We are given the integral \( \int_{-5}^{5} (4 - |x|) \, dx \) and are asked to find the value of the integral.
Since the absolute value function \( |x| \) is piecewise, we need to split the integral at \( x = 0 \), because the function behaves differently for \( x \geq 0 \) and \( x < 0 \).
The integral becomes:
\( \int_{-5}^{5} (4 - |x|) \, dx = \int_{-5}^{0} (4 - (-x)) \, dx + \int_{0}^{5} (4 - x) \, dx \).
Now simplify each integral:
\( \int_{-5}^{0} (4 + x) \, dx \) and \( \int_{0}^{5} (4 - x) \, dx \).
First, compute \( \int_{-5}^{0} (4 + x) \, dx \):
\( \int (4 + x) \, dx = 4x + \frac{x^2}{2} \), so:
\( \left[ 4x + \frac{x^2}{2} \right]_{-5}^{0} = (0 + 0) - (-20 + \frac{25}{2}) = 20 - \frac{25}{2} = \frac{40}{2} - \frac{25}{2} = \frac{15}{2}. \)
Next, compute \( \int_{0}^{5} (4 - x) \, dx \):
\( \int (4 - x) \, dx = 4x - \frac{x^2}{2} \), so:
\( \left[ 4x - \frac{x^2}{2} \right]_{0}^{5} = (20 - \frac{25}{2}) - (0) = 20 - \frac{25}{2} = \frac{40}{2} - \frac{25}{2} = \frac{15}{2}. \)
Adding the results of the two integrals, we get:
\( \frac{15}{2} + \frac{15}{2} = 15. \)
The correct answer is 15.
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
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: