Solved on Feb 26, 2024

Determine if 06e2xdx\int_{0}^{\infty} 6 e^{-2 x} d x converges or diverges. If convergent, calculate its value.

STEP 1

Assumptions
1. We are evaluating the improper integral of the function 6e2x6e^{-2x} from 00 to \infty.
2. An improper integral converges if the limit of its integral as the upper bound approaches infinity exists and is finite.
3. An improper integral diverges if the limit does not exist or is infinite.

STEP 2

To determine convergence or divergence, we need to evaluate the limit of the integral as the upper bound approaches infinity.
limb0b6e2xdx\lim_{b \to \infty} \int_{0}^{b} 6 e^{-2 x} d x

STEP 3

First, we find the antiderivative of the function 6e2x6e^{-2x}.
6e2xdx=3e2x+C\int 6 e^{-2 x} d x = -3 e^{-2 x} + C
Where CC is the constant of integration.

STEP 4

Now, we use the antiderivative to evaluate the integral from 00 to bb.
0b6e2xdx=[3e2x]0b\int_{0}^{b} 6 e^{-2 x} d x = [-3 e^{-2 x}]_{0}^{b}

STEP 5

We plug in the limits of integration.
[3e2x]0b=3e2b(3e20)[-3 e^{-2 x}]_{0}^{b} = -3 e^{-2 b} - (-3 e^{-2 \cdot 0})

STEP 6

Simplify the expression using the property e0=1e^{0} = 1.
3e2b(3e20)=3e2b(31)-3 e^{-2 b} - (-3 e^{-2 \cdot 0}) = -3 e^{-2 b} - (-3 \cdot 1)
=3e2b+3= -3 e^{-2 b} + 3

STEP 7

Now, we take the limit as bb approaches infinity.
limb(3e2b+3)\lim_{b \to \infty} (-3 e^{-2 b} + 3)

STEP 8

Evaluate the limit term by term. The term 3e2b-3 e^{-2 b} approaches 00 as bb approaches infinity because the exponential function decays to zero.
limb(3e2b)=0\lim_{b \to \infty} (-3 e^{-2 b}) = 0

STEP 9

The constant term remains unchanged by the limit.
limb(3)=3\lim_{b \to \infty} (3) = 3

STEP 10

Combine the results of the limits.
limb(3e2b+3)=0+3=3\lim_{b \to \infty} (-3 e^{-2 b} + 3) = 0 + 3 = 3

STEP 11

Since the limit exists and is finite, the improper integral converges.

STEP 12

The value of the convergent improper integral is the result of the limit.
06e2xdx=3\int_{0}^{\infty} 6 e^{-2 x} d x = 3
Therefore, the correct choice is A and the integral evaluates to 3.
A. 06e2xdx=3\int_{0}^{\infty} 6 e^{-2 x} d x= 3
B. The improper integral does not diverge.

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