Step 1: Splitting the integral. The absolute value function \( |x| \) is defined as Splitting the integral at \( x = 0 \): \[ \int_{-1}^1 |x| \, dx = \int_{-1}^0 -x \, dx + \int_0^1 x \, dx \] Step 2: Evaluating the integrals. \[ \int_{-1}^0 -x \, dx = \left[ -\frac{x^2}{2} \right]_{-1}^0 = 0 - \left(-\frac{(-1)^2}{2}\right) = \frac{1}{2} \] \[ \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - 0 = \frac{1}{2} \] Step 3: Summing the results. \[ \int_{-1}^1 |x| \, dx = \frac{1}{2} + \frac{1}{2} = 1 \] Conclusion: Thus, the value of the integral is \( 1 \), which corresponds to option \( \mathbf{(C)} \).
If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to:
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: