Each cross section has side length equal to [tex]x[/tex] satisfying [tex]y=e^x\implies x=\ln y[/tex], where [tex]0\le x\le\ln4[/tex] so that [tex]1\le y\le4[/tex].
The exact volume is given by the definite integral,
[tex]\displaystyle\int_1^4(\ln y)^2\,\mathrm dy[/tex]
Take a slice at any value of [tex]y[/tex] with thickness [tex]\Delta y[/tex]. Then the slice has volume [tex](\ln y)^2\Delta y[/tex].
The approximate total volume of these slices is then given by the Riemann sum,
[tex]\displaystyle\sum_{i=1}^n(\ln y_i)^2\Delta y_i[/tex]
where [tex]y_i[/tex] are chosen however you like from the range above.
Compute the definite integral above for the exact volume: you can do this by parts, taking
[tex]u=(\ln y)^2\implies\mathrm du=\dfrac{2\ln y}y\,\mathrm dy[/tex]
[tex]\mathrm dv=\mathrm dy\implies v=y[/tex]
[tex]\implies\displaystyle\int_1^4(\ln y)^2\,\mathrm dy=y(\ln y)^2\bigg|_1^4-2\int_1^4\ln y\,\mathrm dy[/tex]
The remaining integral can be done by parts again, this time with
[tex]u=\ln y\implies\mathrm du=\dfrac{\mathrm dy}y[/tex]
[tex]\mathrm dv=\mathrm dy\implies v=y[/tex]
[tex]\implies\displaystyle\int_1^4\ln y\,\mathrm dy=y\ln y\bigg|_1^4-\int_1^4\mathrm dy[/tex]
and of course
[tex]\displaystyle\int_1^4\mathrm dy=y\bigg|_1^4[/tex]
So we have
[tex]\displaystyle\int_1^4(\ln y)^2\,\mathrm dy=4(\ln 4)^2-2(4\ln 4-(4-1))[/tex]
[tex]\displaystyle\int_1^4(\ln y)^2\,\mathrm dy=4(\ln 4)^2-8\ln 4+6[/tex]
