The value of $\int_{-5}^{5} |x + 3| dx$ is

The absolute value $|x + 3|$ changes definition at $x = -3$:

When $x < -3$: $|x + 3| = -x - 3$

When $x \geq -3$: $|x + 3| = x + 3$

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Since the function changes at $x = -3$, the integral must be split:

$\int_{-5}^{5} |x + 3| dx = \int_{-5}^{-3} |x + 3| dx + \int_{-3}^{5} |x + 3| dx$

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For $-5 \leq x < -3$, we have $|x + 3| = -x - 3$:

$\int_{-5}^{-3} (-x - 3) dx = \left[-\frac{x^2}{2} - 3x\right]_{-5}^{-3}$

At $x = -3$:

$-\frac{(-3)^2}{2} - 3(-3) = -\frac{9}{2} + 9 = \frac{9}{2}$

At $x = -5$:

$-\frac{(-5)^2}{2} - 3(-5) = -\frac{25}{2} + 15 = \frac{5}{2}$

Result: $\frac{9}{2} - \frac{5}{2} = 2$

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For $-3 \leq x \leq 5$, we have $|x + 3| = x + 3$:

$\int_{-3}^{5} (x + 3) dx = \left[\frac{x^2}{2} + 3x\right]_{-3}^{5}$

At $x = 5$:

$\frac{(5)^2}{2} + 3(5) = \frac{25}{2} + 15 = \frac{55}{2}$

At $x = -3$:

$\frac{(-3)^2}{2} + 3(-3) = \frac{9}{2} - 9 = -\frac{9}{2}$

Result: $\frac{55}{2} - \left(-\frac{9}{2}\right) = \frac{55}{2} + \frac{9}{2} = 32$

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Total: $2 + 32 = 34$

Therefore, $\int_{-5}^{5} |x + 3| dx = 34$