Problem 3
Question
Let \(g(x)=\int_{0}^{x} f(t) d t\), where \(f\) is such that \(\frac{1}{2} \leq
f(t) \leq 1\) for \(t \in[0,1]\) and \(0 \leq f(t) \leq \frac{1}{2}\) for \(t
\in[1,2] .\) Then \(g(2)\) satisfies the inequality:
(A) \(-\frac{3}{2} \leq g(2)<\frac{1}{2}\)
(B) \(0
Step-by-Step Solution
Verified Answer
None of the provided answer choices are correct given the solution.
1Step 1: Express the Integral
The function \( g(x) \) is the integral of \( f(t) \) from 0 to \( x \). For \( g(2) \), we split this into two integrals: \( g(2) = \int_{0}^{1} f(t) dt + \int_{1}^{2} f(t) dt \).
2Step 2: Determine Bounds for \( \int_{0}^{1} f(t) dt \)
Given \( \frac{1}{2} \leq f(t) \leq 1 \) for \( t \in [0, 1] \), the integral \( \int_{0}^{1} f(t) dt \) can be bounded by \( \int_{0}^{1} \frac{1}{2} dt \) and \( \int_{0}^{1} 1 \ dt \), which evaluates to \( \frac{1}{2} \) and 1 respectively.
3Step 3: Determine Bounds for \( \int_{1}^{2} f(t) dt \)
Given \( 0 \leq f(t) \leq \frac{1}{2} \) for \( t \in [1, 2] \), the integral \( \int_{1}^{2} f(t) dt \) can be bounded by \( \int_{1}^{2} 0 \ dt \) and \( \int_{1}^{2} \frac{1}{2} dt \), which evaluates to 0 and \( \frac{1}{2} \) respectively.
4Step 4: Combine the Bounds
By adding the bounds from Steps 2 and 3, we obtain the bounds for \( g(2) \). The lower bound is \( \frac{1}{2} + 0 = \frac{1}{2} \) and the upper bound is \( 1 + \frac{1}{2} = \frac{3}{2} \).
5Step 5: Verify the Answer
The result \( \frac{1}{2} \leq g(2) \leq \frac{3}{2} \) corresponds to answer choice (C), where the correct inequality is \( \frac{3}{2}
Key Concepts
Integral BoundsFunction AnalysisInequalities in Integration
Integral Bounds
When we discuss integral bounds, we are essentially talking about the range of values that an integral can take based on certain conditions or constraints applied to the function being integrated. In this particular exercise, we examine the function \( g(x) = \int_{0}^{x} f(t) dt \). We split this integral into two parts to derive bounds for \( g(2) \): one from 0 to 1 and the other from 1 to 2. By analyzing each part separately, we can determine specific bounds for \( g(2) \).
Combining these calculations, the integral bounds for \( g(2) \) are derived by simply summing both segments: a lower bound of \( \frac{1}{2} + 0 = \frac{1}{2} \) and an upper bound of \( 1 + \frac{1}{2} = \frac{3}{2} \). Thus, \( \frac{1}{2} \leq g(2) \leq \frac{3}{2} \) becomes our integral bound conclusion.
- For the interval \([0, 1]\), we use the constraint \( \frac{1}{2} \leq f(t) \leq 1 \) to evaluate the integral. The bounds for this part of the integral are calculated as interacting over the constant bounds, giving us a range from \( \frac{1}{2} \) to \( 1 \).
- For the interval \([1, 2]\), the constraint \( 0 \leq f(t) \leq \frac{1}{2} \) is used, producing bounds for this integral part from 0 to \( \frac{1}{2} \).
Combining these calculations, the integral bounds for \( g(2) \) are derived by simply summing both segments: a lower bound of \( \frac{1}{2} + 0 = \frac{1}{2} \) and an upper bound of \( 1 + \frac{1}{2} = \frac{3}{2} \). Thus, \( \frac{1}{2} \leq g(2) \leq \frac{3}{2} \) becomes our integral bound conclusion.
Function Analysis
Function analysis is vital as it provides insights into the behavior of the function under study and lends itself to predicting how functions influence integrals. In this context, function analysis helps unravel the behavior of \( f(t) \) as \( t \) varies across the given intervals. Specifically, we analyze the effects of the given constraints for \( f(t) \) on each segment of its domain:
- **Interval [0, 1]:** Here, \( \frac{1}{2} \leq f(t) \leq 1 \), suggesting \( f(t) \) is never smaller than half and may grow up to 1. This interval reveals that in each point within this space, \( f(t) \) contributes consistently no less than \( \frac{1}{2} \) to the integral; hence, it stabilizes the integral's value from the lower bound.
- **Interval [1, 2]:** Now, the function is constrained as \( 0 \leq f(t) \leq \frac{1}{2} \). Here, \( f(t) \) could dip to zero, but not exceed half. The minimum value here thus could bring the integral's outcome down from its greater potential.
By performing this analysis, we've learned that the function \( f(t) \) behaves as a stabilizer in the first segment and varies greatly, potentially reducing values over the second segment. This contrasts help us understand integral bounds better.
- **Interval [0, 1]:** Here, \( \frac{1}{2} \leq f(t) \leq 1 \), suggesting \( f(t) \) is never smaller than half and may grow up to 1. This interval reveals that in each point within this space, \( f(t) \) contributes consistently no less than \( \frac{1}{2} \) to the integral; hence, it stabilizes the integral's value from the lower bound.
- **Interval [1, 2]:** Now, the function is constrained as \( 0 \leq f(t) \leq \frac{1}{2} \). Here, \( f(t) \) could dip to zero, but not exceed half. The minimum value here thus could bring the integral's outcome down from its greater potential.
By performing this analysis, we've learned that the function \( f(t) \) behaves as a stabilizer in the first segment and varies greatly, potentially reducing values over the second segment. This contrasts help us understand integral bounds better.
Inequalities in Integration
Analyzing inequalities in integration is crucial as it aids us in identifying how integrals develop over specified intervals and align with functions’ constraints. The inequalities provided in the problem—through their boundaries—determine the span of values calculated integrals can achieve, based upon behavior constraints of \( f(t) \).
When constructed right, these inequalities help us pick out broad but correct limits within which the integral resides:
- **Conceptualizing Inequalities:** When solving inequalities concerning an integral, it is crucial to analyze the function's input over established split intervals, which help define the range caused by such inequalities.
- **Applying Constraints:** Knowing that inequalities \( \frac{1}{2} \leq g(2) \leq \frac{3}{2} \) offer a narrower and exact look at how \( g(x) \) culminates when evaluated. Here, the calculated inequality highlights even tighter bounds obtained from sequential interval evaluations we've seen.
Understanding inequalities in this problem not only facilitates determining possible output of \( g(2) \) but also shows how such findings emerge step-by-step through comprehension of split inputs.
When constructed right, these inequalities help us pick out broad but correct limits within which the integral resides:
- **Conceptualizing Inequalities:** When solving inequalities concerning an integral, it is crucial to analyze the function's input over established split intervals, which help define the range caused by such inequalities.
- **Applying Constraints:** Knowing that inequalities \( \frac{1}{2} \leq g(2) \leq \frac{3}{2} \) offer a narrower and exact look at how \( g(x) \) culminates when evaluated. Here, the calculated inequality highlights even tighter bounds obtained from sequential interval evaluations we've seen.
Understanding inequalities in this problem not only facilitates determining possible output of \( g(2) \) but also shows how such findings emerge step-by-step through comprehension of split inputs.
Other exercises in this chapter
Problem 1
If \(\int_{0}^{1} e^{x^{3}}(x-\alpha) d x=0\), then (A) \(1
View solution Problem 4
\(\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{r}{n^{2}} \cdot \sec ^{2} \frac{r^{2}}{n^{2}}\) is equal to (A) \(\tan 1\) (B) \(\frac{1}{3} \tan 1\) (C) \(
View solution Problem 5
\(\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \sin ^{2 k} \frac{r \pi}{2 n}\) is equal to (A) \(\frac{2 k !}{2^{2 k}(k !)^{2}}\) (B) \(\frac{2 k !}{
View solution Problem 6
The value of \(\int_{0}^{\pi} \frac{\sin \left(n+\frac{1}{2}\right) x}{\sin \left(\frac{x}{2}\right)} d x\) is (A) \(\frac{\pi}{2}\) (B) 0 (C) \(\pi\) (D) \(2 \
View solution