How can we determine the value of ‘a’ such that the line x = a bisects the area under the curve y = 1/x² for the interval 1 ≤ x ≤ 4?

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To find the value of ‘a’ such that the line x = a bisects the area under the curve y = 1/x² for the interval [1, 4], we need to follow a series of steps:

Step 1: Calculate the total area under the curve.

The area under the curve from x = 1 to x = 4 can be computed using the definite integral:

A = ∫14 (1/x²) dx

Calculating this integral:

A = [-1/x]14 = [-1/4 - (-1/1)] = 1 - 1/4 = 3/4

Step 2: Determine the area to be bisected.

Since we want to find the line x = a that bisects this area, we need half of the total area:

Area_{left} = Area_{right} = 3/4 ÷ 2 = 3/8

Step 3: Set up the equation for the area on the left side of the line.

The area under the curve from x = 1 to x = a using the same integral setup gives:

Area_{left} = ∫1a (1/x²) dx = [-1/x]1a = -1/a + 1

Step 4: Set the area equation equal to 3/8.

Now, we set the area from x = 1 to x = a equal to 3/8:

-1/a + 1 = 3/8

Step 5: Solve for ‘a’.

Rearranging and solving for a:

-1/a = 3/8 - 1 → -1/a = 3/8 - 8/8 = -5/8 → a = 8/5

Conclusion:

Thus, the value of a that bisects the area under the curve y = 1/x² from x = 1 to x = 4 is:

a = 8/5 = 1.6

Therefore, a = 1.6 is the point where the line x = a divides the area equally.

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