The volume is #\frac{81\sqrt{3}}{40}\approx 3.5074#.

The curve #y=-x^2+5x-4=-(x-1)(x-4)# has #x#-intercepts at #x=1# and #x=4# (and is above the #x#-axis for #1<x<4#).

For #1< x< 4#, let #b(x)=-x^2+5x-4#. This will be the base of the equilateral triangle cross-section at #x#.

The solid itself looks something like this:

Now draw the cross-section equilateral triangle, label the sides #b(x)#, and draw a vertical line for the height #h(x)# from the top vertex down perpendicular to the base.

By the Pythagorean Theorem, #(b(x))^2=(\frac{b(x)}{2})^2+(h(x))^2#.

so that #h(x)=\sqrt{(b(x))^2-(\frac{b(x)}{2})^2}=\sqrt{\frac{3}{4}(b(x))^2}=\frac{\sqrt{3}}{2}b(x)#.

The cross-sectional area is #A(x)=\frac{1}{2}b(x)h(x)=\frac{\sqrt{3}}{4}(b(x))^2=\frac{\sqrt{3}}{4}(x^{4}-10x^{3}+33x^{2}-40x+16)#, which means the volume of the solid is #V=\int_{1}^{4}A(x)dx=\frac{\sqrt{3}}{4}\int_{1}^{4}(x^{4}-10x^{3}+33x^{2}-40x+16)dx=\frac{\sqrt{3}}{4}(\frac{x^{5}}{5}-\frac{5x^{4}}{2}+11x^{3}-20x^{2}+16x)|_{x=1}^{x=4}#.

Continuing to simplify gives #V=\frac{\sqrt{3}}{4}((\frac{1024}{5}-\frac{1280}{2}+704-320+64)-(\frac{1}{5}-\frac{5}{2}+11-20+16))=\frac{\sqrt{3}}{4}(\frac{64}{5}-\frac{47}{10})=\frac{81\sqrt{3}}{40}#.