Find the area bounded by the curve y = ln (sqrt(x²+1)
Find the area bounded by the curve $y = \ln\frac{\sqrt{x^2+1} - x}{\sqrt{x^2+1} + x}$ and the lines $x = 1$, $y = 0$. Based on this, prove the inequality: $\ln(\sqrt{2} - 1) < 1 - \sqrt{2}$.
Explanation of Method
Simplify the logarithmic expression using the identity $(\sqrt{x^2+1}-x)(\sqrt{x^2+1}+x)=1$, which lets the curve be written as $y=-2\ln(\sqrt{x^2+1}+x)$. Locate where it crosses the axis, integrate by parts to find the enclosed area, and then use the fact that a geometric area must be positive to derive the stated inequality.
We will simplify the curve, integrate by parts, and turn the resulting positivity of the area into an algebraic inequality.
Step 1: Simplify the curve equation
\[y = \ln\frac{\sqrt{x^2+1} - x}{\sqrt{x^2+1} + x}\]
Multiply numerator and denominator by $\sqrt{x^2+1} - x$:
\[\frac{\sqrt{x^2+1} - x}{\sqrt{x^2+1} + x} \cdot \frac{\sqrt{x^2+1} - x}{\sqrt{x^2+1} - x} = \frac{(\sqrt{x^2+1} - x)^2}{(x^2+1) - x^2} = (\sqrt{x^2+1} - x)^2\]
So:
\[y = \ln(\sqrt{x^2+1} - x)^2 = 2\ln(\sqrt{x^2+1} - x)\]
Equivalently, using $\ln\frac{a}{b} = \ln a - \ln b$:
\[y = \ln(\sqrt{x^2+1} - x) - \ln(\sqrt{x^2+1} + x)\]
Step 2: Find where the curve intersects $y = 0$
When $y = 0$:
\[\frac{\sqrt{x^2+1} - x}{\sqrt{x^2+1} + x} = 1\]
\[\sqrt{x^2+1} - x = \sqrt{x^2+1} + x \implies -x = x \implies x = 0\]
So the curve intersects $y = 0$ at $x = 0$.
Step 3: Determine the sign of $y$
For $x > 0$: $\sqrt{x^2+1} > x$, so $\sqrt{x^2+1} - x > 0$ and $\sqrt{x^2+1} + x > \sqrt{x^2+1} - x$. Therefore:
\[\frac{\sqrt{x^2+1} - x}{\sqrt{x^2+1} + x} < 1\]
which means $y = \ln(\text{something} < 1) < 0$ for $x > 0$. So the curve is below the $x$-axis for $x \in (0, 1]$.
Step 4: Set up the integral
The area is:
\[A = \int_0^1 |y| \, dx = -\int_0^1 y \, dx = -2\int_0^1 \ln(\sqrt{x^2+1} - x) \, dx\]
Step 5: Evaluate the integral
Let $I = \int_0^1 \ln(\sqrt{x^2+1} - x) \, dx$. Since $(\sqrt{x^2+1} - x)(\sqrt{x^2+1} + x) = x^2 + 1 - x^2 = 1$, we have $\sqrt{x^2+1} - x = \frac{1}{\sqrt{x^2+1} + x}$, so:
\[\ln(\sqrt{x^2+1} - x) = -\ln(\sqrt{x^2+1} + x)\]
Therefore:
\[I = -\int_0^1 \ln(\sqrt{x^2+1} + x) \, dx\]
By parts, with $u = \ln(\sqrt{x^2+1}+x)$ and $dv = dx$:
\[du = \frac{x/\sqrt{x^2+1} + 1}{\sqrt{x^2+1}+x}\,dx = \frac{1}{\sqrt{x^2+1}}\,dx, \qquad v = x\]
\[\int \ln(\sqrt{x^2+1} + x) \, dx = x\ln(\sqrt{x^2+1} + x) - \int \frac{x}{\sqrt{x^2+1}} \, dx = x\ln(\sqrt{x^2+1} + x) - \sqrt{x^2+1} + C\]
Evaluating from $0$ to $1$:
\[\int_0^1 \ln(\sqrt{x^2+1} + x) \, dx = \left[x\ln(\sqrt{x^2+1} + x) - \sqrt{x^2+1}\right]_0^1\]
At $x=1$: $\ln(\sqrt{2}+1) - \sqrt{2}$. At $x=0$: $0 - 1 = -1$.
\[\int_0^1 \ln(\sqrt{x^2+1} + x) \, dx = \ln(\sqrt{2}+1) - \sqrt{2} + 1\]
Therefore:
\[I = -[\ln(\sqrt{2}+1) - \sqrt{2} + 1] = \sqrt{2} - 1 - \ln(\sqrt{2}+1)\]
And the area is:
\[A = -2I = 2\ln(\sqrt{2}+1) + 2 - 2\sqrt{2}\]
\[\boxed{A = 2\ln(\sqrt{2} + 1) + 2 - 2\sqrt{2} \text{ square units}}\]
Step 6: Prove the inequality
Since the area $A>0$:
\[2\ln(\sqrt{2} + 1) + 2 - 2\sqrt{2} > 0 \implies \ln(\sqrt{2} + 1) > \sqrt{2} - 1\]
Since $(\sqrt2-1)(\sqrt2+1)=1$, we have $\sqrt2+1=\frac{1}{\sqrt2-1}$, so:
\[\ln(\sqrt{2} + 1) = \ln\frac{1}{\sqrt{2} - 1} = -\ln(\sqrt{2} - 1)\]
Substituting into $\ln(\sqrt2+1) > \sqrt2-1$:
\[-\ln(\sqrt{2} - 1) > \sqrt{2} - 1 \implies \ln(\sqrt{2} - 1) < 1 - \sqrt{2}\]
\[\boxed{\ln(\sqrt{2} - 1) < 1 - \sqrt{2}}\]
Summary
\[\boxed{A = 2\ln(\sqrt{2} + 1) + 2 - 2\sqrt{2} \approx 0.329 \text{ square units}}\]
\[\boxed{\ln(\sqrt{2} - 1) < 1 - \sqrt{2}}\]
Key insight: A purely geometric fact — that an area must be positive — becomes an algebraic inequality once the area is expressed in closed form.
Result
Final answer:
\[\boxed{A = 2\ln(\sqrt{2} + 1) + 2 - 2\sqrt{2} \approx 0.329 \text{ square units}, \qquad \ln(\sqrt{2}-1) < 1-\sqrt2}\]