Questions: Determine whether the improper integral diverges or converges. ∫ from 0 to ∞ of (1 / (e^(4x) + e^(-4x))) dx converges diverges Evaluate the integral if it converges. (If the quantity diverges, enter DIVERGES.)

Determine whether the improper integral diverges or converges.

∫ from 0 to ∞ of (1 / (e^(4x) + e^(-4x))) dx

converges
diverges

Evaluate the integral if it converges. (If the quantity diverges, enter DIVERGES.)
Transcript text: Determine whether the improper integral diverges or converges. \[ \int_{0}^{\infty} \frac{1}{e^{4 x}+e^{-4 x}} d x \] converges diverges Evaluate the integral if it converges. (If the quantity diverges, enter DIVERGES.) $\square$
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Solution

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Solution Steps

To determine whether the improper integral converges or diverges, we can analyze the behavior of the integrand as x x approaches the limits of integration (0 and \infty). If the integral converges, we can evaluate it using numerical integration methods.

Step 1: Analyze the Integral

We need to evaluate the improper integral

01e4x+e4xdx. \int_{0}^{\infty} \frac{1}{e^{4x} + e^{-4x}} \, dx.

To determine whether this integral converges or diverges, we will analyze the behavior of the integrand as x x approaches 0 and \infty .

Step 2: Behavior as x0 x \to 0

As x0 x \to 0 , we can approximate the integrand:

e4x1ande4x1. e^{4x} \approx 1 \quad \text{and} \quad e^{-4x} \approx 1.

Thus,

e4x+e4x2, e^{4x} + e^{-4x} \approx 2,

leading to

1e4x+e4x12. \frac{1}{e^{4x} + e^{-4x}} \approx \frac{1}{2}.

The integral near 0 behaves like

0ϵ12dx=ϵ2, \int_{0}^{\epsilon} \frac{1}{2} \, dx = \frac{\epsilon}{2},

which converges as ϵ0 \epsilon \to 0 .

Step 3: Behavior as x x \to \infty

As x x \to \infty , we have

e4xande4x0. e^{4x} \to \infty \quad \text{and} \quad e^{-4x} \to 0.

Thus,

e4x+e4xe4x, e^{4x} + e^{-4x} \approx e^{4x},

leading to

1e4x+e4x1e4x. \frac{1}{e^{4x} + e^{-4x}} \approx \frac{1}{e^{4x}}.

The integral behaves like

M1e4xdx, \int_{M}^{\infty} \frac{1}{e^{4x}} \, dx,

which converges since

1e4xdx=14e4x. \int \frac{1}{e^{4x}} \, dx = -\frac{1}{4e^{4x}}.

Step 4: Evaluate the Integral

Since the integral converges, we can evaluate it:

01e4x+e4xdx. \int_{0}^{\infty} \frac{1}{e^{4x} + e^{-4x}} \, dx.

Using the substitution u=e4x u = e^{4x} , we have du=4e4xdx du = 4e^{4x} \, dx or dx=du4u dx = \frac{du}{4u} . The limits change from x=0 x = 0 to x= x = \infty which corresponds to u=1 u = 1 to u= u = \infty :

11u+1udu4u=1411u2+1du. \int_{1}^{\infty} \frac{1}{u + \frac{1}{u}} \cdot \frac{du}{4u} = \frac{1}{4} \int_{1}^{\infty} \frac{1}{u^2 + 1} \, du.

The integral

1u2+1du=tan1(u), \int \frac{1}{u^2 + 1} \, du = \tan^{-1}(u),

evaluated from 1 to \infty gives:

[tan1(u)]1=π2π4=π4. \left[ \tan^{-1}(u) \right]_{1}^{\infty} = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}.

Thus,

01e4x+e4xdx=14π4=π16. \int_{0}^{\infty} \frac{1}{e^{4x} + e^{-4x}} \, dx = \frac{1}{4} \cdot \frac{\pi}{4} = \frac{\pi}{16}.

Final Answer

The integral converges and its value is

π16. \boxed{\frac{\pi}{16}}.

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