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Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

$\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]$

The left and right endpoints of the $i$ -th subinterval, respectively, are

$\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8$

$r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8$

Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

$T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}$

Midpoint rule

We approximate the area for each subinterval by

$M_i=f(m_i)(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}$

Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

3 0

3 years ago

Find the greatest common factor of 10, 30,10,30,10, comma, 30, comma and 454545.

borishaifa [10]

Answer:

GCF = 5

Step-by-step explanation:

Given:

The numbers are given as:

10, 30 and 45

A factor of a number is a number by which the given number is evenly divisible.

Example: 2 is a factor of 4 as $4\div 2 =2$ . So, 4 is evenly divisible by 2 and hence 2 is a factor of 4.

Now, factors of 10 = 1, 2, 5, 10.

Factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30.

Factors of 45 = 1, 3, 5, 9, 15, 45.

So, the greatest common factor in all the three numbers is 5.

Hence, the greatest common factor of 10, 30 and 45 is 5.

5 0

3 years ago

Match each expression on the left with an equivalent expression on the right

Lana71 [14]

Answer: See explanation
Step-by-step explanation:
You didn't give the expressions but here are some expressions
a. ✓4x²y^4. 1. 2x✓y
b. ✓8x²y. 2. 2y✓2x
c. ✓4x²y. 3. 2xy²
d. ✓16xy². 4. 2x✓2y
e. ✓8xy². 5. 4y✓x
a. ✓4x²y^4 = ✓4 × ✓x² × ✓y^4
= 2 × x × y²
= 2xy²
Therefore, ✓4x²y^4 = 2xy²
b. ✓8x²y = ✓8 × ✓x² × ✓y
= ✓4 × ✓2 × ✓x² × ✓y
= 2 × ✓2 × x × ✓y
= 2x✓2y
Therefore, ✓8x²y = 2x✓2y
c. ✓4x²y = ✓4 × ✓x² × ✓y
= 2 × x × ✓y
= 2x✓y
Therefore, ✓4x²y = 2x✓y
d. ✓16xy² = ✓16 × ✓x × ✓y²
= 4 × ✓x × y
= 4y✓x
Therefore, ✓16xy² = 4y✓x
e. ✓8xy² = ✓8 × ✓x × ✓y²
= ✓4 × ✓2 × ✓x × ✓y²

8 0

2 years ago

Read 2 more answers

PLEASE 10 PTS!!

ira [324]

Its c because first 14 he got -42 the four remaining he got 8=-34

6 0

3 years ago

Read 2 more answers

Will give brainliest. :)

riadik2000 [5.3K]

Answer: