Problem 34
Question
Find an approximate value of \(1^{1 / 3}+2^{1 / 3}+\cdots+1000^{1 / 3}\).
Step-by-Step Solution
Verified Answer
The approximate value of \(1^{1/3} + 2^{1/3} + \cdots + 1000^{1/3}\) is approximately \(74,999,999.25\).
1Step 1: Understand the summation we are given
Given the summation \(1^{1/3} + 2^{1/3} + \cdots + 1000^{1/3}\), we are asked to find the approximate value of this sum. We can express the summation as \(\sum_{n=1}^{1000} n^{1/3}\).
2Step 2: Represent the summation as an integral
Since the function \(f(x) = x^{1/3}\) is continuous and well-behaved, we can use the Riemann sum to approximate this sum using the integral:
\[
\sum_{n=1}^{1000} n^{1/3} \approx \int_{1}^{1000} x^{1/3} dx
\]
3Step 3: Evaluate the integral
Now we can evaluate the integral. To do this, we apply the power rule for integration. The general formula for integrating a function of the form \(x^n\) is:
\[
\int x^n dx = \frac{x^{n+1}}{n+1} + C
\]
Where C is the constant of integration which is not needed when we are calculating definite integrals. Applying this formula to our integral:
\[
\int_{1}^{1000} x^{1/3} dx = \left[\frac{x^{4/3}}{(4/3)}\right]_1^{1000}
\]
4Step 4: Calculate the values at the limits of integration
We can now find the values of the function at the limits of integration:
\[
\frac{1000^{4/3}}{(4/3)} - \frac{1^{4/3}}{(4/3)} = \frac{100^{4(4/3)}}{(4/3)} - \frac{1}{(4/3)}
\]
5Step 5: Calculate the approximate value of the sum
Simplify the expression:
\[
\frac{1000^{4/3}-1}{(4/3)} \approx \frac{100^4 - 1}{(4/3)} = \frac{100000000 - 1}{(4/3)} = 75000000 - \frac{3}{4}
\]
So, the approximate value of \(1^{1/3} + 2^{1/3} + \cdots + 1000^{1/3}\) is approximately \(75000000 - \frac{3}{4}= 75,000,000 - 0.75 = 74,999,999.25\).
Key Concepts
Definite IntegralPower Rule for IntegrationSummation Notation
Definite Integral
When we speak of a definite integral, we are referring to the calculation of the area under a curve bounded by certain limits, typically defined on the x-axis. This concept is foundational in calculus and serves numerous practical purposes such as physics for computing distances, areas, and volumes.
For instance, to find the area under a curve of the function from, say, x=1 to x=3, we would use a definite integral that would be denoted as \[ \int_{1}^{3} f(x) dx \]. The key idea here is that the definite integral gives us a precise answer, not an approximation, for the total accumulated quantity between those two points on the function. This concept was applied in the exercise when we translated the summation of cubes' roots into an integral form to efficiently calculate its approximate value.
For instance, to find the area under a curve of the function from, say, x=1 to x=3, we would use a definite integral that would be denoted as \[ \int_{1}^{3} f(x) dx \]. The key idea here is that the definite integral gives us a precise answer, not an approximation, for the total accumulated quantity between those two points on the function. This concept was applied in the exercise when we translated the summation of cubes' roots into an integral form to efficiently calculate its approximate value.
Power Rule for Integration
The power rule for integration is a shortcut that greatly simplifies the process of integrating polynomials. It states that for a function in the form of \(f(x) = x^n\), where n is a real number and not equal to -1, the integral is given by \[ \int x^n dx = \frac{x^{n+1}}{n+1} + C \] where C represents a constant of integration.
This rule was directly applied in the solution to our exercise, where the function \(f(x) = x^{1/3}\) was integrated over the given interval. Understanding this rule is a major step in solving many calculus problems, as it helps eliminate the often lengthy process of summing the infinitesimally small pieces used to represent the area under a curve.
This rule was directly applied in the solution to our exercise, where the function \(f(x) = x^{1/3}\) was integrated over the given interval. Understanding this rule is a major step in solving many calculus problems, as it helps eliminate the often lengthy process of summing the infinitesimally small pieces used to represent the area under a curve.
Summation Notation
When faced with the task of adding up a long series of numbers that follow a specific pattern, we use summation notation, also known as sigma notation, to write the series in a compact form. This mathematical shorthand greatly aids in simplifying the representation of sums and in setting up problems for further analysis or solving.
The summation notation is represented by the Greek letter sigma (\(\Sigma\)) and is structured as \[\sum_{n=a}^{b} f(n)\], where n is the index of summation, a is the lower bound, b is the upper bound, and f(n) is the function of n that we're summing. In the context of the exercise, the task was to sum the cubes' roots from 1 to 1000, which using summation notation became \[\sum_{n=1}^{1000} n^{1/3}\]. Summation notation provides a powerful tool for handling series and sequences, particularly when we integrate functions over discrete intervals, as seen with the Riemann sum approximation in the exercise.
The summation notation is represented by the Greek letter sigma (\(\Sigma\)) and is structured as \[\sum_{n=a}^{b} f(n)\], where n is the index of summation, a is the lower bound, b is the upper bound, and f(n) is the function of n that we're summing. In the context of the exercise, the task was to sum the cubes' roots from 1 to 1000, which using summation notation became \[\sum_{n=1}^{1000} n^{1/3}\]. Summation notation provides a powerful tool for handling series and sequences, particularly when we integrate functions over discrete intervals, as seen with the Riemann sum approximation in the exercise.
Other exercises in this chapter
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