## Task Condition

Calculate the area of the figure bounded by the graphs of the functions:

$$
y=\operatorname{arctg} x, y=0, x=\sqrt{3}
$$
## Solution

Find the abscissas of the points of intersection of the graphs of the functions

$y=\operatorname{arctg} x, y=0$ :

$$
\operatorname{arctg} x=0
$$

$$
x=0
$$

In the case of $y=\operatorname{arctg} x, x=\sqrt{3}$ it is obvious:

$$
x=\sqrt{3}
$$

Calculate the area:

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-27.jpg?height=1028&width=1034&top_left_y=934&top_left_x=908)

$$
S=\int_{0}^{\sqrt{3}}(\operatorname{arctg} x-0) d x=\int_{0}^{\sqrt{3}} \operatorname{arctg} x d x=
$$

Integrate by parts $\int u d v=u v-\int v d u$ :

$$
\begin{aligned}
& u=\operatorname{arctg} x \Rightarrow d u=\frac{d x}{x^{2}+1} \\
& d v=d x \Rightarrow v=x
\end{aligned}
$$

We get:

$$
=\left.\operatorname{arctg} x \cdot x\right|_{0} ^{\sqrt{3}}-\int_{0}^{\sqrt{3}} \frac{x}{x^{2}+1} d x=
$$

Substitution:

$$
\begin{aligned}
& t=x^{2}+1 \Rightarrow d t=2 x d x \\
& x=0 \Rightarrow t=0^{2}+1=1 \\
& x=\sqrt{3} \Rightarrow t=(\sqrt{3})^{2}+1=4
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =(\operatorname{arctg} \sqrt{3} \cdot \sqrt{3}-\operatorname{arctg} 0 \cdot 0)-\frac{1}{2} \int_{1}^{4} \frac{d t}{t}=\left(\frac{\pi}{3} \cdot \sqrt{3}-0\right)-\left.\frac{1}{2} \cdot \ln t\right|_{1} ^{4}= \\
& =\frac{\pi}{\sqrt{3}}-\frac{1}{2} \cdot(\ln 4-\ln 1)=\frac{\pi}{\sqrt{3}}-\frac{\ln 4}{2}=\frac{\pi}{\sqrt{3}}-\ln 2
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-15$ »

Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals

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