Question

Use the Midpoint Rule with the given value of n to approximate the integral. Round the answer to four decimal places. 64 0 sin( x ) dx, n = 4

Answer 1

Answer:

$\int\limits^{64}_0 {sin(x)} \, dx\approx 4.9168$

Step-by-step explanation:

We want to approximate the integral

$\int\limits^{64}_0 {sin(x)} \, dx$

using n = 4.

The Midpoint Rule is given by

$\int_{a}^{b}f(x)dx\approx\Delta{x}\left(f\left(\frac{x_0+x_1}{2}\right)+f\left(\frac{x_1+x_2}{2}\right)+f\left(\frac{x_2+x_3}{2}\right)+...+f\left(\frac{x_{n-2}+x_{n-1}}{2}\right)+f\left(\frac{x_{n-1}+x_{n}}{2}\right)\right)$

where $\Delta{x}=\frac{b-a}{n}$

We know that a = 0, b = 64 and n = 4. Therefore, $\Delta{x}=\frac{64-0}{4}=16$.

We need to divide the interval [0, 64] into 4 sub-intervals of length $\Delta x =16$; 0, 16, 32, 48, 64.

Now, we just evaluate the function at these endpoints:

$f\left(\frac{x_{0}+x_{1}}{2}\right)=f\left(\frac{\left(0\right)+\left(16\right)}{2}\right)=f\left(8\right)=\sin{\left(8 \right)}=0.9894$

$f\left(\frac{x_{1}+x_{2}}{2}\right)=f\left(\frac{\left(16\right)+\left(32\right)}{2}\right)=f\left(24\right)=\sin{\left(24 \right)}=-0.9056$

$f\left(\frac{x_{2}+x_{3}}{2}\right)=f\left(\frac{\left(32\right)+\left(48\right)}{2}\right)=f\left(40\right)=\sin{\left(40 \right)}=0.7451$

$f\left(\frac{x_{3}+x_{4}}{2}\right)=f\left(\frac{\left(48\right)+\left(64\right)}{2}\right)=f\left(56\right)=\sin{\left(56 \right)}=-0.5216$

Finally, just sum up the above values and multiply by $\Delta x =16$

$\int\limits^{64}_0 {sin(x)} \, dx \approx 16(0.9894-0.9056+0.7451-0.5216)=4.9168$

Answer 2

The mathematical problem above is solved using one of the important theorems in geometry called the Mid Point Theorem. The solution is given as $\int_{0}^{64}$sin(x) dx ≈ 5

What is the Midpoint Theorem?

The midpoint theorem indicates that the center point of the line segment is an average of the endpoints and it requires coordinates "X" and "Y" to be known before the equation can be solved.

The solution to the problem above

From the information given, we can state that:

$\int_{0}^{64}$ sin (x) dx; where n = 4

Recall that the midpoint formula states:

$\int_{a}^{b}$ f(x) dx ≈ Δx (f (x₀ + x₁)/2) + (f (x₁ + x₂)/2) + ... (f (xₙ₋₂ + xₙ₋₁)/2) + (f (xₙ₋₁ + xₙ)/2))

Where Δx = (b-a)/n

Recall that

a= 0

b= 64 and

n=4, therefore,

Δx = (64-0)/4 = 16

The next step requires that the interval [0,64] be divided into 4 sub-intervals with lenght = Δx =16

Hence, we've got, 0, 16, 32, 48, 64.

From this point, we calibrate the respective functions as follows:

f (x₀ + x₁)/2) = f ((0+16)/2) = f(8) = Sin (8) = 0.98935824662

(f (x₁ + x₂)/2) = f((16+32)/2) = f(24) =Sin (24) = -0.905578362

(f (x₂ + x₃)/2) = f((32+48)/2) = f(40) = Sin (40) = 0.74511316047

(f (x₃ + x₄)/2) = f((48+64)/2) = f(56) = Sin (56) = -0.52155100208

At this point, we sum up the above values derive the product of the total and Δx = 16

$\int_{0}^{64}$sin (x) dx = 16(0.98935824662 - 0.905578362 +  0.74511316047 - 0.52155100208)

= 16 (0.30734204301)

= 4.91747268816

≈ 5

Learn more about the midpoint theorem at:
https://brainly.com/question/14655028