Consider `I(alpha)=int_(alpha)^(alpha^(2))(dx)/(x)` (where `alpha gt 0`), then the value of `Sigma_(r=2)^(5)I(r )+Sigma_(k=2)^(5)I((1)/(k))` is

Options

A. 0
B. 1
C. ln 2
D. ln 4

Answer

To solve the problem, we need to evaluate the expression:

\[ \Sigma_{r=2}^{5} I(r) + \Sigma_{k=2}^{5} I\left(\frac{1}{k}\right) \]

where

\[ I(\alpha) = \int_{\alpha}^{\alpha^2} \frac{dx}{x} \]

### Step 1: Evaluate \( I(\alpha) \)

We start with the integral:

\[ I(\alpha) = \int_{\alpha}^{\alpha^2} \frac{dx}{x} \]

The integral of \( \frac{1}{x} \) is \( \log x \). Therefore, we can evaluate the integral:

\[ I(\alpha) = \left[ \log x \right]_{\alpha}^{\alpha^2} = \log(\alpha^2) - \log(\alpha) \]

Using the properties of logarithms:

\[ I(\alpha) = 2 \log \alpha - \log \alpha = \log \alpha \]

### Step 2: Evaluate \( \Sigma_{r=2}^{5} I(r) \)

Now we calculate \( \Sigma_{r=2}^{5} I(r) \):

\[ \Sigma_{r=2}^{5} I(r) = I(2) + I(3) + I(4) + I(5) \]

Substituting the values of \( I(r) \):

\[ = \log 2 + \log 3 + \log 4 + \log 5 \]

### Step 3: Evaluate \( \Sigma_{k=2}^{5} I\left(\frac{1}{k}\right) \)

Next, we evaluate \( \Sigma_{k=2}^{5} I\left(\frac{1}{k}\right) \):

\[ I\left(\frac{1}{k}\right) = \log\left(\frac{1}{k}\right) = -\log k \]

Thus,

\[ \Sigma_{k=2}^{5} I\left(\frac{1}{k}\right) = -\log 2 - \log 3 - \log 4 - \log 5 \]

### Step 4: Combine the results

Now we combine both sums:

\[ \Sigma_{r=2}^{5} I(r) + \Sigma_{k=2}^{5} I\left(\frac{1}{k}\right) = (\log 2 + \log 3 + \log 4 + \log 5) + (-\log 2 - \log 3 - \log 4 - \log 5) \]

This simplifies to:

\[ 0 \]

### Final Answer

Thus, the final answer is:

\[ \boxed{0} \]