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 \(I(\alpha)\)

First, we compute \(I(\alpha)\):

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

The integral of \(\frac{1}{x}\) is \(\log x\). Thus, we have:

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

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

Now we compute the first summation:

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

Calculating each term:

\[ I(2) = \log(2), \quad I(3) = \log(3), \quad I(4) = \log(4), \quad I(5) = \log(5) \]

Thus,

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

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

Next, we compute the second summation:

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

Calculating each term:

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

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 summations:

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

This simplifies to:

\[ 0 \]

### Final Answer

Thus, the value of

\[ \Sigma_{r=2}^{5} I(r) + \Sigma_{k=2}^{5} I\left(\frac{1}{k}\right) = 0 \]