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2 years ago

I really don’t get this one

Physics

1 answer:

worty [1.4K]2 years ago

4 0

Compounds are molecules with 2 or more elements

So the answer would be the third one

CO2;H2O

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1.(16 pts.) Find the volume of the solid obtained by revolving the region enclosed by y = xex , y = 0 and x = 1 about the x-axis

MrRa [10]

Answer:

The Volume is 5.018 cubic units

Explanation:

Volume Of A Solid Of Revolution

Let f(x) be a continuous function defined in an interval [a,b], if we take the area enclosed by f(x) between x=a, x=b and revolve it around the x-axis, we get a solid whose volume can be computed as

$\displaystyle V=\pi \int_a^bf^2(x)dx$

It's called the disk method. There are other available methods to compute the volume.

We have

$f(x)=xe^x$

And the boundaries defined as x=1, y=0 and revolved around the x-axis. The left endpoint of the integral is easily identified as x=0, because it defines the beginning of the region to revolve. So we need to compute

$\displaystyle V=\pi \int_0^1(xe^x)^2dx=\pi \int_0^1x^2e^{2x}dx$

We need to first determine the antiderivative

$\displaystyle I=\int x^2e^{2x}dx$

Let's integrate by parts using the formula

$\displaystyle \int u.dv=u.v-\int v.du$

We pick $u=x^2,\ dv=e^{2x}dx$

Then $du=2xdx,\ v=\frac{e^{2x}}{2}$

Applying by parts:

$\displaystyle I=x^2\frac{e^{2x}}{2}-\int 2x\frac{e^{2x}}{2}dx$

$\displaystyle I=\frac{x^2e^{2x}}{2}-\int xe^{2x}dx$

Now we solve

$\displaystyle I_1=\int xe^{2x}dx$

Making $u=x,\ dv=e^{2x}dx$

$\displaystyle du=dx,\ v=\frac{e^{2x}}{2}$

Applying by parts again:

$\displaystyle I_1=x\frac{e^{2x}}{2}-\int \frac{e^{2x}}{2}dx$

$\displaystyle I_1=\frac{xe^{2x}}{2}-\frac{1}{2}\int e^{2x}dx$

The last integral is directly computed

$\displaystyle \int e^{2x}dx=\frac{e^{2x}}{2}$

Replacing every integral computed above

$\displaystyle I=\frac{x^2e^{2x}}{2}-\left(\frac{xe^{2x}}{2}-\frac{1}{2}\frac{e^{2x}}{2}\right)$

Simplifying

$\displaystyle I=\dfrac{\left(2x^2-2x+1\right)\mathrm{e}^{2x}}{4}$

Now we compute the definite integral as the volume

$V=\pi \left[\dfrac{\left(2(1)^2-2(1)+1\right)\mathrm{e}^{2(1)}-\left(2(0)^2-2(0)+1\right)\mathrm{e}^{2(0)}}{4}\right]$

Finally

$V=\pi \dfrac{\mathrm{e}^2-1}{4}=5.018$

The Volume is 5.018 cubic units

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A body floats at a volume of 2/5 when it is placed in oil, how dense the object is ....s=900km/m3

Nookie1986 [14]

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