Question

1. $int_{1}^{2}int_{0}^{2z}int_{0}^{ln x}xe^{-y}dydxdz$

Explanation:

Step1: Integrate with respect to $y$

We know that $\int xe^{-y}dy=-xe^{-y}+C$. Evaluating the definite - integral $\int_{0}^{\ln x}xe^{-y}dy=-xe^{-y}\big|_{0}^{\ln x}=-x\cdot x^{-1}+x\cdot1=- 1 + x$.

Step2: Integrate the result with respect to $x$

Now we integrate $\int_{0}^{2z}(-1 + x)dx=\int_{0}^{2z}-1dx+\int_{0}^{2z}xdx=-x\big|_{0}^{2z}+\frac{1}{2}x^{2}\big|_{0}^{2z}=-2z + \frac{1}{2}(2z)^{2}=-2z + 2z^{2}$.

Step3: Integrate the result with respect to $z$

Finally, we integrate $\int_{1}^{2}(-2z + 2z^{2})dz=\int_{1}^{2}-2zdz+\int_{1}^{2}2z^{2}dz=-z^{2}\big|_{1}^{2}+\frac{2}{3}z^{3}\big|_{1}^{2}=(-4 + 1)+\frac{2}{3}(8 - 1)=-3+\frac{14}{3}=\frac{-9 + 14}{3}=\frac{5}{3}$.

Answer:

$\frac{5}{3}$