## Task Condition

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

$$
y=\frac{x}{1+\sqrt{x}}, y=0, x=1
$$
## Solution

$$
\begin{aligned}
S= & \int_{0}^{1} \frac{x}{1+\sqrt{x}} d x=\left|\begin{array}{c}
x=t^{2} \\
d x=2 t d t \\
x=0 ; t=0 \\
x=1 ; t=1
\end{array}\right|= \\
& =\int_{0}^{1} \frac{t^{2} \cdot 2 t}{1+t} d t= \\
& =2 \int_{0}^{1} \frac{t^{3}}{1+t} d t= \\
& =2 \int_{0}^{1}\left(t^{2}-t+1-\frac{1}{1+t}\right) d t= \\
& =\left.2\left(\frac{t^{3}}{3}-\frac{t^{2}}{2}+t-\ln |1+t|\right)\right|_{0} ^{1}= \\
& =2\left(\frac{1}{3}-0\right)-2\left(\frac{1}{2}-0\right)+2(1-0)-2(\ln 2-\ln 1)= \\
& =\frac{2}{3}-\frac{2}{2}+2-2(\ln 2-0)=\frac{5}{3}-2 \ln 2
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-36.jpg?height=945&width=797&top_left_y=1078&top_left_x=1155)

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-19$ "

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

- Last edited on this page: 07:49, 11 June 2010.
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## Problem Kuznetsov Integrals 14-20

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