Math / Calculus

Question6. The improper integral $\int_{1}^{\infty} \frac{1}{\sqrt{x^{2}-0.1}} d x$ A. Converges to $\sin ^{-1}\left(\frac{1}{10}\right)$ B. Converges to $\sin ^{-1}(5)$ C. Converges to $\sec ^{-1}\left(\frac{1}{5}\right)$ D. Converges to $\sec ^{-1}(2)$ E. Converges to $\tan ^{-1}\left(\frac{1}{2}\right)$ F. Diverges

Studdy Solution

STEP 1

1. The integral is improper because it has an infinite upper limit.
2. We need to determine if the integral converges or diverges.
3. If it converges, we need to find the value it converges to.

STEP 2

1. Analyze the behavior of the integrand as $x \to \infty$ .
2. Determine if the integral converges or diverges.
3. Evaluate the integral if it converges.

STEP 3

First, consider the behavior of the integrand $\frac{1}{\sqrt{x^2 - 0.1}}$ as $x \to \infty$ .
As $x \to \infty$ , the term $\sqrt{x^2 - 0.1}$ behaves like $\sqrt{x^2} = x$ . Therefore, the integrand behaves like $\frac{1}{x}$ .

STEP 4

The integral $\int_{1}^{\infty} \frac{1}{x} \, dx$ diverges because it is a $p$ -integral with $p = 1$ .
Since the behavior of the integrand $\frac{1}{\sqrt{x^2 - 0.1}}$ is similar to $\frac{1}{x}$ as $x \to \infty$ , the given integral also diverges.
The improper integral diverges.
$\boxed{\text{F. Diverges}}$

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