If you integrate (I'd rather say, antidifferentiate) $f(x)$, you don't get $g(x)$, you get $g(x)+C$ for some arbitrary constant $C$. So it's wrong to think of $g(b)$ as "the area up to $x=b$."
Let's look at an example. When you antidifferentiate $\sin x$, you could get $-\cos x$, or you could get $-\cos x+42$ (or any of a number of other correct answers, but these two will do for now). So, the area up to $\pi/2$: is it $-\cos(\pi/2)$, which is $0$? or is it $42$?
Well, neither of these is a particularly good answer, which illustrates the difficulties you get into if you think of $g(b)$ as the area up to $x=b$.
Rather, you should think of $\int_a^xf(t)\,dt=g(x)-g(a)$ as the area from $a$ to $x$; it's only definite integrals that have an unambiguous interpretation as areas.
Notice that if we use a new symbol, say, $h(x)=\int_0^xf(t)\,dt$, for the area from zero to $x$ then if $f$ is always positive and $a$ is negative we'll have $h(a)$ negative, so in that sense area to the left of the $y$-axis is negative. But what this really shows is that you can't always interpret $\int_a^bf(x)\,dx$ as an area; you can always calculate the definite integral, but the number that comes out will not be what we generally accept as the area unless $a\le b$ and $f(x)\ge0$ for $a\le x\le b$.
You should express $y$ first. $y=x^{3/2}$. Then
$$
y' = \frac32 x^{1/2}
$$
and
$$
\int_0^{5/9} \sqrt{1 + y'^2}\,dx = \int_0^{5/9} \sqrt{1 + \frac94 x}\,dx = \left.\frac{8}{27}\left(1 + \frac94x\right)^{\!3/2}\right|_{x=0}^{x=5/9} = \frac{19}{27}
$$
Best Answer
Equating$y=\frac{x^2}{4}$ and $y=4$ we get $x=4$ and $x=-4$ you have to integrate $y$ from $-4$ to $4$ $$\int_{-4}^{4} \frac{x^2}{4}dx=\frac{x^3}{12}\mid_{-4}^{4}$$ and you will get the answer.In integrating with respect to $y$ you only computed the half of the area.