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Chapter 2: Problem 23

Find the greatest and the least value of the function \(F(x)=\int_{1}^{x}|t|\) dt on the interval \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)

Short Answer

Expert verified
The greatest value of the function is \(-\frac{1}{4}\), and the least value is \(-\frac{1}{2}\).

Step by step solution

01

Identify the region of integration

Given that we have to find the extreme values of the function \(F(x) = \int_{1}^{x}|t|\) dt, first, we should check the region of integration. On the interval \(\left[-\frac{1}{2}, \frac{1}{2}\right]\), we have that \(1>x\geq-\frac{1}{2}\), so the integral will be over a region where the function inside the integral is always positive or zero. Since \(|t|\geq0\), the integal will always be well-defined.
02

Express |t| in terms of t

The function inside the integral is given in terms of absolute value: \(|t|\). We need to rewrite it in terms of \(t\) for our given range. For the given interval \(\left[-\frac{1}{2}, \frac{1}{2}\right]\), we have: For \(t\in \left[-\frac{1}{2}, 0\right]\), \(|t|=-t\). For \(t\in(0,\frac{1}{2}]\), \(|t|=t\). We can now split the integral into two parts for these sections.
03

Split the integral and evaluate

Now, we split the integral according to the intervals we found in Step 2: For \(x\in \left[-\frac{1}{2}, 0\right]\), \(F(x) = \int_{1}^{x}{-t}\) dt, and For \(x\in(0,\frac{1}{2}]\), \(F(x) = \int_{1}^{x}{t}\) dt. By integrating, we get: For \(x\in \left[-\frac{1}{2}, 0\right]\), \(F(x) = \left[-\frac{t^2}{2}\right]_{1}^{x} = \frac{x^2}{2} -\frac{1}{2}\) For \(x\in(0,\frac{1}{2}]\), \(F(x) = \left[\frac{t^2}{2}\right]_{1}^{x} = \frac{x^2}{2} -\frac{1}{2}\).
04

Identify the extreme values

Now, we want to find the extreme values of the function \(F(x) = \frac{x^2}{2} -\frac{1}{2}\) when \(x\in \left[-\frac{1}{2}, \frac{1}{2}\right]\). The minimum value of \(F(x)\) occurs when \(x=0\): \(F(0) = \frac{0^2}{2}-\frac{1}{2} = -\frac{1}{2}\). The maximum value of \(F(x)\) occurs at the endpoints of the interval: \(F\left(-\frac{1}{2}\right) = F\left(\frac{1}{2}\right) = \frac{\left(\frac{1}{2}\right)^2}{2}-\frac{1}{2} = -\frac{1}{4}\). So, the greatest and least value of the function \(F(x)\) on the interval \(\left[-\frac{1}{2},\frac{1}{2}\right]\) are \(-\frac{1}{4}\) and \(-\frac{1}{2}\), respectively.

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