Math / Calculus

Question(b) (5 points) $\int_{0}^{2} \int_{0}^{1} \int_{0}^{y} e^{x} d x d y d z$

Studdy Solution

STEP 1

1. We are dealing with a triple integral in the form $\int \int \int e^x \, dx \, dy \, dz$ .
2. The limits of integration are given for $x$ , $y$ , and $z$ .
3. The function $e^x$ is continuous and integrable over the given domain.

STEP 2

1. Integrate with respect to $x$ .
2. Integrate the resulting expression with respect to $y$ .
3. Integrate the resulting expression with respect to $z$ .

STEP 3

First, integrate the innermost integral with respect to $x$ . The limits for $x$ are from $0$ to $y$ :
$\int_{0}^{y} e^{x} \, dx$
The antiderivative of $e^x$ is $e^x$ , so we evaluate it from $0$ to $y$ :
$\left[ e^x \right]_{0}^{y} = e^y - e^0 = e^y - 1$

STEP 4

Next, integrate the result with respect to $y$ . The limits for $y$ are from $0$ to $1$ :
$\int_{0}^{1} (e^y - 1) \, dy$
Integrate each term separately:
$\int e^y \, dy = e^y \quad \text{and} \quad \int 1 \, dy = y$
Evaluate from $0$ to $1$ :
$\left[ e^y - y \right]_{0}^{1} = (e^1 - 1) - (e^0 - 0) = (e - 1) - (1 - 0) = e - 2$

STEP 5

Finally, integrate the result with respect to $z$ . The limits for $z$ are from $0$ to $2$ :
$\int_{0}^{2} (e - 2) \, dz$
Since $e - 2$ is a constant, the integral is straightforward:
$(e - 2) \int_{0}^{2} 1 \, dz = (e - 2) \cdot [z]_{0}^{2} = (e - 2) \cdot (2 - 0) = 2(e - 2)$
The value of the triple integral is:
$\boxed{2(e - 2)}$

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