Yes, as you guessed, a good way to do the integral is to
find the intersection of the lines $y=-x+10$ and $y=4x$
(this occurs at $x=2$),
and divide the area into the portion to the left of the line $x=2$
(bounded by $y=4x$ above and $y=2x$ below)
and the portion to the right of the line $x=2$
(bounded by $y=-x+10$ above and $y=2x$ below).
You can then solve this in two integrals.
Another answer shows how to find the area using three integrals,
again using the line $x=2$ to divide a single region into two.
The methods are pretty much the same if the bounds are curves rather than
straight lines, unless there are additional intersections of the bounds
to be concerned about. (Two distinct lines can intersect only once;
two distinct curves can intersect multiple times.)
But suppose (for example) you had the bound $y=\frac{16}{x}$
instead of $y=-x+10$; you'd still be able to use two integrals,
one to the left of $x=2$ and one to the right.
Just for fun, let's consider what happens if we apply the
transformation $(x,y) \mapsto (x,y-2x)$ to this figure.
This transformation happens to preserve area, so the resulting
triangle, bounded by the lines $y=-3x+10$, $y=2x$, and $y=0$,
has the same area as the original triangle.
The first two bounds intersect at $(2,4)$.
Let's put the equations of those two lines in terms of $y$ and
integrate over $y$ from $0$ to $4$:
$$y=-x+10 \implies x = -\frac13(y-10) = \frac{10}{3} - \frac13 y;$$
$$y=2x \implies x = \frac12 y;$$
$$\begin{eqnarray}
A &=&
\int_0^4 \left(\left(\frac{10}{3} - \frac13 y\right) - \frac12 y\right)\,dy \\
&=& \int_0^4 \left(\frac{10}{3} - \frac56 y\right)\,dy \\
&=& \left.\left(\frac{10}{3}y - \frac{5}{12} y^2\right) \right|_0^4 \\
&=& \frac{20}{3}.
\end{eqnarray}$$
Best Answer
Let's observe picture bellow.Total area $A$ is represented as sum of blue and yellow area:
$A=A_1+A_2$ , where $A_1=\int\limits_{-1}^{0} \sqrt{1+x}\, dx$ ,and $A_2=\int\limits_{0}^{1} (1-x) \, dx$
So we have that:
$A=\frac{2}{3}+\frac{1}{2}=\frac{7}{6}$