Step 1: Set up the area.
The area under the curve from $x=1$ to $x=2$ is
\[ \int_1^2 x^5 e^{x^3}\,dx \]
Step 2: Pick a smart substitution.
Let $u=x^3$, so $du=3x^2\,dx$. Notice $x^5\,dx=x^3\cdot x^2\,dx=u\cdot\dfrac{du}{3}$.
Step 3: Shift the limits.
When $x=1$, $u=1$. When $x=2$, $u=8$. The integral turns into
\[ \frac{1}{3}\int_1^8 u\,e^{u}\,du \]
Step 4: Integrate by parts.
A handy fact is $\int u e^{u}\,du=e^{u}(u-1)$. So we get
\[ \frac{1}{3}\Big[e^{u}(u-1)\Big]_1^8 \]
Step 5: Put in the limits.
At $u=8$ the bracket is $7e^8$, and at $u=1$ it is $0$. So the value is
\[ \frac{1}{3}(7e^8)=\frac{7}{3}e^8 \]
\[ \boxed{\dfrac{7}{3}e^8} \]