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3 years ago

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A region is bounded by y=e−3x, the x-axis, the y-axis and the line x = 3. If the region is the base of a solid such that each cr

oss section perpendicular to the x-axis is an equilateral triangle, set up the integral that would find the volume of that solid.

1 answer:

grigory [225]3 years ago

8 0

Answer:

0.333 $\pi$ units³

Step-by-step explanation:

Think process:

The equation is given as y = $e^{-3x}$

Let, y = f (x)

Therefore, $f (x) = e^{-3x}$

We know that the limits are y-axis and x= 3

Y-axis: x= 0

then limits are given as x= 0 and x = 3

Integrating gives:

$\int\limits^3_0 {e^{-3x} } \, dx$ = $\frac{-1}{3}e^{-3x}$ + C

calculating from x= 0 to x = 3, we know volume is given by $\pi \int\limits^a_b {f(x)} \, dx$

= $\pi$ [ $\frac{-1}{3} e^{-9} - (\frac{-1}{3} e^{0})$ ]

= $\pi$ [0.000041136 + 1/3]

= 0.333 $\pi$ units³

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Use Hooke's Law to determine the work done by the variable force in the spring problem. A force of 450 newtons stretches a sprin

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Answer:

The work done is 202.50Nm

Step-by-step explanation:

Given

$F =450N$

$x_1 = 30cm$

$x_2 = 60cm$

Required

The work done

First, we calculate the spring constant (k)

$F = kx_1$

$450N = k *30cm$

$k = \frac{450N}{30cm}$

$k =15N/cm$

So:

$F = kx_1$

$F(x) = 15x$

The work done using Hooke's law is:

$W =\int\limits^a_b {F(x)} \, dx$

This gives:

$W =\int\limits^{60}_{30} {15x} \, dx$

Rewrite as:

$W =15\int\limits^{60}_{30} {x} \, dx$

Integrate

$W =15 \frac{x^2}{2}|\limits^{60}_{30}$

This gives:

$W =15 *\frac{60^2 - 30^2}{2}$

$W =15 *\frac{2700}{2}$

$W =15 *1350$

$W =20250N-cm$

Convert to Nm

$W =\frac{20250Nm}{100}$

$W =202.50Nm$

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Step-by-step explanation:

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2,17,82,257,626,1297 next one please ?

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The easy thing to do is notice that 1^4 = 1, 2^4 = 16, 3^4 = 81, and so on, so the sequence follows the rule $n^4+1$ . The next number would then be fourth power of 7 plus 1, or 2402.

And the harder way: Denote the <em>n</em>-th term in this sequence by $a_n$ , and denote the given sequence by $\{a_n\}_{n\ge1}$ .

Let $b_n$ denote the <em>n</em>-th term in the sequence of forward differences of $\{a_n\}$ , defined by

$b_n=a_{n+1}-a_n$

for <em>n</em> ≥ 1. That is, $\{b_n\}$ is the sequence with

$b_1=a_2-a_1=17-2=15$

$b_2=a_3-a_2=82-17=65$

$b_3=a_4-a_3=175$

$b_4=a_5-a_4=369$

$b_5=a_6-a_5=671$

and so on.

Next, let $c_n$ denote the <em>n</em>-th term of the differences of $\{b_n\}$ , i.e. for <em>n</em> ≥ 1,

$c_n=b_{n+1}-b_n$

so that

$c_1=b_2-b_1=65-15=50$

$c_2=110$

$c_3=194$

$c_4=302$

etc.

Again: let $d_n$ denote the <em>n</em>-th difference of $\{c_n\}$ :

$d_n=c_{n+1}-c_n$

$d_1=c_2-c_1=60$

$d_2=84$

$d_3=108$

etc.

One more time: let $e_n$ denote the <em>n</em>-th difference of $\{d_n\}$ :

$e_n=d_{n+1}-d_n$

$e_1=d_2-d_1=24$

$e_2=24$

etc.

The fact that these last differences are constant is a good sign that $e_n=24$ for all <em>n</em> ≥ 1. Assuming this, we would see that $\{d_n\}$ is an arithmetic sequence given recursively by

$\begin{cases}d_1=60\\d_{n+1}=d_n+24&\text{for }n>1\end{cases}$

and we can easily find the explicit rule:

$d_2=d_1+24$

$d_3=d_2+24=d_1+24\cdot2$

$d_4=d_3+24=d_1+24\cdot3$

and so on, up to

$d_n=d_1+24(n-1)$

$d_n=24n+36$

Use the same strategy to find a closed form for $\{c_n\}$ , then for $\{b_n\}$ , and finally $\{a_n\}$ .

$\begin{cases}c_1=50\\c_{n+1}=c_n+24n+36&\text{for }n>1\end{cases}$

$c_2=c_1+24\cdot1+36$

$c_3=c_2+24\cdot2+36=c_1+24(1+2)+36\cdot2$

$c_4=c_3+24\cdot3+36=c_1+24(1+2+3)+36\cdot3$

and so on, up to

$c_n=c_1+24(1+2+3+\cdots+(n-1))+36(n-1)$

Recall the formula for the sum of consecutive integers:

$1+2+3+\cdots+n=\displaystyle\sum_{k=1}^nk=\frac{n(n+1)}2$

$\implies c_n=c_1+\dfrac{24(n-1)n}2+36(n-1)$

$\implies c_n=12n^2+24n+14$

$\begin{cases}b_1=15\\b_{n+1}=b_n+12n^2+24n+14&\text{for }n>1\end{cases}$

$b_2=b_1+12\cdot1^2+24\cdot1+14$

$b_3=b_2+12\cdot2^2+24\cdot2+14=b_1+12(1^2+2^2)+24(1+2)+14\cdot2$

$b_4=b_3+12\cdot3^2+24\cdot3+14=b_1+12(1^2+2^2+3^2)+24(1+2+3)+14\cdot3$

and so on, up to

$b_n=b_1+12(1^2+2^2+3^2+\cdots+(n-1)^2)+24(1+2+3+\cdots+(n-1))+14(n-1)$

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$\implies b_n=15+\dfrac{12(n-1)n(2(n-1)+1)}6+\dfrac{24(n-1)n}2+14(n-1)$

$\implies b_n=4n^3+6n^2+4n+1$

$\begin{cases}a_1=2\\a_{n+1}=a_n+4n^3+6n^2+4n+1&\text{for }n>1\end{cases}$

$a_2=a_1+4\cdot1^3+6\cdot1^2+4\cdot1+1$

$a_3=a_2+4(1^3+2^3)+6(1^2+2^2)+4(1+2)+1\cdot2$

$a_4=a_3+4(1^3+2^3+3^3)+6(1^2+2^2+3^2)+4(1+2+3)+1\cdot3$

$\implies a_n=a_1+4\displaystyle\sum_{k=1}^3k^3+6\sum_{k=1}^3k^2+4\sum_{k=1}^3k+\sum_{k=1}^{n-1}1$

$\displaystyle\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4$

$\implies a_n=2+\dfrac{4(n-1)^2n^2}4+\dfrac{6(n-1)n(2n)}6+\dfrac{4(n-1)n}2+(n-1)$

$\implies a_n=n^4+1$

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