Math / Calculus

QuestionFind the area under the curve $y=\frac{1}{x}$ over the interval $[1,6]$ . $\square$ Find the average value of $y=\frac{1}{x}$ over the interval $[1,6]$ . $\square$

Studdy Solution

STEP 1

1. We are given the function $y = \frac{1}{x}$ .
2. We need to find the area under the curve over the interval $[1, 6]$ .
3. We also need to find the average value of the function over the same interval.

STEP 2

1. Set up the definite integral to find the area under the curve.
2. Evaluate the definite integral.
3. Use the result to find the average value of the function over the interval.

STEP 3

Set up the definite integral for the area under the curve:
$\int_{1}^{6} \frac{1}{x} \, dx$

STEP 4

Evaluate the definite integral:
The antiderivative of $\frac{1}{x}$ is $\ln|x|$ .
$\int_{1}^{6} \frac{1}{x} \, dx = \left[ \ln|x| \right]_{1}^{6}$

STEP 5

Apply the limits of integration:
$\left[ \ln|x| \right]_{1}^{6} = \ln|6| - \ln|1|$
Since $\ln|1| = 0$ , the expression simplifies to:
$\ln|6| - 0 = \ln 6$

STEP 6

To find the average value of the function over the interval $[1, 6]$ , use the formula for the average value of a function:
$\text{Average value} = \frac{1}{b-a} \int_{a}^{b} f(x) \, dx$
For $f(x) = \frac{1}{x}$ , $a = 1$ , and $b = 6$ :
$\text{Average value} = \frac{1}{6-1} \int_{1}^{6} \frac{1}{x} \, dx$

STEP 7

Substitute the result from the definite integral:
$\text{Average value} = \frac{1}{5} \ln 6$
The area under the curve is:
$\boxed{\ln 6}$
The average value of the function over the interval is:
$\boxed{\frac{1}{5} \ln 6}$

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QuestionFind the area under the curve $y=\frac{1}{x}$ over the interval $[1,6]$ . $\square$ Find the average value of $y=\frac{1}{x}$ over the interval $[1,6]$ . $\square$

Studdy Solution

STEP 1

1. We are given the function $y = \frac{1}{x}$ .
2. We need to find the area under the curve over the interval $[1, 6]$ .
3. We also need to find the average value of the function over the same interval.

STEP 2

1. Set up the definite integral to find the area under the curve.
2. Evaluate the definite integral.
3. Use the result to find the average value of the function over the interval.

STEP 3

Set up the definite integral for the area under the curve:
$\int_{1}^{6} \frac{1}{x} \, dx$

STEP 4

Evaluate the definite integral:
The antiderivative of $\frac{1}{x}$ is $\ln|x|$ .
$\int_{1}^{6} \frac{1}{x} \, dx = \left[ \ln|x| \right]_{1}^{6}$

STEP 5

Apply the limits of integration:
$\left[ \ln|x| \right]_{1}^{6} = \ln|6| - \ln|1|$
Since $\ln|1| = 0$ , the expression simplifies to:
$\ln|6| - 0 = \ln 6$

STEP 6

STEP 7

Substitute the result from the definite integral:
$\text{Average value} = \frac{1}{5} \ln 6$
The area under the curve is:
$\boxed{\ln 6}$
The average value of the function over the interval is:
$\boxed{\frac{1}{5} \ln 6}$

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QuestionFind the area under the curve y=1xy=\frac{1}{x}y=x1​ over the interval [1,6][1,6][1,6]. □\square□ Find the average value of y=1xy=\frac{1}{x}y=x1​ over the interval [1,6][1,6][1,6]. □\square□

Studdy Solution

STEP 1

1. We are given the function y=1x y = \frac{1}{x} y=x1​.2. We need to find the area under the curve over the interval [1,6][1, 6][1,6].3. We also need to find the average value of the function over the same interval.

STEP 2

1. Set up the definite integral to find the area under the curve.2. Evaluate the definite integral.3. Use the result to find the average value of the function over the interval.

STEP 3

Set up the definite integral for the area under the curve:∫161x dx \int_{1}^{6} \frac{1}{x} \, dx ∫16​x1​dx

STEP 4

Evaluate the definite integral:The antiderivative of 1x\frac{1}{x}x1​ is ln⁡∣x∣\ln|x|ln∣x∣.∫161x dx=[ln⁡∣x∣]16 \int_{1}^{6} \frac{1}{x} \, dx = \left[ \ln|x| \right]_{1}^{6} ∫16​x1​dx=[ln∣x∣]16​

STEP 5

Apply the limits of integration:[ln⁡∣x∣]16=ln⁡∣6∣−ln⁡∣1∣ \left[ \ln|x| \right]_{1}^{6} = \ln|6| - \ln|1| [ln∣x∣]16​=ln∣6∣−ln∣1∣Since ln⁡∣1∣=0\ln|1| = 0ln∣1∣=0, the expression simplifies to:ln⁡∣6∣−0=ln⁡6 \ln|6| - 0 = \ln 6 ln∣6∣−0=ln6

STEP 6

STEP 7

Substitute the result from the definite integral:Average value=15ln⁡6 \text{Average value} = \frac{1}{5} \ln 6 Average value=51​ln6 The area under the curve is:ln⁡6 \boxed{\ln 6} ln6​The average value of the function over the interval is:15ln⁡6 \boxed{\frac{1}{5} \ln 6} 51​ln6​

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QuestionFind the area under the curve $y=\frac{1}{x}$ over the interval $[1,6]$ . $\square$ Find the average value of $y=\frac{1}{x}$ over the interval $[1,6]$ . $\square$

1. We are given the function $y = \frac{1}{x}$ .
2. We need to find the area under the curve over the interval $[1, 6]$ .
3. We also need to find the average value of the function over the same interval.

1. Set up the definite integral to find the area under the curve.
2. Evaluate the definite integral.
3. Use the result to find the average value of the function over the interval.

Set up the definite integral for the area under the curve:
$\int_{1}^{6} \frac{1}{x} \, dx$

Evaluate the definite integral:
The antiderivative of $\frac{1}{x}$ is $\ln|x|$ .
$\int_{1}^{6} \frac{1}{x} \, dx = \left[ \ln|x| \right]_{1}^{6}$

Apply the limits of integration:
$\left[ \ln|x| \right]_{1}^{6} = \ln|6| - \ln|1|$
Since $\ln|1| = 0$ , the expression simplifies to:
$\ln|6| - 0 = \ln 6$

Substitute the result from the definite integral:
$\text{Average value} = \frac{1}{5} \ln 6$
The area under the curve is:
$\boxed{\ln 6}$
The average value of the function over the interval is:
$\boxed{\frac{1}{5} \ln 6}$