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3 years ago

Integral Calculation

Mathematics

1 answer:

Tatiana [17]3 years ago

8 0

Answer:

$\frac{4}{3(e^2+1)}$

Step-by-step explanation:

We want to evaluate:

$\int\limits^1_{-1} {\frac{x^2+1}{e^2+1} } \, dx$

This is the same as:

$\frac{2}{e^2+1} \int\limits^1_{0} {x^2+1} \, dx$

We integrate to obtain:

$\frac{2}{e^2+1} (\frac{x^3}{3}+x)|_0^1$

We evaluate to obtain:

$\frac{2}{e^2+1} (\frac{1^3}{3}+1)=\frac{4}{3(e^2+1)}$

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Step-by-step explanation:

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3 years ago

The current population of a town is 10,000 and its growth in years can be represented by P(t) = 10,000(0.2)^t, where t is the ti

DiKsa [7]

Answer:

1) 20%

2) Choice a.

Step-by-step explanation:

$P(t)=10000(0.2)^t$

1) $P(0)$ is the population initially.

$P(1)$ is the population after a year.

$\frac{P(1)}{P(0)}$ represents the population increase factor.

So let's evaluate that fraction:

$\frac{P(1)}{P(0)}$

$\frac{10000(0.2)^1}{10000(0.2)^0}$

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2) Let's figure out the population growth in terms of months instead of years.

$P(t)=10000(0.2)^{t}$

We want t to represent months.

A full year is 12 months, in a full year we have that $P(1)=10000(0.2)^1=10000(0.2)=2000$

So we want a new P such that $P(12)=2000$ since 12 months equals a year.

Let's look at the functions given to see which gives us this:

a) $P(12)=10000(0.87449)^{12}=2000 \text{approximately}$

b) $P(12)=10000(0.87449)^{12(12)}=0 \text{ approximately}$

c) $P(12)=10000(0.87449)^{\frac{1}{12}}=9889 \text{approximately}$

d) $P(12)=10000(0.87449)^{12+12}=400 \text{approximately}$

So a is the function we want.

Also another way to look at this:

$P(t)=10000(.2)^t$ where $t$ is in years.

$P(t)=10000(.2^\frac{1}{12})^t$ where $t$ is in months.

And $.2^\frac{1}{12}=0.874485 \text{approximately}$

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