Questions: ∫ from -2 to 5 ∫ from 0 to 3x ∫ from y to x+2 4 dz dy dx =

Transcript text: \(\int_{-2}^{5} \int_{0}^{3 x} \int_{y}^{x+2} 4 d z d y d x=\)

Solution

Step 1: Evaluate the Innermost Integral

We start with the innermost integral:

\[ \int_{y}^{x+2} 4 \, dz = 4z \bigg|_{y}^{x+2} = 4(x + 2) - 4y = 4x - 4y + 8 \]

Step 2: Evaluate the Middle Integral

Next, we integrate the result with respect to \( y \):

\[ \int_{0}^{3x} (4x - 4y + 8) \, dy = \left(4xy - 2y^2 + 8y\right) \bigg|_{0}^{3x} = 4x(3x) - 2(3x)^2 + 8(3x) = 12x^2 - 18x^2 + 24x = -6x^2 + 24x \]

Step 3: Evaluate the Outermost Integral

Finally, we integrate the result with respect to \( x \):

\[ \int_{-2}^{5} (-6x^2 + 24x) \, dx = \left(-2x^3 + 12x^2\right) \bigg|_{-2}^{5} \]

Calculating this gives:

\[ \left(-2(5)^3 + 12(5)^2\right) - \left(-2(-2)^3 + 12(-2)^2\right) = (-250 + 300) - (16 + 48) = 50 - 64 = -14 \]

The value of the triple integral is

\[ \boxed{-14} \]

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