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Nadya [2.5K]
3 years ago
14

A wheel has a diameter of 3.5ft. Approximately how far does it travel if it makes 20 complete revolutions? use 22/7 for pi.

Mathematics
1 answer:
Anna [14]3 years ago
8 0
If the wheel doesn't slip or skid, then every time it makes
a revolution, it travels the circumference of the wheel.

If its diameter is 3.5 ft, then its circumference is  (3.5 pi)

                   = (3.5 ft) x (22/7)  =  77/7 ft  =  11 ft .

In 20 complete revolutions, it travels

                   (20) x (11 ft)  =  220 ft .

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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.
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Split up the integration interval into 4 subintervals:

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

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

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

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

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We approximate the area for each subinterval by

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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.

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