Example. Find the area of the region bounded by the graphs of the functions

$$
y=x^{2}-4 x+3, \quad y=-x^{2}+2 x+3
$$
## Solution.

1. Find the abscissas $a$ and $b$ of the points of intersection of the graphs. For this, solve the equation

$$
x^{2}-4 x+3=-x^{2}+2 x+3
$$

We get $a=0, \quad b=3$.

2. Investigate the sign of the function $\varphi=x^{2}-4 x+3-\left(-x^{2}+2 x+3\right)$ on the interval $[a, b]=[0,3]$. For this, assign $x$ any value from $(0,3)$, for example $x=1$. We get that $\varphi(1)=-4$. Therefore, $\varphi<0$ when $x \in(0,3)$. Hence, $x^{2}-4 x+3 \leq-x^{2}+2 x+3$ when $x \in[0,3]$ and the region $D$ is defined by the system of inequalities

$$
\left\{\begin{aligned}
0 & \leq x \leq 3 \\
x^{2}-4 x+3 & \leq y \leq-x^{2}+2 x+3
\end{aligned}\right.
$$

3. Apply the formula (1) with $v(x)=-x^{2}+2 x+3, \quad u(x)=x^{2}-4 x+3, a=0$ and $b=3$:

$$
S_{D}=\int_{0}^{3}\left(-x^{2}+2 x+3-x^{2}+4 x-3\right) d x=\int_{0}^{3}\left(-2 x^{2}+6 x\right) d x=9
$$

Answer. $S=9(\text{units of length})^{2}$.

Conditions of the Problems. Calculate the areas of regions bounded by the graphs of the given functions.

1. $y=32-x^{2}, \quad y=-4 x$.
2. $y=3 \sqrt{x}, \quad y=3 / x, \quad x=4$.
3. $x=5-y^{2}, \quad x=-4 y$.
4. $y=\sqrt{e^{x}-1}, \quad y=0, \quad x=\ln 4$.
5. $y=\sin x, \quad y=\cos x, \quad x=0(x \geq 0)$.
6. $y=\sqrt{x}, \quad y=1 / x, \quad x=16$.
7. $x=27-y^{2}, \quad x=-6 y$.
8. $y=\sin x, \quad y=\cos x, \quad x=0(x \leq 0)$.
9. $y=\sqrt{9-x^{2}}, \quad y=0, \quad x=0, \quad x=3 / 2$.
10. $y=2 / x, \quad y=5 e^{x}, \quad y=2, \quad y=5$.

Answers. 1. 288. $2.14-3 \ln 4$. 3. 36. 4. $(6 \sqrt{3}-2 \pi) / 3$. 5. $\sqrt{2}-1$.
6. $42-\ln 16$.
7. 288.
8. $1+\sqrt{2}$.
9. $(3 \pi+\sqrt{3}) / 4$. 10. 3.

8.8. Calculation of arc lengths $y=f(x)$

PROBLEM STATEMENT. Calculate the length of the curve given by the equation

$$
y=f(x)
$$

and bounded by points with abscissas $x=a$ and $x=b$.

SOLUTION PLAN. The length $l$ of a piecewise smooth curve $y=f(x)$, bounded by points with abscissas $x=a$ and $x=b$, is equal to

$$
l=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x
$$

1. Find $y^{\prime}=f^{\prime}(x)$.
2. Compute the differential of the arc length

$$
d l=\sqrt{1+\left(y^{\prime}\right)^{2}} d x
$$

3. Find the arc length by evaluating the definite integral (1). Write the answer, not forgetting the dimension.