## Task Condition

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

$$
y=\arccos x, y=0, x=0
$$
## Solution

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-19.jpg?height=1422&width=1100&top_left_y=1028&top_left_x=615)

$S=\int_{0}^{1} \arccos x d x$

We will use the integration by parts formula:

$$
\begin{aligned}
& \int_{a}^{b} u d v=\left.u \cdot v\right|_{a} ^{b}-\int_{a}^{b} v d u \\
& \text{where} \\
& u=\arccos x ; v=x ; d u=-\frac{d x}{\sqrt{1-x^{2}}} ; d v=d x
\end{aligned}
$$

Then we get

$$
\begin{aligned}
& S=\int_{0}^{1} \arccos x d x= \\
& =\left.x \cdot \arccos x\right|_{0} ^{1}-\left(\int_{0}^{1} x \frac{-d x}{\sqrt{1-x^{2}}}\right)= \\
& =\left.x \cdot \arccos x\right|_{0} ^{1}+\int_{0}^{1} \frac{x d x}{\sqrt{1-x^{2}}}= \\
& =\left.x \cdot \arccos x\right|_{0} ^{1}-\frac{1}{2} \cdot \int_{0}^{1} \frac{d\left(1-x^{2}\right)}{\sqrt{1-x^{2}}}= \\
& =\left.x \cdot \arccos x\right|_{0} ^{1}-\left.\sqrt{1-x^{2}}\right|_{0} ^{1}= \\
& =\left(1 \cdot 0-0 \cdot \frac{\pi}{2}\right)-(\sqrt{0}-\sqrt{1})=1
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-10$ » Categories: Kuznetsov's Problem Book Integrals Problem 14 | Integrals

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## Problem Kuznetsov Integrals 14-11

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