Solved on Feb 26, 2024

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

STEP 1

Assumptions
1. We are evaluating the improper integral of the function $6e^{-2x}$ from $0$ 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.
$\lim_{b \to \infty} \int_{0}^{b} 6 e^{-2 x} d x$

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

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

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

STEP 1

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

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

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Determine if ∫0∞6e−2xdx\int_{0}^{\infty} 6 e^{-2 x} d x∫0∞​6e−2xdx converges or diverges. If convergent, calculate its value.

STEP 1

STEP 2

To determine convergence or divergence, we need to evaluate the limit of the integral as the upper bound approaches infinity.lim⁡b→∞∫0b6e−2xdx\lim_{b \to \infty} \int_{0}^{b} 6 e^{-2 x} d xb→∞lim​∫0b​6e−2xdx

STEP 3

First, we find the antiderivative of the function 6e−2x6e^{-2x}6e−2x.∫6e−2xdx=−3e−2x+C\int 6 e^{-2 x} d x = -3 e^{-2 x} + C∫6e−2xdx=−3e−2x+CWhere CCC is the constant of integration.

STEP 4

Now, we use the antiderivative to evaluate the integral from 000 to bbb.∫0b6e−2xdx=[−3e−2x]0b\int_{0}^{b} 6 e^{-2 x} d x = [-3 e^{-2 x}]_{0}^{b}∫0b​6e−2xdx=[−3e−2x]0b​

STEP 5

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

STEP 6

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

STEP 7

Now, we take the limit as bbb approaches infinity.lim⁡b→∞(−3e−2b+3)\lim_{b \to \infty} (-3 e^{-2 b} + 3)b→∞lim​(−3e−2b+3)

STEP 8

Evaluate the limit term by term. The term −3e−2b-3 e^{-2 b}−3e−2b approaches 000 as bbb approaches infinity because the exponential function decays to zero.lim⁡b→∞(−3e−2b)=0\lim_{b \to \infty} (-3 e^{-2 b}) = 0b→∞lim​(−3e−2b)=0

STEP 9

The constant term remains unchanged by the limit.lim⁡b→∞(3)=3\lim_{b \to \infty} (3) = 3b→∞lim​(3)=3

STEP 10

Combine the results of the limits.lim⁡b→∞(−3e−2b+3)=0+3=3\lim_{b \to \infty} (-3 e^{-2 b} + 3) = 0 + 3 = 3b→∞lim​(−3e−2b+3)=0+3=3

STEP 11

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

STEP 12

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

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

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

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

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

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

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

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

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

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