## Task Condition

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

$$
y=\sqrt{e^{x}-1}, y=0, x=\ln 2
$$
## Solution

Integral $14-8$

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-15.jpg?height=1411&width=1422&top_left_y=1042&top_left_x=451)

The desired area $S$ is:

$S=\int_{0}^{\ln x} \sqrt{e^{x}-1} d x$

We make a substitution of the variable:

$$
\sqrt{e^{x}-1}=t, \text { then } x=\ln \left(t^{2}+1\right), d x=\frac{2 t d t}{t^{2}+1}, 0 \leq t \leq 1
$$

## Then we get

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

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

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

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