities into the expression, we get:

=12[5cos⁡(π8)+5+522+5cos⁡(π8)]= \frac{1}{2} \left[ 5 \cos \left(\frac{\pi}{8}\right) + 5 + \frac{5\sqrt{2}}{2} + 5 \cos \left(\frac{\pi}{8}\right) \right]

=12[10cos⁡(π8)+5+522]= \frac{1}{2} \left[ 10 \cos \left(\frac{\pi}{8}\right) + 5 + \frac{5\sqrt{2}}{2} \right]

Evaluating the Trigonometric Functions

To evaluate the trigonometric functions, we can use the following values:

cos⁡(π8)=2+22\cos \left(\frac{\pi}{8}\right) = \frac{\sqrt{2+\sqrt{2}}}{2}

Substituting this value into the expression, we get:

=12[10⋅2+22+5+522]= \frac{1}{2} \left[ 10 \cdot \frac{\sqrt{2+\sqrt{2}}}{2} + 5 + \frac{5\sqrt{2}}{2} \right]

=12[52+2+5+522]= \frac{1}{2} \left[ 5 \sqrt{2+\sqrt{2}} + 5 + \frac{5\sqrt{2}}{2} \right]

Simplifying the Expression

To simplify the expression, we can combine the terms:

=12[52+2+522+5]= \frac{1}{2} \left[ 5 \sqrt{2+\sqrt{2}} + \frac{5\sqrt{2}}{2} + 5 \right]

= \frac{1}{2} \left<br/> # **Midpoint Rule Approximation: Frequently Asked Questions** ===========================================================

Q: What is the Midpoint Rule?

A: The Midpoint Rule is a numerical integration technique used to approximate the value of a definite integral. It is based on the idea of approximating the area under a curve by dividing the area into small rectangles and summing the areas of these rectangles.

Q: How does the Midpoint Rule work?

A: The Midpoint Rule works by dividing the interval [a,b][a,b] into nn subintervals, each of width Δx=b−an\Delta x = \frac{b-a}{n}. The function is then evaluated at the midpoint of each subinterval, and the sum of these function values is multiplied by Δx\Delta x to obtain the Midpoint Rule approximation.

Q: What are the advantages of the Midpoint Rule?

A: The advantages of the Midpoint Rule include:

  • It is easy to implement and understand.
  • It is a simple and efficient method for approximating definite integrals.
  • It can be used to approximate integrals with a wide range of functions.

Q: What are the disadvantages of the Midpoint Rule?

A: The disadvantages of the Midpoint Rule include:

  • It can be less accurate than other numerical integration methods, such as the Trapezoidal Rule or Simpson's Rule.
  • It can be sensitive to the choice of subinterval width.

Q: When should I use the Midpoint Rule?

A: You should use the Midpoint Rule when:

  • You need a simple and efficient method for approximating a definite integral.
  • You are working with a function that is easy to evaluate at the midpoint of each subinterval.
  • You are looking for a rough estimate of the value of a definite integral.

Q: How do I choose the number of subintervals (n) for the Midpoint Rule?

A: The choice of nn depends on the desired level of accuracy and the complexity of the function being integrated. A larger value of nn will generally result in a more accurate approximation, but will also increase the computational cost.

Q: Can I use the Midpoint Rule to approximate integrals with infinite limits?

A: No, the Midpoint Rule is not suitable for approximating integrals with infinite limits. The Midpoint Rule requires a finite interval [a,b][a,b] to work, and infinite limits are not compatible with this requirement.

Q: Can I use the Midpoint Rule to approximate integrals with discontinuous functions?

A: Yes, the Midpoint Rule can be used to approximate integrals with discontinuous functions. However, the accuracy of the approximation may be affected by the presence of discontinuities.

Q: How do I implement the Midpoint Rule in a programming language?

A: The implementation of the Midpoint Rule in a programming language will depend on the specific language and the desired level of accuracy. However, the basic steps are:

  1. Define the function to be integrated.
  2. Choose the number of subintervals (n).
  3. Calculate the width of each subinterval (Δx\Delta x).
  4. Evaluate the function at the midpoint of each subinterval.
  5. Multiply the sum of the function values by Δx\Delta x to obtain the Midpoint Rule approximation.

Q: What are some common applications of the Midpoint Rule?

A: The Midpoint Rule has a wide range of applications, including:

  • Approximating definite integrals in physics and engineering.
  • Solving differential equations.
  • Approximating areas under curves in statistics and data analysis.
  • Approximating volumes of solids in geometry and calculus.

Q: What are some common mistakes to avoid when using the Midpoint Rule?

A: Some common mistakes to avoid when using the Midpoint Rule include:

  • Choosing a value of nn that is too small, resulting in a poor approximation.
  • Failing to evaluate the function at the midpoint of each subinterval.
  • Multiplying the sum of the function values by the wrong value of Δx\Delta x.
  • Failing to check for discontinuities in the function being integrated.