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Magazine Chapter 6: Problem 8
Find the average value of the function on the given interval. $$ h(u)=(\ln u) / u, \quad[1,5] $$
Short Answer
Expert verified
The average value of the function is \( \frac{5 \ln 5 - 4}{4} \).
Step by step solution
01
Understand the formula for average value
The average value of a function \( f(x) \) over the interval \([a, b]\) is given by the formula \( \frac{1}{b-a} \int_a^b f(x) \, dx \). We will use this formula to find the average value of \( h(u) = \frac{\ln u}{u} \) over \([1,5]\).
02
Set up the integral
Substitute the function \( h(u) = \frac{\ln u}{u} \) and the interval \([1,5]\) into the average value formula: \[ \frac{1}{5-1} \int_1^5 \frac{\ln u}{u} \, du \]. The integral we need to solve is \( \int_1^5 \frac{\ln u}{u} \, du \).
03
Solve the integral
To solve \( \int \frac{\ln u}{u} \, du \), use integration by parts. Let \( v = \ln u \) and \( dw = \frac{1}{u} \, du \), so \( dv = \frac{1}{u} \, du \) and \( w = u \). The integration by parts formula gives \[ \int v \, dw = vw - \int w \, dv \]. Thus: \[ \int \frac{\ln u}{u} \, du = u \ln u - \int 1 \, du = u \ln u - u + C \].
04
Evaluate the definite integral
Now, evaluate \( u \ln u - u \) from 1 to 5: \[ \left[ u \ln u - u \right]_1^5 = (5 \ln 5 - 5) - (1 \ln 1 - 1) \]. Since \( \ln 1 = 0 \), this simplifies to \( 5 \ln 5 - 5 + 1 = 5 \ln 5 - 4 \).
05
Find the average value
Use the result from Step 4 in the average value formula: \[ \frac{1}{4} (5 \ln 5 - 4) = \frac{5 \ln 5 - 4}{4} \]. This terminates as the average value of the function over the interval.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Average Value of a Function
The average value of a function provides insight into the function's behavior over a specific interval. Consider it as the constant height a rectangle must have to produce the same area under a curve over a given range.
To find the average value of function \( f(x) \) over interval \([a, b]\), the formula is:
Let's apply this to \( h(u) = \frac{\ln u}{u} \) over the interval \([1, 5]\). Evaluating the definite integral \( \int_1^5 \frac{\ln u}{u} \, du \), and then dividing by \(4\), gives us the average value of the function across the interval.
To find the average value of function \( f(x) \) over interval \([a, b]\), the formula is:
- \( \text{Average value} = \frac{1}{b-a} \int_a^b f(x) \, dx \)
Let's apply this to \( h(u) = \frac{\ln u}{u} \) over the interval \([1, 5]\). Evaluating the definite integral \( \int_1^5 \frac{\ln u}{u} \, du \), and then dividing by \(4\), gives us the average value of the function across the interval.
Integration by Parts
Integration by parts is a technique used to integrate products of functions. It's essentially the reverse of the product rule used in differentiation.
The formula is:
Substituting these values into the integration by parts formula allows us to decompose \( \int \frac{\ln u}{u} \, du = u \ln u - \int 1 \, du \). This simplifies further to yield terms that are easier to integrate.
Integration by parts turns complex integration into a more manageable process, elegantly allowing us to handle logarithmic and exponential functions.
The formula is:
- \( \int u \, dv = uv - \int v \, du \)
Substituting these values into the integration by parts formula allows us to decompose \( \int \frac{\ln u}{u} \, du = u \ln u - \int 1 \, du \). This simplifies further to yield terms that are easier to integrate.
Integration by parts turns complex integration into a more manageable process, elegantly allowing us to handle logarithmic and exponential functions.
Definite Integral
The definite integral is used to calculate the total accumulation of quantities, often the area under a curve, between two bounds.
For a function \( f(x) \), the definite integral from \( a \) to \( b \) is expressed as \( \int_a^b f(x) \, dx \).
After applying the antiderivative, the boundary values are substituted, and their difference is calculated, giving the area under \( h(u) = \frac{\ln u}{u} \) from 1 to 5.
Definite integrals reduce expression complexity by providing a singular solution over an interval.
For a function \( f(x) \), the definite integral from \( a \) to \( b \) is expressed as \( \int_a^b f(x) \, dx \).
- It gives the precise difference between the values of the antiderivative at \( b \) and \( a \).
After applying the antiderivative, the boundary values are substituted, and their difference is calculated, giving the area under \( h(u) = \frac{\ln u}{u} \) from 1 to 5.
Definite integrals reduce expression complexity by providing a singular solution over an interval.
Natural Logarithm
The natural logarithm \( \ln x \) is the logarithm to the base \( e \) (Euler's number, approximately 2.718). It connects exponential and logarithmic functions in a straightforward manner.
Some important properties include:
For the function \( h(u) = \frac{\ln u}{u} \), understanding \( \ln u \) helps in setting up the integration by parts approach.
Its derivative \( \frac{1}{u} \) plays a crucial role in both differential and integral calculus, making it a constant presence in solving varied mathematical problems.
Some important properties include:
- \( \ln 1 = 0 \)
- \( \ln e = 1 \)
- \( \ln(ab) = \ln a + \ln b \)
For the function \( h(u) = \frac{\ln u}{u} \), understanding \( \ln u \) helps in setting up the integration by parts approach.
Its derivative \( \frac{1}{u} \) plays a crucial role in both differential and integral calculus, making it a constant presence in solving varied mathematical problems.
