Question

Let $I = \int_{2}^{2.5} (\frac{e^x - e^{-x}}{e^x + e^{-x}} + 2) \, dx$ and $J = \int_{2 + \frac{e^{4}-1}{e^{4}+1}}^{2 + \frac{e^{5}-1}{e^{5}+1}} \ln \sqrt{\frac{x-1}{3-x}} \, dx$.

If $I + J = a - 5(\frac{1}{e^5 + 1}) + 4(\frac{1}{e^4 + 1})$, then $a$ is (where $a$ is a rational number)

Answer 1

3/2

Answer 2

To evaluate $I+J$, we will evaluate $I$ and $J$ separately.

1. Evaluate Integral $I$: $I = \int_{2}^{2.5} \left(\frac{e^x - e^{-x}}{e^x + e^{-x}} + 2\right) \, dx$ The term $\frac{e^x - e^{-x}}{e^x + e^{-x}}$ is the definition of $\tanh x$. So, $I = \int_{2}^{2.5} (\tanh x + 2) \, dx$ We know that $\int \tanh x \, dx = \ln|\cosh x| + C$. Since $\cosh x = \frac{e^x + e^{-x}}{2}$ is always positive, we can write $\ln(\cosh x)$. $I = [\ln(\cosh x) + 2x]_{2}^{2.5}$ $I = (\ln(\cosh 2.5) + 2(2.5)) - (\ln(\cosh 2) + 2(2))$ $I = \ln(\cosh 2.5) + 5 - \ln(\cosh 2) - 4$ $I = \ln(\cosh 2.5) - \ln(\cosh 2) + 1$

2. Evaluate Integral $J$: $J = \int_{2 + \frac{e^{4}-1}{e^{4}+1}}^{2 + \frac{e^{5}-1}{e^{5}+1}} \ln \sqrt{\frac{x-1}{3-x}} \, dx$ Let's use the substitution $x = 2 + \tanh y$. Then $dx = \text{sech}^2 y \, dy$.

Now, let's change the limits of integration: For the lower limit, $x = 2 + \frac{e^{4}-1}{e^{4}+1}$. So, $2 + \tanh y = 2 + \frac{e^{4}-1}{e^{4}+1} \Rightarrow \tanh y = \frac{e^{4}-1}{e^{4}+1}$. We know that $\tanh y = \frac{e^y - e^{-y}}{e^y + e^{-y}} = \frac{e^{2y} - 1}{e^{2y} + 1}$. Comparing $\frac{e^{2y} - 1}{e^{2y} + 1}$ with $\frac{e^{4}-1}{e^{4}+1}$, we get $e^{2y} = e^4$, which implies $2y = 4$, so $y = 2$.

For the upper limit, $x = 2 + \frac{e^{5}-1}{e^{5}+1}$. Similarly, $\tanh y = \frac{e^{5}-1}{e^{5}+1}$, which implies $e^{2y} = e^5$, so $2y = 5$, and $y = 2.5$.

Now, let's transform the integrand: $\ln \sqrt{\frac{x-1}{3-x}} = \ln \sqrt{\frac{(2 + \tanh y) - 1}{3 - (2 + \tanh y)}} = \ln \sqrt{\frac{1 + \tanh y}{1 - \tanh y}}$ We know that $\frac{1 + \tanh y}{1 - \tanh y} = \frac{1 + \frac{\sinh y}{\cosh y}}{1 - \frac{\sinh y}{\cosh y}} = \frac{\cosh y + \sinh y}{\cosh y - \sinh y} = \frac{e^y}{e^{-y}} = e^{2y}$. So, $\ln \sqrt{\frac{1 + \tanh y}{1 - \tanh y}} = \ln \sqrt{e^{2y}} = \ln(e^y) = y$.

Substituting these into the integral $J$: $J = \int_{2}^{2.5} y \cdot \text{sech}^2 y \, dy$ We use integration by parts, $\int u \, dv = uv - \int v \, du$. Let $u = y$ and $dv = \text{sech}^2 y \, dy$. Then $du = dy$ and $v = \int \text{sech}^2 y \, dy = \tanh y$. $J = [y \tanh y]_{2}^{2.5} - \int_{2}^{2.5} \tanh y \, dy$ $J = (2.5 \tanh 2.5 - 2 \tanh 2) - [\ln(\cosh y)]_{2}^{2.5}$ $J = (2.5 \tanh 2.5 - 2 \tanh 2) - (\ln(\cosh 2.5) - \ln(\cosh 2))$ $J = 2.5 \tanh 2.5 - 2 \tanh 2 - \ln(\cosh 2.5) + \ln(\cosh 2)$

3. Calculate $I + J$: $I + J = (\ln(\cosh 2.5) - \ln(\cosh 2) + 1) + (2.5 \tanh 2.5 - 2 \tanh 2 - \ln(\cosh 2.5) + \ln(\cosh 2))$ Notice that the $\ln(\cosh x)$ terms cancel out. $I + J = 1 + 2.5 \tanh 2.5 - 2 \tanh 2$

Now, express $\tanh x$ in terms of exponentials: $\tanh x = \frac{e^{2x} - 1}{e^{2x} + 1}$. $I + J = 1 + 2.5 \left(\frac{e^{2(2.5)} - 1}{e^{2(2.5)} + 1}\right) - 2 \left(\frac{e^{2(2)} - 1}{e^{2(2)} + 1}\right)$ $I + J = 1 + 2.5 \left(\frac{e^5 - 1}{e^5 + 1}\right) - 2 \left(\frac{e^4 - 1}{e^4 + 1}\right)$ To match the given form $a - 5(\frac{1}{e^5 + 1}) + 4(\frac{1}{e^4 + 1})$, we manipulate the terms: $I + J = 1 + \frac{5}{2} \left(\frac{e^5 + 1 - 2}{e^5 + 1}\right) - 2 \left(\frac{e^4 + 1 - 2}{e^4 + 1}\right)$ $I + J = 1 + \frac{5}{2} \left(1 - \frac{2}{e^5 + 1}\right) - 2 \left(1 - \frac{2}{e^4 + 1}\right)$ $I + J = 1 + \frac{5}{2} - \frac{5}{2} \cdot \frac{2}{e^5 + 1} - 2 + 2 \cdot \frac{2}{e^4 + 1}$ $I + J = 1 + \frac{5}{2} - 2 - \frac{5}{e^5 + 1} + \frac{4}{e^4 + 1}$ $I + J = \left(1 + \frac{5}{2} - 2\right) - \frac{5}{e^5 + 1} + \frac{4}{e^4 + 1}$ $I + J = \left(\frac{2+5-4}{2}\right) - \frac{5}{e^5 + 1} + \frac{4}{e^4 + 1}$ $I + J = \frac{3}{2} - \frac{5}{e^5 + 1} + \frac{4}{e^4 + 1}$

4. Determine $a$: Comparing our result with the given expression $I + J = a - 5(\frac{1}{e^5 + 1}) + 4(\frac{1}{e^4 + 1})$, we find: $a = \frac{3}{2}$