Answer
$$\int_{1}^{4} \int_{0}^{1} \int_{0}^{x} 2 z e^{-x^{2}} d y d x d z =\frac{15}{2}\left(1-\frac{1}{e}\right)$$
Work Step by Step
Given $$\int_{1}^{4} \int_{0}^{1} \int_{0}^{x} 2 z e^{-x^{2}} d y d x d z $$
So, we have
\begin{aligned}
I&=\int_{1}^{4} \int_{0}^{1} \int_{0}^{x} 2 z e^{-x^{2}} d y d x d z \\&=\int_{1}^{4} \int_{0}^{1}\left[\left(2 z e^{-x^{2}}\right) y\right]_{0}^{x} d x d z\\
&=\int_{1}^{4} \int_{0}^{1} \left(2 z e^{-x^{2}}\right) (x-0) d x d z\\
&=\int_{1}^{4} \int_{0}^{1} 2 z x e^{-x^{2}} d x d z \\ &=\int_{1}^{4}\left[-z e^{-x^{2}}\right]_{0}^{1} d z\\
&=-\int_{1}^{4} z\left(e^{-1}-1\right) d z\\
&=\int_{1}^{4} z\left(1-e^{-1}\right) d z\\
&=\left[\left(1-e^{-1}\right) \frac{z^{2}}{2}\right]_{1}^{4}\\
&= \left(1-e^{-1}\right) \frac{4^{2}-1^2}{2} \\
&=\frac{15}{2}\left(1-\frac{1}{e}\right) \end{aligned}
