If the line x=a bisects the area under the curve `y=(1)/(x^(2)), 1 le x le 9` , then a is equal to

Options

A. `4/9`
B. `9/5`
C. `5/9`
D. `9/4`

Explanation

To solve the problem of finding the value of \( a \) such that the line \( x = a \) bisects the area under the curve \( y = \frac{1}{x^2} \) from \( x = 1 \) to \( x = 9 \), we can follow these steps:

### Step 1: Calculate the total area under the curve from \( x = 1 \) to \( x = 9 \)

We need to find the total area under the curve \( y = \frac{1}{x^2} \) from \( x = 1 \) to \( x = 9 \). This can be done using integration:

\[ \text{Total Area} = \int_{1}^{9} \frac{1}{x^2} \, dx \]

### Step 2: Evaluate the integral

The integral \( \int \frac{1}{x^2} \, dx \) can be computed as follows:

\[ \int \frac{1}{x^2} \, dx = -\frac{1}{x} \]

Now, we evaluate it from 1 to 9:

\[ \int_{1}^{9} \frac{1}{x^2} \, dx = \left[-\frac{1}{x}\right]_{1}^{9} = -\frac{1}{9} - \left(-\frac{1}{1}\right) = -\frac{1}{9} + 1 = 1 - \frac{1}{9} = \frac{8}{9} \]

### Step 3: Set up the equation for the area bisected by \( x = a \)

Since the line \( x = a \) bisects the area, the area from \( x = 1 \) to \( x = a \) must equal half of the total area:

\[ \int_{1}^{a} \frac{1}{x^2} \, dx = \frac{1}{2} \cdot \frac{8}{9} = \frac{4}{9} \]

### Step 4: Evaluate the left-hand side integral

Now we compute the left-hand side:

\[ \int_{1}^{a} \frac{1}{x^2} \, dx = \left[-\frac{1}{x}\right]_{1}^{a} = -\frac{1}{a} + 1 = 1 - \frac{1}{a} \]

### Step 5: Set up the equation

Now we set the two expressions equal to each other:

\[ 1 - \frac{1}{a} = \frac{4}{9} \]

### Step 6: Solve for \( a \)

Rearranging gives:

\[ 1 - \frac{4}{9} = \frac{1}{a} \]

Calculating the left side:

\[ \frac{9}{9} - \frac{4}{9} = \frac{5}{9} \]

Thus, we have:

\[ \frac{1}{a} = \frac{5}{9} \]

Taking the reciprocal gives:

\[ a = \frac{9}{5} \]

### Conclusion

The value of \( a \) that bisects the area under the curve \( y = \frac{1}{x^2} \) from \( x = 1 \) to \( x = 9 \) is:

\[ \boxed{\frac{9}{5}} \]