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# Find the number a such that the line x = a bisects the area under the curve y = 1/x² for 1 ≤ x ≤ 4.

**Solution:**

We will use the limit of the function.

y = 1/x²……... 1 ≤ x ≤ 4

1/x² can be written as x^{-2}

∵ The number 'a' bisects the area under the curve,

Therefore, \(\int\limits_1^4\)[x^{-2}]dx

⇒ \(\int\limits_1^a\)[x^{-2}]dx = \(\int\limits_a^4\)[x^{-2}]dx

Integrating both sides with respect to x.

⇒ [x^{-1}/ -1]\(^{a}_1\)= [x^{-1}/ -1]\(^{4}_a\)

Using the limits, we get

⇒ -1 [ 1/a - 1] = -1 [1/4 - 1/a]

⇒ 1 - 1/a = 1/a - 1/4

Take LCM

⇒ (a - 1) / a = (4 - a)/ 4a

Solve by cross multiplying the denominators.

⇒ 4a (a - 1) = a(4 - a)

⇒ 4a² - 4a = 4a - a²

⇒ 5a² = 8a

Divide both the sides by 5a

a = 8/5

## Find the number a such that the line x = a bisects the area under the curve y = 1/x² for 1 ≤ x ≤ 4.

**Summary: **

The value of number a such that the line x = a bisects the area under the curve y = 1/x² for 1 ≤ x ≤ 4 is 8/5.

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