Example 6. Given points $A(4,5), B(4,0)$ and $C(0,5)$. Calculate the second type integral $\int_{L}(4 x+8 y+5) d x+(9 x+8) d y$, where $L:$ a) segment $O A$; b) broken line $O C A$; c) parabola $y=k x^{2}$, passing through points $O$ and $A$.
Solution. The calculation boils down to formulating equations for $L$ and reducing the line integral to a definite integral. For clarity, we will use a diagram (Fig. 3.21).
a) $O A: y=\frac{5}{4} x, x \in[0 ; 4], d y=\frac{5}{4} d x$.

## $I(O A)=$

$=\int_{0}^{4}(4 x+10 x+5) d x+\left(9 x \cdot \frac{5}{4}+10\right) d x=$

$$
\begin{aligned}
& =\int_{0}^{4}\left(\frac{101}{4} x+15\right) d x= \\
& \quad=\left.\left(\frac{101 \cdot x^{2}}{8}+15 x\right)\right|_{0} ^{4}=262
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-068.jpg?height=373&width=375&top_left_y=225&top_left_x=753)

Fig. 3.21 b) $O C A=O C+O A$. We will first compute the integral along the segment $O C$, then along the segment $C A$, and sum the results.

$$
\begin{gathered}
O C: x=0, d x=0, y \in[0 ; 5] \\
I(O C)=\int_{0}^{5} 8 d y=\left.8 y\right|_{0} ^{5}=40 \\
C A: y=5, d y=0, x \in[0,4] \\
I(C A)=\int_{0}^{4}(4 x+45) d x=\left.\left(2 x^{2}+45 x\right)\right|_{0} ^{4}=212
\end{gathered}
$$

Thus, $I(O C A)=40+212=252$.

c) $O A: y=k x^{2}$. Since the parabola passes through the point $A(4,5)$, we have $5=k \cdot 16$, hence $k=\frac{5}{16}$. Therefore, the equation of the parabola is $y=\frac{5}{16} x^{2}$. In this case, $d y=\frac{5 x}{8} d x, x \in[0 ; 4]$. Therefore,

$$
\begin{gathered}
I(O A)=\int_{0}^{4}\left(4 x+\frac{5}{2} x^{2}+5\right) d x+(9 x+8) \cdot \frac{5 x}{8} d x= \\
=\int_{0}^{4}\left(\frac{65}{8} x^{2}+9 x+5\right) d x=\left.\left(\frac{65 x^{3}}{24}+\frac{9 x^{2}}{2}+5 x\right)\right|_{0} ^{4}= \\
=\frac{520}{3}+72+20=\frac{796}{3}
\end{gathered}
$$

Answer
a) 262 ; b) 252 ; c) $\frac{796}{3}$