## Task Condition

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

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

## Integrals 14-17

![](https://cdn.mathpix.com/cropped/2024_05_22_aa6951b1e92815ea34b6g-31.jpg?height=1153&width=1105&top_left_y=1031&top_left_x=790)

$$
x=\sqrt{e^{y}-1} \Rightarrow y=\ln \left(x^{2}+1\right)
$$

Let's find the limits of integration:

$$
\begin{aligned}
& x_{1}=0 \text { - given in the problem statement } \\
& x_{2}=\sqrt{e^{y}-1} ; y=\ln 2 ; \Rightarrow x_{2}=\sqrt{e^{\ln 2}-1}=\sqrt{2-1}=1
\end{aligned}
$$

We get:

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

To compute the second integral, we use integration by parts:

$\int_{a}^{b} u d v=\left.u \cdot v\right|_{a} ^{b}-\int_{a}^{b} v d u$

Where

$$
u=\ln \left(x^{2}+1\right) ; d u=\frac{2 x}{x^{2}+1} ; v=x ; d v=d x
$$

Then we get

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

In the end:

$$
S=\ln 2-\left(\ln 2-2+\frac{\pi}{2}\right)=2-\frac{\pi}{2} \approx 0.429
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{\%} 0 \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+14-17$ » Categories: Kuznetsov Problem Book Integrals Problem 14 | Integrals

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## Problem Kuznetsov Integrals 14-18

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