Math / Calculus

QuestionFind $\int_{1}^{e^{2}} \frac{6}{x} d x$

Studdy Solution

STEP 1

1. We are asked to evaluate the definite integral $\int_{1}^{e^{2}} \frac{6}{x} \, dx$ .
2. The function $\frac{6}{x}$ is continuous and integrable over the interval $[1, e^2]$ .

STEP 2

1. Identify the antiderivative of the integrand $\frac{6}{x}$ .
2. Evaluate the definite integral using the Fundamental Theorem of Calculus.

STEP 3

Identify the antiderivative of $\frac{6}{x}$ :
The antiderivative of $\frac{1}{x}$ is $\ln |x|$ . Therefore, the antiderivative of $\frac{6}{x}$ is:
$\int \frac{6}{x} \, dx = 6 \ln |x| + C$
where $C$ is the constant of integration.

STEP 4

Apply the Fundamental Theorem of Calculus to evaluate the definite integral:
$\int_{1}^{e^{2}} \frac{6}{x} \, dx = \left[ 6 \ln |x| \right]_{1}^{e^{2}}$

STEP 5

Evaluate the expression at the upper limit $e^2$ :
$6 \ln |e^2| = 6 \ln (e^2) = 6 \cdot 2 \ln e = 12 \cdot 1 = 12$

STEP 6

Evaluate the expression at the lower limit $1$ :
$6 \ln |1| = 6 \ln (1) = 6 \cdot 0 = 0$

STEP 7

Subtract the evaluated lower limit from the evaluated upper limit:
$\int_{1}^{e^{2}} \frac{6}{x} \, dx = 12 - 0 = 12$
The value of the definite integral is:
$\boxed{12}$

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QuestionFind $\int_{1}^{e^{2}} \frac{6}{x} d x$

Studdy Solution

STEP 1

1. We are asked to evaluate the definite integral $\int_{1}^{e^{2}} \frac{6}{x} \, dx$ .
2. The function $\frac{6}{x}$ is continuous and integrable over the interval $[1, e^2]$ .

STEP 2

1. Identify the antiderivative of the integrand $\frac{6}{x}$ .
2. Evaluate the definite integral using the Fundamental Theorem of Calculus.

STEP 3

Identify the antiderivative of $\frac{6}{x}$ :
The antiderivative of $\frac{1}{x}$ is $\ln |x|$ . Therefore, the antiderivative of $\frac{6}{x}$ is:
$\int \frac{6}{x} \, dx = 6 \ln |x| + C$
where $C$ is the constant of integration.

STEP 4

Apply the Fundamental Theorem of Calculus to evaluate the definite integral:
$\int_{1}^{e^{2}} \frac{6}{x} \, dx = \left[ 6 \ln |x| \right]_{1}^{e^{2}}$

STEP 5

Evaluate the expression at the upper limit $e^2$ :
$6 \ln |e^2| = 6 \ln (e^2) = 6 \cdot 2 \ln e = 12 \cdot 1 = 12$

STEP 6

Evaluate the expression at the lower limit $1$ :
$6 \ln |1| = 6 \ln (1) = 6 \cdot 0 = 0$

STEP 7

Subtract the evaluated lower limit from the evaluated upper limit:
$\int_{1}^{e^{2}} \frac{6}{x} \, dx = 12 - 0 = 12$
The value of the definite integral is:
$\boxed{12}$

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QuestionFind ∫1e26xdx\int_{1}^{e^{2}} \frac{6}{x} d x∫1e2​x6​dx

Studdy Solution

STEP 1

1. We are asked to evaluate the definite integral ∫1e26x dx\int_{1}^{e^{2}} \frac{6}{x} \, dx∫1e2​x6​dx.2. The function 6x\frac{6}{x}x6​ is continuous and integrable over the interval [1,e2][1, e^2][1,e2].

STEP 2

1. Identify the antiderivative of the integrand 6x\frac{6}{x}x6​.2. Evaluate the definite integral using the Fundamental Theorem of Calculus.

STEP 3

STEP 4

Apply the Fundamental Theorem of Calculus to evaluate the definite integral:∫1e26x dx=[6ln⁡∣x∣]1e2 \int_{1}^{e^{2}} \frac{6}{x} \, dx = \left[ 6 \ln |x| \right]_{1}^{e^{2}} ∫1e2​x6​dx=[6ln∣x∣]1e2​

STEP 5

Evaluate the expression at the upper limit e2e^2e2:6ln⁡∣e2∣=6ln⁡(e2)=6⋅2ln⁡e=12⋅1=12 6 \ln |e^2| = 6 \ln (e^2) = 6 \cdot 2 \ln e = 12 \cdot 1 = 12 6ln∣e2∣=6ln(e2)=6⋅2lne=12⋅1=12

STEP 6

Evaluate the expression at the lower limit 111:6ln⁡∣1∣=6ln⁡(1)=6⋅0=0 6 \ln |1| = 6 \ln (1) = 6 \cdot 0 = 0 6ln∣1∣=6ln(1)=6⋅0=0

STEP 7

Subtract the evaluated lower limit from the evaluated upper limit:∫1e26x dx=12−0=12 \int_{1}^{e^{2}} \frac{6}{x} \, dx = 12 - 0 = 12 ∫1e2​x6​dx=12−0=12 The value of the definite integral is:12 \boxed{12} 12​

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QuestionFind $\int_{1}^{e^{2}} \frac{6}{x} d x$

1. We are asked to evaluate the definite integral $\int_{1}^{e^{2}} \frac{6}{x} \, dx$ .
2. The function $\frac{6}{x}$ is continuous and integrable over the interval $[1, e^2]$ .

1. Identify the antiderivative of the integrand $\frac{6}{x}$ .
2. Evaluate the definite integral using the Fundamental Theorem of Calculus.

Apply the Fundamental Theorem of Calculus to evaluate the definite integral:
$\int_{1}^{e^{2}} \frac{6}{x} \, dx = \left[ 6 \ln |x| \right]_{1}^{e^{2}}$

Evaluate the expression at the upper limit $e^2$ :
$6 \ln |e^2| = 6 \ln (e^2) = 6 \cdot 2 \ln e = 12 \cdot 1 = 12$

Evaluate the expression at the lower limit $1$ :
$6 \ln |1| = 6 \ln (1) = 6 \cdot 0 = 0$

Subtract the evaluated lower limit from the evaluated upper limit:
$\int_{1}^{e^{2}} \frac{6}{x} \, dx = 12 - 0 = 12$
The value of the definite integral is:
$\boxed{12}$