The line integral along the given positively oriented curve is mathematically given as
[tex]=3\left[e^{4}-1\right][/tex]
This is further explained below.
Generally,
[tex]\int M d x+N d y=\iint\left(\frac{\partial N}{\partial x}-\frac{\partial M}{2 y}\right) d y d x[/tex]
M=y e^{x}
Therefore
x -->0 to 4
y --> 0 to 3
[tex]\end{aligned}\\&=\int_{0}^{4} \int_{0}^{3}\left(2 e^{x}-e^{x}\right) d y d x\\&=\int_{0}^{4} \int_{0}^{3} e^{x} d y d x\\&=3 \int_{0}^{4} e^{x} d x\\&=3\left[e^{x}\right]_{0}^{4}\\&=3\left[e^{4}-e^{0}\right]\\&=3\left[e^{4}-1\right]\end{aligned}[/tex]
In conclusion, the line integral along the given positively oriented curve.
[tex]=3\left[e^{4}-1\right][/tex]
Read more about line integral
https://brainly.com/question/15177673
#SPJ4
