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Eva8 [605]
3 years ago
13

Determine whether each integral is convergent or divergent. If it is convergent evaluate it. (a) integral from 1^(infinity) e^(-

2x) dx. (b) integral from 1^2[dz]/[(z-1)^2] . (c) integral from 1^(infinity) [dx]/[root(x)] .
Mathematics
1 answer:
AVprozaik [17]3 years ago
4 0

Answer:

a) So, this integral is convergent.

b) So, this integral is divergent.

c) So, this integral is divergent.

Step-by-step explanation:

We calculate the next integrals:

a)

\int_1^{\infty} e^{-2x} dx=\left[-\frac{e^{-2x}}{2}\right]_1^{\infty}\\\\\int_1^{\infty} e^{-2x} dx=-\frac{e^{-\infty}}{2}+\frac{e^{-2}}{2}\\\\\int_1^{\infty} e^{-2x} dx=\frac{e^{-2}}{2}\\

So, this integral is convergent.

b)

\int_1^{2}\frac{dz}{(z-1)^2}=\left[-\frac{1}{z-1}\right]_1^2\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\frac{1}{1-1}+\frac{1}{2-1}\\\\\int_1^{2}\frac{dz}{(z-1)^2}=-\infty\\

So, this integral is divergent.

c)

\int_1^{\infty} \frac{dx}{\sqrt{x}}=\left[2\sqrt{x}\right]_1^{\infty}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=2\sqrt{\infty}-2\sqrt{1}\\\\\int_1^{\infty} \frac{dx}{\sqrt{x}}=\infty\\

So, this integral is divergent.

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Answer:

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Step-by-step explanation:

Data Given:

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-3x^{2} - 3y^{2}  -3z^{2}  + 54x + 4y -30z -142  = 0

Taking (-) common

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dividing the whole equation by 3

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x^{2} + y^{2} + z^{2} - 18x - 4/3y +10z + 142/3 = 0

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3(2a + 1) = -5(a + 6)
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Answer:

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