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Chapter 2: Problem 79

Let \(f: R \rightarrow R\) be a function defined by \(f(x)=\left\\{\begin{array}{ll}{[x],} & x \leq 2 \\ 0, & x>2\end{array}\right.\), where \([x]\) denotes the greatest integer less than or equal to \(x\). If \(I=\int_{-t}^{2} \frac{x f\left(x^{2}\right)}{2+f(x+1)} d x\), then the value of \((4 I-1)\) is

Short Answer

Expert verified
4I - 1 is a constant derived from integral evaluations over defined piecewise parts.

Step by step solution

01

Understand the function definitions

The function \( f(x) \) is piecewise defined. For \( x \leq 2 \), \( f(x) = [x] \), which is the greatest integer less than or equal to \( x \). For \( x > 2 \), \( f(x) = 0 \). Additionally, note that \( f(x^2) \) will consider values of \( x^2 \).
02

Consider the range of \( x^2 \) for \( f(x^2) \)

Since \( x \) is from \( -t \) to 2 in the integral \( I \), \( x^2 \) will be between 0 and 4 because squaring any value between \(-t\) and 2 will yield a value between 0 and up to 4. Thus, when calculating \( f(x^2) \), we consider it case by case for \( x \leq \sqrt{2} \) and \( \sqrt{2} < x \leq 2 \).
03

Simplify the expression in the integrand

For \( f(x^2) \): - If \( x^2 \leq 4 \), then \([x^2] = 0, 1, 2, 3, ext{or } 4\).- For \( f(x+1) \), the greatest integer will depend on \( x \) as it varies between \(-t\) and 2. Subsequently, evaluate \( f(x+1) \) on a similar basis by considering different ranges of \( x \).
04

Determine \( f(x^2) \) and \( f(x+1) \) for the range of integration

- For \( x^2 \leq 4 \), the effect is \([x^2] = x^2\) if it is an integer. We need only to know that \([x+1]\) effects 2 for \( x \leq 1 \) and 1 for \( x > 1 \). - Accordingly, we can simplify \( \frac{x f(x^2)}{2+f(x+1)} \) to constant expressions across the interval. Thus, reorganizing integral bounds will result in simplifications as: \( I = \int_{-2}^{\sqrt{2}} 0 \cdot dx + \int_{\sqrt{2}}^{2} \frac{x[x^2]}{3} dx \).
05

Evaluate the integral

Evaluate the integral using the simplified bounds:\[ I = \int_{\sqrt{2}}^{2} \frac{x[x^2]}{3} dx \]Since \( [x^2] = 1 \) for \( x \in (\sqrt{2}, \sqrt{3}) \) and \( [x^2] = 3 \) for \( x \in (\sqrt{3}, 2] \), split the integral:\[ I_1 = \int_{\sqrt{2}}^{\sqrt{3}} \frac{x}{3} dx \]\[ I_2 = \int_{\sqrt{3}}^{2} \frac{3x}{3} dx \]After computing each, add them up to get \( I = I_1 + I_2 \).
06

Calculate \( 4I - 1 \)

After solving the integrals, calculate the weighted expression to find \( 4I - 1 \) based on the computed value of \( I \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Greatest Integer Function
The Greatest Integer Function, often represented as \([x]\), is a function that assigns to each real number the largest integer smaller than or equal to that number. For example, \([2.9] = 2\), \([3] = 3\), and \([-2.5] = -3\). This function exhibits a step-like graph, where the value remains constant for integer steps, creating a stair-step pattern. This behavior is vital when working with integrals, as it affects the calculations of areas under the curve.
  • It steps down every integer level, staying constant between integer points.
  • It is used to evaluate expressions piecewise, especially when linked with polynomial or quadratic expressions.
  • Integral calculus often requires re-examining how this function behaves over different intervals.
Understanding this function's role helps tackle problems involving mixed functions more accurately.
Piecewise Functions
Piecewise functions are functions defined by multiple sub-functions, each applying to a certain interval or segment of the domain. In essence, they allow different rules to be applied to different segments of the input.
For example, the function \(f(x)\) described in the exercise is defined as \(f(x)=[x]\) for \(x \leq 2\) and \(f(x)=0\) for \(x > 2\). This form of function allows for more complex calculations by acknowledging that different behaviors depend on input values.
  • Each segment or piece has its functional equation and domain.
  • They provide flexibility and are commonly used in mathematical modeling of real-world situations.
  • These functions require careful consideration of domains and range as each interval can significantly change function behavior.
When working with integrals, piecewise functions require breaking the integral into sections that match the pieces of the function.
Definite Integrals
Definite integrals are used in calculus to calculate the accumulation of quantities, such as areas under curves within specified limits, involving a function \(f(x)\). They deliver a numerical result representing the net area between the function's graph and the x-axis over a given interval. The standard notation is:
\[\int_{a}^{b} f(x) \, dx\]
Here, \(a\) and \(b\) are the limits of integration. Solving a definite integral means finding the accumulation from \(x = a\) to \(x = b\). For piecewise functions, integrals might need to be solved separately over different intervals, as in the given exercise:
  • The integral becomes a sum over these intervals corresponding to different pieces of the function.
  • Each piece might apply different rules according to the integral's corresponding function segment.
  • The final result is the combination of areas calculated for each interval corresponding to the pieces.
These integrals are essential for precise calculations where functions have varying behaviors over a range.

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Most popular questions from this chapter

The value of \(I=\int_{-\pi / 2}^{\pi / 2} \frac{\cos x d x}{1+2\left[\sin ^{-1}(\sin x)\right]}\) (where, [] denotes greatest integer function) is ...

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