Find the number a such that the line x = a bisects the area under the curve y = 1/x2 for 1 ≤ x ≤ 4. 8 5 (b) find the number b such that the line y = b bisects the area in part (a).

The area under a curve is function of the definite integral between the points under the curve.

(a) Find a

To do this, we make use of:

Rewrite as:

Integrate

Expand

Collect like terms

Take LCM

Inverse both sides

Multiply both sides by 2

Simplify

(b) Find b

In (a), we have:

Multiply both sides by x^2

Divide through by y

Take square roots

So, the definite integral is:

Rewrite as:

Integrate

Expand

Divide through by 2

Collect like terms

Divide through by 2

Square both sides

Hence, the value of b is 1/4

Read more about areas under curves at:

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a. We're looking for such that

b. Integrating with respect to will make things easier.

(where we take the positive square root because we know )

Now we want to find such that

b

step-by-step explanation: