When the water was 4 feet from the end of the hose, what was its height above the ground? 3.2 feet 4.8 feet 5.6 feet 6.8 feet

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

The population of Tinyville is 16.2 Tinys per acre. Exactly 405 Tinys live in Tinyville. How many acres is Tinyville? Only enter

Mice21 [21]

The answer is 25 because 405/16.2=25.

8 0

3 years ago

What is the value of this expression when c = -4 and d = 10?

jok3333 [9.3K]

Where is the expression?

3 0

3 years ago

Calculate the area of quadrilateral ABCD.<br> State the units of your answer.

cluponka [151]

Answer:

158.76 cm²

Step-by-step explanation:

Quadrilateral ABCD is composed of 2 right triangles

Construct the line BD, then

Δ ABD and Δ BCD are the 2 right triangles.

The area (A) of a triangle is calculated using

A = $\frac{1}{2}$ bh ( b is the base and h the perpendicular height )

Δ ABD with b = 18 and h = 7.5

A = $\frac{1}{2}$ × 18 × 7.5 = 9 × 7.5 = 67.5 cm²

Δ BCD with b = 15.6 and h = 11.7

A = $\frac{1}{2}$ × 15.6 × 11.7 = 7.8 × 11.7 = 91.26 cm²

Thus

area of ABCD = 67.5 + 91.26 = 158.76 cm²

3 0

3 years ago

Height of a table tennis net: 6 in. height of a tennis net: 3ft

astraxan [27]

The table is 3.5 ft tall

7 0

3 years ago