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poizon [28]
2 years ago
6

Im not really sure ugh

Mathematics
2 answers:
Alexxandr [17]2 years ago
5 0

Answer:

If you have any questions I can help I just don't see a pic or anything for this one.

Step-by-step explanation:

Have a good day.

tekilochka [14]2 years ago
4 0

Answer:

I am not sure either

Step-by-step explanation:

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a) So, this integral is convergent.

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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}\\

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\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.

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\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\\

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3 years ago
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