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Chapter 13: 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 \(\int_{0}^{\ln2}\int_{0}^{\ln3}\int_{0}^{\ln4} e^{-x+y+z} dz\,dy\,dx\). Answer: -8.

Step by step solution

01

Integrate with respect to x

To begin, we will integrate the function \(e^{-x+y+z}\) with respect to x over the interval [0, ln(2)]. The integral with respect to x is: $$\int_{0}^{\ln 2} e^{-x+y+z} dx$$ To evaluate this integral, we use the antiderivative of \(e^{-x+y+z}\) with respect to x, which is \(-e^{-x+y+z}\): $$-e^{-x+y+z} \biggr|_{0}^{\ln 2} = -(e^{\ln 0.5+y+z} - e^{y+z}) = e^{y+z} - e^{\ln 0.5+y+z}$$
02

Integrate with respect to y

Next, we will integrate the result from Step 1 with respect to y over the interval [0, ln(3)]: $$\int_{0}^{\ln 3} (e^{y+z} - e^{\ln 0.5+y+z}) dy$$ To evaluate this integral, we use the antiderivative of \(e^{y+z} - e^{\ln 0.5+y+z}\) with respect to y, which is \((e^{y+z} - e^{\ln 0.5+y+z})\): $$\left(e^{y+z} - e^{\ln 0.5+y+z}\right) \biggr|_{0}^{\ln 3} = ((e^{z+\ln 3} - e^{\ln 0.5+z+\ln 3}) - (e^{z} - e^{\ln 0.5+z}))$$ Simplifying this expression, we get: $$= 2e^{z} - 0.5e^{z+\ln 6} - e^{z} + 0.5e^{\ln 0.5+z} = e^{z} - 0.5(e^{z+\ln 6} + e^{\ln 0.5+z})$$
03

Integrate with respect to z

Finally, we will integrate the result from Step 2 with respect to z over the interval [0, ln(4)]: $$\int_{0}^{\ln 4} (e^{z} - 0.5(e^{z+\ln 6} + e^{\ln 0.5+z})) dz$$ To evaluate this integral, we use the antiderivative of \(e^{z} - 0.5(e^{z+\ln 6} + e^{\ln 0.5+z})\) with respect to z, which is \((e^{z} - 0.5(e^{z+\ln 6} + e^{\ln 0.5+z}))\): $$\left(e^{z} - 0.5(e^{z+\ln 6} + e^{\ln 0.5+z})\right) \biggr|_{0}^{\ln 4} = (e^{\ln 4} - 0.5(e^{\ln 4+\ln 6} + e^{\ln 0.5+\ln 4}) - (1 - 0.5(6 + 0.5)))$$ Simplifying this expression, we get: $$ = 3 - 0.5(e^{\ln 24} + e^{\ln 2}) - 10/4 = 3 - 0.5(24 + 2) + 10/4 = 3 - 0.5(26) + 2.5 = 5 - 13 = -8$$ So, the value of the given triple integral is -8.

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