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2 years ago

Solve for x to the nearest tenth!!!!

Mathematics

1 answer:

Sedaia [141]2 years ago

6 0

Answer:

√2685 is your answer

hope this will help you

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Use Simpson's Rule with n = 10 to estimate the arc length of the curve. Compare your answer with the value of the integral produ

SOVA2 [1]

$y=\ln(6+x^3)\implies y'=\dfrac{3x^2}{6+x^3}$

The arc length of the curve is

$\displaystyle\int_0^5\sqrt{1+\frac{9x^4}{(6+x^3)^2}}\,\mathrm dx$

which has a value of about 5.99086.

Let $f(x)=\sqrt{1+\frac{9x^4}{(6+x^3)^2}}$ . Split up the interval of integration into 10 subintervals,

[0, 1/2], [1/2, 1], [1, 3/2], ..., [9/2, 5]

The left and right endpoints are given respectively by the sequences,

$\ell_i=\dfrac{i-1}2$

$r_i=\dfrac i2$

with $1\le i\le10$ .

These subintervals have midpoints given by

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

Over each subinterval, we approximate $f(x)$ with the quadratic polynomial

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

so that the integral we want to find can be estimated as

$\displaystyle\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It turns out that

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

so that the arc length is approximately

$\displaystyle\sum_{i=1}^{10}\frac{f(\ell_i)+4f(m_i)+f(r_i)}6\approx5.99086$

5 0

3 years ago

For the function given below, find a formula for the Riemann sum obtained by dividing the interval (0, 3) into n equal subinterv

Viktor [21]

Splitting up [0, 3] into $n$ equally-spaced subintervals of length $\Delta x=\frac{3-0}n = \frac3n$ gives the partition

$\left[0, \dfrac3n\right] \cup \left[\dfrac3n, \dfrac6n\right] \cup \left[\dfrac6n, \dfrac9n\right] \cup \cdots \cup \left[\dfrac{3(n-1)}n, 3\right]$

where the right endpoint of the $i$ -th subinterval is given by the sequence

$r_i = \dfrac{3i}n$

for $i\in\{1,2,3,\ldots,n\}$ .

Then the definite integral is given by the infinite Riemann sum

$\displaystyle \int_0^3 2x^2 \, dx = \lim_{n\to\infty} \sum_{i=1}^n 2{r_i}^2 \Delta x \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac6n \sum_{i=1}^n \left(\frac{3i}n\right)^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3} \sum_{i=1}^n i^2 \\\\ ~~~~~~~~ = \lim_{n\to\infty} \frac{54}{n^3}\cdot\frac{n(n+1)(2n+1)}6 = \boxed{18}$