Problem 135 - Calculus Volume 1

Question

Suppose that for each \(i\) such that \(1 \leq i \leq N\) one has \(\int_{i-1}^{i} f(t) d t=i^{2}\) show that \(\int_{0}^{N} f(t) d t=\frac{N(N+1)(2 N+1)}{6}\).

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{N(N+1)(2N+1)}{6} \).

Step 1: Understanding the Problem

We have been given integrals of a function over individual intervals from \(i-1\) to \(i\), and we need to find the integral over the entire interval from 0 to \(N\). Specifically, the integral for each interval \( [i-1, i] \) is equal to \(i^2\).

Step 2: Sum the Individual Integrals

The integral over the entire interval from 0 to \(N\) can be found by summing the integrals over each subinterval [\(i-1\), \(i\)], for \(i\) ranging from 1 to \(N\). This results in the sum: \[\int_{0}^{N} f(t) \, dt = \sum_{i=1}^{N} \int_{i-1}^{i} f(t) \, dt = \sum_{i=1}^{N} i^2.\]

Step 3: Use the Formula for the Sum of Squares

The sum \(\sum_{i=1}^{N} i^2\) can be evaluated using the formula for the sum of the squares of the first \(N\) natural numbers, given by:\[\sum_{i=1}^{N} i^2 = \frac{N(N+1)(2N+1)}{6}.\]

Step 4: Combine and Conclude

Substituting the sum formula into our earlier expression for the integral gives:\[\int_{0}^{N} f(t) \, dt = \frac{N(N+1)(2N+1)}{6}.\]Thus, this is the value of the integral over the entire interval from 0 to \(N\).

Key Concepts

Definite Integrals

A definite integral allows us to calculate the area under a curve for a specific interval. It is an essential tool for measuring the total accumulation of a quantity. In the problem, the integral of the function \( f(t) \) is calculated over each interval \([i-1, i]\), from 1 to \( N \). The notation \( \int_{a}^{b} f(t) \, dt \) signifies this calculation from \( a \) to \( b \), where \( f(t) \) is a continuous function on that interval. Think of it as slicing the area under the curve into smaller sections, and then summing these sections for the total area from \( t=0 \) to \( t=N \). This total integral, \( \int_{0}^{N} f(t) \, dt \), reflects all individual areas accumulated together. Each piece represents the integral of \( f(t) \) over a small segment, \([i-1, i]\), resulting in \( i^2 \) as given.

Summation of Series

When you deal with integrals over intervals, they often involve a series of numbers. In this problem, we look at the sum of several integrals, each equal to a square of a number, \( i^2 \). To find the total integral \( \int_{0}^{N} f(t) \, dt \), you sum the results of each smaller integral:

• Sum all \( i^2 \) from 1 to \( N \),
• Represented by \( \sum_{i=1}^{N} i^2 \).

This process of summation helps consolidate smaller sections into a single answer. Applying this idea echoes what happens in definite integrals, which summarize accumulations over specific intervals. In mathematics, summing series is a powerful concept that is used for simplifying long strings of computations.

Sum of Squares

The sum of squares is a standard mathematical calculation used in various fields, particularly in calculus and statistics. The sum of the squares of the first \( N \) natural numbers is given by the formula: \[\sum_{i=1}^{N} i^2 = \frac{N(N+1)(2N+1)}{6}. \]This formula allows you to swiftly calculate the sum without manually adding each squared term. It is highly useful when dealing with large sequences, as in this problem, where the sum must be turned into a single number representing a definite integral. In our exercise, substituting \( \sum_{i=1}^{N} i^2 \) using this formula is crucial to arriving at the result for the entire interval integral \( \int_{0}^{N} f(t) \, dt \).

Integral Properties

Integrals possess several useful properties which can simplify the computation process. These properties make integrals versatile tools in calculus. One important property utilized in this exercise is the additivity of integrals. This property states that the integral over a large interval can be expressed as the sum of integrals over smaller, contiguous subintervals.
Another useful property is linearity, which allows for the separation and individual integration of terms. However, the main property in our context is using the known formula for the sum of squares to express a streamlined result.
Recognizing these properties enables efficient problem-solving, especially when dealing with complex or lengthy expressions. Integrals often leverage such properties to simplify and calculate areas, accumulations, or averages efficiently.