Problem 146
Question
If \(x>0\) and \(\int_{0}^{x}[x] d x=[x]\left(\frac{1}{2} A+B\right)\), where \([.]\) denotes the greatest integer function, then (A) \(A=[x]-1\) (B) \(B=x-[x]\) (C) \(A=[x]+1\) (D) \(B=x+[x]\)
Step-by-Step Solution
Verified Answer
A = [x] + 1 and B = x - [x].
1Step 1: Understand the Problem
The problem involves an integral expression and the greatest integer function, denoted by \([x]\), which represents the greatest integer less than or equal to \(x\). We are tasked with finding the values for constants \(A\) and \(B\) that satisfy the given condition.
2Step 2: Evaluate the Integral
The integral given is \(\int_{0}^{x}[x] \, dx\). Since \([x]\) is a constant with respect to \(x\) on the interval where \([x]\) is constant, we can evaluate this integral as \([x](x - 0) = [x]x\).
3Step 3: Set Up the Equation
The equation given in the problem is \([x]x = [x]\left(\frac{1}{2} A + B\right)\). This indicates a relationship involving \([x]\) as a common factor on both sides.
4Step 4: Simplify and Compare Coefficients
Divide both sides of the equation \([x]x = [x]\left(\frac{1}{2} A + B\right)\) by \([x]\) assuming \([x]eq 0\), resulting in \(x = \frac{1}{2}A + B\). Since both sides should match, this equation will help us determine \(A\) and \(B\).
5Step 5: Determine A and B
From \(x = \frac{1}{2}A + B\), and by comparing options for \(A\) and \(B\), note the choices: (A) and (C) suggest \(A\) as \([x] \pm 1\), and (B) and (D) suggest \(B\) as combinations of \(x\) and \([x]\). The expression \( \frac{1}{2}([x] + 1) + (x - [x]) = \frac{1}{2}[x] + \frac{1}{2} + x - [x]\) simplifies to \(x\), indicating that \(A = [x] + 1\) and \(B = x - [x]\).
6Step 6: Verify the Solution
Substitute \(A = [x] + 1\) and \(B = x - [x]\) back into \([x](\frac{1}{2}A + B) = [x]x\):\([x](\frac{1}{2}([x]+1) + (x-[x])) = [x](\frac{1}{2}[x] + \frac{1}{2} + x - [x]) = [x]x\).The equation holds, confirming the values of \(A\) and \(B\) are correct.
Key Concepts
Greatest Integer FunctionIntegral CalculusDefinite Integrals
Greatest Integer Function
The Greatest Integer Function, often denoted by \([x]\), gives the largest integer less than or equal to a given number \x\. This is also known as the floor function. It plays a unique role in various mathematical applications by simplifying expressions, especially when dealing with integrals or functions where piecewise definitions are involved.
- For example, if \x = 2.7\, then \([x] = 2\), because 2 is the largest integer less than 2.7.
- Similarly, if \x = -1.4\, then \([x] = -2\), because -2 is the largest integer less than -1.4.
Integral Calculus
Integral Calculus is one of the two main branches of calculus. It focuses on finding the total size or value, such as areas under curves or the accumulation of quantities. Integrals can be thought of as the "sum over a continuum," providing a way to calculate quantities that accumulate continuously rather than discretely.
- The symbol for integration is \(\int\), and it requires an integrand or function, along with the variable of integration.
- Definite integrals have limits of integration, which define the interval over which the integration is performed. They yield a numerical value representing the area under a curve between those limits.
- Indefinite integrals, on the other hand, do not have limits and usually require an additional constant of integration.
Definite Integrals
Definite Integrals are a type of integral calculus where the limits of integration are specified. They compute the accumulation of quantities that can be represented geometrically as the area under a curve within certain bounds. For instance, if we're looking at an integral \(\int_{a}^{b} f(x) \, dx\), \a\ and \b\ are the limits of integration.
- The significance of definite integrals in mathematical analysis is crucial as they provide exact values rather than expressions with constants of integration.
- By evaluating the area under the curve \f(x)\ between \a\ and \b\, definite integrals can solve numerous practical problems ranging from physics to statistics.
- They are often solved using the properties of antiderivatives and applying the Fundamental Theorem of Calculus, which relates differentiation and integration.
Other exercises in this chapter
Problem 144
If \(A_{n}=\int_{0}^{\pi / 2} \frac{\sin (2 n-1) x}{\sin x} d x ;\) \(B_{n}=\int_{0}^{\pi / 2}\left(\frac{\sin n x}{\sin x}\right)^{2} d x ;\) for \(n \in N\),
View solution Problem 145
If \(g(x)=\int_{0}^{x} \cos 4 t d t\), then \(g(x+\pi)\) equals (A) \(\frac{g(x)}{g(\pi)}\) (B) \(g(x)+g(\pi)\) (C) \(g(x)-g(\pi)\) (D) \(g(x) \cdot g(\pi)\)
View solution Problem 147
If \(\int_{1}^{4}|x-3| d x=2 A+B\), Then (A) \(A=3 / 2, B=4\) (B) \(A=1, B=1 / 2\) (C) \(A=2, B=-3 / 2\) (D) \(A=1 / 2, B=3 / 2\)
View solution Problem 148
If \(\int_{a}^{b}|\sin x| d x=8\) and \(\int_{a}^{a+b}|\cos x| d x=\frac{9}{2}\), then (A) \(a=\frac{\pi}{2}\) (B) \(b=\frac{11 \pi}{4}\) (C) \(a=\frac{\pi}{4}\
View solution