∫∞68(x+7)3/2dx

We are tasked with evaluating the improper integral:

$\int_{6}^{\infty} \frac{1}{(x + 7)^{3/2}} \, dx$

Step 1: Check for convergence

To determine whether the integral is convergent, we need to examine the behavior of the integrand as $x \to \infty$ and near the lower bound $x = 6$.

Behavior as $x \to \infty$:

As $x$ approaches infinity, the integrand behaves like: $\frac{1}{(x + 7)^{3/2}} \sim \frac{1}{x^{3/2}}$ The integral of $\frac{1}{x^{3/2}}$ from 6 to infinity is known to converge because the exponent $\frac{3}{2} > 1$. So, the integral converges at infinity.

Behavior near $x = 6$:

At $x = 6$, the integrand is finite since $\frac{1}{(x + 7)^{3/2}}$ is well-behaved. There are no issues near $x = 6$, meaning the integral does not diverge at the lower bound.

Since the integral converges at both bounds, the overall integral is convergent.

Step 2: Evaluate the integral

We now proceed to evaluate the integral:

$\int_{6}^{\infty} \frac{1}{(x + 7)^{3/2}} \, dx$

To solve this, we perform a substitution. Let:

$u = x + 7$

Then:

$du = dx$ When $x = 6$, $u = 13$. When $x \to \infty$, $u \to \infty$. So the integral becomes:

$\int_{13}^{\infty} \frac{1}{u^{3/2}} \, du$

This is a standard integral. The antiderivative of $u^{-3/2}$ is:

$\int u^{-3/2} \, du = -2 u^{-1/2}$

Thus, the integral evaluates to:

$-2 \left[ u^{-1/2} \right]_{13}^{\infty}$

At the upper limit $u \to \infty$, $u^{-1/2} \to 0$. At the lower limit $u = 13$, we have:

$-2 \left( 0 - \frac{1}{\sqrt{13}} \right) = \frac{2}{\sqrt{13}}$

Thus, the value of the integral is:

$\frac{2}{\sqrt{13}}$

Final Answer:

The integral is convergent and evaluates to $\frac{2}{\sqrt{13}}$.

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Would you like further details on any of the steps? Here are some related questions you might find interesting:

1. How do improper integrals differ from definite integrals?
2. What is the significance of the exponent in terms of convergence for improper integrals?
3. Can we generalize this integral to other exponents or functions?
4. What other methods could we use to check for convergence?
5. How can we compute this integral using numerical methods?

Tip: When dealing with improper integrals, always analyze both ends of the integration range to confirm whether the integral converges or diverges.