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Chapter 16: Problem 10

Evaluate the following integrals. A sketch of the region of integration may be useful. $$\int_{0}^{\ln 4} \int_{0}^{\ln 3} \int_{0}^{\ln 2} e^{-x+y+z} d x d y d z$$

Short Answer

Expert verified
Question: Evaluate the triple integral of the function \(e^{-x+y+z}\) over the rectangular box with coordinate intervals \(0 \le x \le \ln 4\), \(0 \le y \le \ln 3\), and \(0 \le z \le \ln 2\). Answer: The value of the triple integral is \(\frac{3}{2}\).

Step by step solution

01

Understand the Region of Integration

The region of integration is a rectangular box with the following coordinate intervals: - \(0 \le x \le \ln 4\) - \(0 \le y \le \ln 3\) - \(0 \le z \le \ln 2\)
02

Integrate Over x

We will first integrate the given function with respect to x: $$\int_{0}^{\ln 4} e^{-x+y+z} dx$$ Integrating with respect to x, we get: $$-\int_{0}^{\ln 4} de^{-x} = -[e^{-x}]_{0}^{\ln 4} = -(e^{-\ln4} - e^{0}) = -(e^{\ln(1/4)} - 1) = 1 - \frac{1}{4}$$
03

Integrate Over y

Now we need to integrate the result obtained in step 2 with respect to y: $$\int_{0}^{\ln 3} \left(\frac{3}{4}\right) e^{y+z} dy$$ Integrating with respect to y, we get: $$\left(\frac{3}{4}\right)\int_{0}^{\ln 3} e^{y} dy = \left(\frac{3}{4}\right)[e^{y}]_{0}^{\ln 3} = \left(\frac{3}{4}\right)(e^{\ln3} - e^{0}) = \left(\frac{3}{4}\right)(3 - 1)$$
04

Integrate Over z

Lastly, we need to integrate the result obtained in step 3 with respect to z: $$\int_{0}^{\ln 2} \left(\frac{3}{2}\right) e^{z} dz$$ Integrating with respect to z, we get: $$\left(\frac{3}{2}\right)\int_{0}^{\ln 2} e^{z} dz = \left(\frac{3}{2}\right)[e^{z}]_{0}^{\ln 2} = \left(\frac{3}{2}\right)(e^{\ln2} - e^{0}) = \left(\frac{3}{2}\right)(2 - 1)$$
05

Find the Value of the Integral

After integrating with respect to all three variables, we get the final value of the triple integral: $$\int_{0}^{\ln 4} \int_{0}^{\ln 3} \int_{0}^{\ln 2} e^{-x+y+z} d x d y d z = \left(\frac{3}{2}\right) = \frac{3}{2}$$ Thus, the value of the integral is \(\frac{3}{2}\).

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