Math Problem Statement

Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, state your answer as "divergent".

∫∞68(x+7)3/2dx

Solution

We are tasked with determining whether the following improper integral is divergent or convergent, and if convergent, we will evaluate it:

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

Step 1: Determine Convergence or Divergence

The integral is improper because the upper limit is infinity. To analyze the behavior of the integral, let's start by checking whether the integrand behaves properly at infinity.

We need to examine the convergence of:

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

Analyze the integrand at infinity:

For large values of xx, the term (x+7)3/2(x + 7)^{3/2} behaves like x3/2x^{3/2}. So, asymptotically, the integrand behaves like:

1x3/2\frac{1}{x^{3/2}}

To determine if this integral converges, we compare it to the standard p-integral:

a1xpdx\int_{a}^{\infty} \frac{1}{x^p} \, dx

This integral converges if p>1p > 1. In our case, p=32p = \frac{3}{2}, which is greater than 1, so the integral converges.

Step 2: Evaluate the Integral

Now, let's proceed with the evaluation of the integral:

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

We will make a substitution to simplify the integral.

Let:

u=x+7du=dxu = x + 7 \quad \Rightarrow \quad du = dx

When x=6x = 6, u=13u = 13. When xx \to \infty, uu \to \infty.

Thus, the integral becomes:

131u3/2du\int_{13}^{\infty} \frac{1}{u^{3/2}} \, du

Now, evaluate this integral:

13u3/2du\int_{13}^{\infty} u^{-3/2} \, du

The antiderivative of u3/2u^{-3/2} is:

u1/21/2=2u1/2=2u\frac{u^{-1/2}}{-1/2} = -2u^{-1/2} = -\frac{2}{\sqrt{u}}

Now, apply the limits:

[2u]13\left[ -\frac{2}{\sqrt{u}} \right]_{13}^{\infty}

At u=u = \infty, 2u0\frac{2}{\sqrt{u}} \to 0. At u=13u = 13, we have:

213-\frac{2}{\sqrt{13}}

Thus, the value of the integral is:

0(213)=2130 - \left( -\frac{2}{\sqrt{13}} \right) = \frac{2}{\sqrt{13}}

Conclusion

The integral is convergent, and the evaluated result is:

213\frac{2}{\sqrt{13}}


Do you want more details or have any questions? Here are some additional related questions for practice:

  1. Evaluate 11(x+2)2dx\int_1^\infty \frac{1}{(x+2)^2} dx.
  2. Determine if 21x3dx\int_2^\infty \frac{1}{x^3} dx is convergent or divergent.
  3. Find the value of 31x5/2dx\int_3^\infty \frac{1}{x^{5/2}} dx.
  4. Is 0dx(x+1)3\int_0^\infty \frac{dx}{(x+1)^3} convergent or divergent?
  5. Evaluate 01x3/2dx\int_0^\infty \frac{1}{x^{3/2}} dx.

Tip: When solving improper integrals, checking the behavior at infinity using the pp-integral test is very useful!

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Math Problem Analysis

Mathematical Concepts

Calculus
Improper Integrals
p-Integral Test
Convergence and Divergence

Formulas

∫ 1/(x + 7)^(3/2) dx
p-Integral Test: ∫ (1/x^p) dx for p > 1

Theorems

p-Integral Convergence Theorem
Power Rule for Integration

Suitable Grade Level

Grades 11-12 or introductory university calculus