Question

Show that \(\int_{0}^{x}|t| d t=\frac{1}{2} x|x|\) for all real \(x\) and express \(\mathrm{F}(\mathrm{x})=\int_{-1}^{\mathrm{x}}|\mathrm{t}| \mathrm{dt}\) in a piecewise form that does not involve an integral.

Short answer

Using the step-by-step solution above, we were able to prove that \(\int_{0}^{x}|t| d t=\frac{1}{2} x|x|\) for all real \(x\). Furthermore, we expressed \(F(x)=\int_{-1}^{x}|t| dt\) in a piecewise form without any integral as: $$F(x) = \begin{cases} \frac{1}{2}(1-x^2) & \text{if } -1 \le x \le 0 \\ \frac{1}{2}(1+x^2) & \text{if } 0 < x \le 1 \end{cases}$$

Step-by-step solution

Rewrite the integral using the definition of absolute value

We can rewrite \(|t| dt\) as: - \(t dt\) if \(t \ge 0\) - \(-t dt\) if \(t < 0\) Now, we will consider two cases: \(x \ge 0\) and \(x < 0\).

Evaluate the integral for \(x \ge 0\)

When \(x \ge 0\), we have: $$\int_{0}^{x}|t| dt = \int_{0}^{x}t dt$$ Now evaluate: $$\int_{0}^{x}t dt = \frac{1}{2}t^2 \Big|_0^x = \frac{1}{2}x^2$$

Evaluate the integral for \(x < 0\)

When \(x < 0\), we have: $$\int_{0}^{x}|t| dt = \int_{x}^{0}(-t) dt$$ Now evaluate: $$\int_{x}^{0}(-t) dt = \frac{1}{2}(-t^2) \Big|_x^0 = \frac{1}{2}x^2$$

Combine the results for both cases

Combine the results from steps 2 and 3, we have: $$\int_{0}^{x}|t| dt = \frac{1}{2} x|x|$$ for all real \(x\) as required. Now, let's find a piecewise form for \(F(x)=\int_{-1}^{x}|t| dt\).

Rewrite the integral \(F(x)\) using the definition of absolute value

We can rewrite \(|t| dt\) as: - \(t dt\) if \(t \ge 0\) - \(-t dt\) if \(t < 0\) Now, we have to evaluate the integral \(F(x)\) in the intervals \((-1,0]\) and \((0,1]\).

Evaluate the integral for \(x \in (-1,0]\)

In this case, we have: $$F(x) = \int_{-1}^{x}(-t) dt$$ Now evaluate: $$F(x) = \frac{1}{2}(-t^2) \Big|_{-1}^x = \frac{1}{2}(1-x^2)$$

Evaluate the integral for \(x \in (0,1]\)

In this case, we have: $$F(x) = \int_{-1}^{0}(-t) dt + \int_{0}^{x}(t) dt$$ Now evaluate: $$F(x) = \frac{1}{2}(-t^2) \Big|_{-1}^0 + \frac{1}{2}(t^2) \Big|_{0}^x = \frac{1}{2}(1-x^2) + \frac{1}{2}x^2 = \frac{1}{2}(1+x^2)$$

Combine the results in piecewise form

Combine the results from steps 6 and 7, we have the following piecewise form for \(F(x)\): $$F(x) = \begin{cases} \frac{1}{2}(1-x^2) & \text{if } -1 \le x \le 0 \\ \frac{1}{2}(1+x^2) & \text{if } 0 < x \le 1 \end{cases}$$ This is the required piecewise form for \(F(x)=\int_{-1}^{x}|t| dt\) without any integral.