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schepotkina [342]
2 years ago
8

Find the value of the integral that converges. ∫^-5_-[infinity] x^-2 dx.

Mathematics
1 answer:
Bingel [31]2 years ago
8 0

Answer:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

Step-by-step explanation:

For this case we want to find the following integral:

\int_{-\infty}^{-5} x^{-2}dx

And we can solve the integral on this way:

\int_{-\infty}^{-5} x^{-2}dx= \frac{x^{-2+1}}{-2+1} \Big|_{-\infty}^{-5}

\int_{-\infty}^{-5} x^{-2}dx= -\frac{1}{x} \Big|_{-\infty}^{-5}

And if we evaluate the integral using the fundamental theorem of calculus we got:

\int_{-\infty}^{-5} x^{-2}dx= \frac{1}{5} + \lim_{x\to -\infty} \frac{1}{x} =\frac{1}{5}

Because the \lim_{x\to -\infty} \frac{1}{x} =0

The integral converges to \frac{1}{5}

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