1284. Calculate the area of the figure bounded by the following lines:
1) the coordinate axes, the line $x=3$ and the parabola $y=x^{2}+1$;
2) the ordinate axis, the lines $y=-2, y=3$ and the parabola $x=$ $=\frac{1}{2} y^{2}$
3) the parabolas $y=x^{2}$ and $x=y^{2}$;
4) the parabolas $y=x^{2}+1, y=\frac{1}{2} x^{2}$ and the line $y=5$.
Solution. 1) The area of the curvilinear trapezoid $O C B A$ (Fig. $79, a$) will be found using formula (1):

$$
S=\int_{0}^{3}\left(x^{2}+1\right) d x=\left.\left(\frac{x^{3}}{3}+x\right)\right|_{0} ^{3}=9+3=12
$$

2) The area of the curvilinear trapezoid $A B C D$ (Fig. 79, b) will be found using formula (2):

$$
S=\frac{1}{2} \int_{-2}^{3} y^{2} d y=\left.\frac{y^{3}}{6}\right|_{-2} ^{3}=\frac{27}{6}-\left(-\frac{8}{6}\right)=\frac{35}{6}
$$

3) Let's find the points of intersection of the given curves (Fig. 79, c). For this, we solve the system of equations:

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-246.jpg?height=120&width=688&top_left_y=408&top_left_x=243)

Thus, the parabolas intersect at points $O(0 ; 0)$ and $B(1 ; 1)$. Now it is easy to determine the area of the figure $O A B C O$ by considering it as the difference in the areas of two curvilinear trapezoids $O D B C O$ and $O D B A O$. Using formula (3), we get

$$
S=\int_{0}^{1}\left(\sqrt{x}-x^{2}\right) d x=\left.\left(\frac{2 x^{\frac{3}{2}}}{3}-\frac{x^{3}}{3}\right)\right|_{0} ^{1}=\frac{1}{3}
$$

4) The figure is symmetric with respect to the $O y$ axis (Fig. 79, d), so we can calculate half the area and then double the result. The area of the figure $O A B C D$ is the difference in the areas of the curvilinear trapezoids $O A B E$ and $D C E$. We will use formula (2). From the equations of the curves, we find $x=\sqrt{2 y}$ and $x=\sqrt{y-1}$. The intervals of integration are easily determined (see Fig. 79,2 ).

Thus, half of the desired area is

$$
\begin{aligned}
\frac{1}{2} S= & \int_{0}^{5} \sqrt{2 y} d y-\int_{1}^{5} \sqrt{y-1} d y=\left.\frac{2 \sqrt{2} y^{\frac{3}{2}}}{3}\right|_{0} ^{5}- \\
& -\left.\frac{2(y-1)^{\frac{3}{2}}}{3}\right|_{1} ^{5}=\frac{2}{3}(5 \sqrt{10}-8),
\end{aligned}
$$

from which

$$
S=\frac{4}{3}(5 \sqrt{10}-8) \approx 10.41
$$

Let's calculate the area in another way. We will integrate with respect to the variable $x$. First, we find half the area, for which we calculate the areas of $O A C D$ and $A B C$. The sum of these areas will constitute half of the desired area. Solving the systems of equations

$$
\left\{\begin{array} { l } 
{ y = 5 } \\
{ y = x ^ { 2 } + 1 }
\end{array} \quad \text { and } \left\{\begin{array}{l}
y=5 \\
y=\frac{1}{2} x^{2}
\end{array}\right.\right.
$$

we find the coordinates of points $C(2 ; 5)$ and $B(\sqrt{10} ; 5)$. Thus, the intervals of integration are now known.

Applying formula (3), we get

$$
\begin{gathered}
\frac{1}{2} S=\int_{0}^{2}\left(x^{2}+1-\frac{1}{2} x^{2}\right) d x+\int_{2}^{\sqrt{10}}\left(5-\frac{1}{2} x^{2}\right) d x= \\
=\left.\left(\frac{x^{3}}{6}+x\right)\right|_{0} ^{2}+\left.\left(5 x-\frac{x^{3}}{6}\right)\right|_{2} ^{\sqrt{10}}= \\
=\frac{4}{3}+2+5 \sqrt{10}-\frac{5 \sqrt{10}}{3}-\left(10-\frac{4}{3}\right)=\frac{2}{3}(5 \sqrt{10}-8)
\end{gathered}
$$

and

$$
S=\frac{4}{3}(5 \sqrt{10}-8) \approx 10.41
$$

It is not difficult to see that integration with respect to the variable $y$ leads to the desired result more quickly.