Problem 12 - Calculus Early Transcendentals

Question

Evaluate the following integrals. A sketch of the region of integration may be useful. $$\int_{0}^{2} \int_{1}^{2} \int_{0}^{1} y z e^{x} d x d z d y$$

Step-by-Step Solution

Answer

In this problem, the student is asked to evaluate the given triple integral, which represents the volume of a region. After following the steps to evaluate each integral, the final value of the triple integral is 3(e-1).

Step 1: Evaluate the integral with respect to x

Integrate the expression yze^x with respect to x over the interval [0, 1]: $$\int_{0}^{1} y z e^{x} dx.$$ Using the integration rule for exponential functions, we get: $$[yze^x]_{0}^{1} = yze^1 - yze^0$$ The result is: $$yz(e-1)$$

Step 2: Evaluate the integral with respect to z

Now, integrate the expression yz(e-1) with respect to z over the interval [1, 2]: $$\int_{1}^{2} yz(e-1) dz$$ Using the power rule for integration, we get: $$\left[\frac{1}{2}yz^2(e-1)\right]_{1}^{2} = y\left(\frac{1}{2}(2^2 - 1^2)(e-1)\right) = \frac{3}{2}y(e-1)$$

Step 3: Evaluate the integral with respect to y

Finally, integrate the expression \(\frac{3}{2}y(e-1)\) with respect to y over the interval [0, 2]: $$\int_{0}^{2} \frac{3}{2}y(e-1) dy$$ Using the power rule again, we get: $$\left[\frac{1}{2}\cdot\frac{3}{2}y^2(e-1)\right]_{0}^{2} = \frac{3}{4}(4)(e-1) = 3(e-1)$$ The value of the triple integral is: $$\int_{0}^{2} \int_{1}^{2} \int_{0}^{1} y z e^{x} d x d z d y = 3(e-1)$$