Multivariable Calculus Integral volume question 
Volume is:
$$
\int_{x=0}^{1}\int_{z=4x^2}^{z=5-x^2}(1-x)\;dz\;dx
$$
The picture above is the solution for the question that I need to find the volume of the region bounded by $z=5-x^2, z=4x^2,$ and the planes $y=0, x+y=1$. Is the integral set up correctly?
I'm not sure what it means by the shaded area is part of the vertical plane $x+y=1$. Also how come x cannot be less than 0? ie what is it bounded by?

Those limits don't work unless your teacher meant something else or I have misunderstood the question. 
$z=4x^2$ and $z=5-x^2$ define a pair of parabolic cylinders with the y-axis as the axis. These intersect at$ x=\pm 1, z=4$. 
The lower face of this volume is capped by the xz plane ($y=0$). 
If the upper face had been another constant $y=c$, notice that you would use $x=-1$ in your limit of integration - this is the minimum x-value always. However, since the plane $x+y=1$ bounds the volume, the upper x-limit is always $1-y$. The lower x-limit is always -1. When $y=2$, the upper and lower limits coincide (no more of the volume to integrate) so your volume is
$$
\int_{y=0}^{y=2} \int_{x=-1}^{x=1-y} \int_{z=4x^2}^{5-x^2} dz dx dy
$$
Edited to add: The limits you posted would work if you took $x=0$ as another bounding plane.
Then you would have
$$
\int_{x=0}^{x=1} \int_{y=0}^{y=1-x} \int_{z=4x^2}^{5-x^2} dz dy dx
$$
