Calculus question?

Question

On the interval [1, 7], find the value of x that divides the area under y = (1/x) and above the x-axis, into 2 equal areas. Write an integral expression or equation that supports your answer.

Please help!!

Answer 1

In order to find the value of x that divides the area of y = 1/x into two equal areas, we need to find the area in our interval where the area from 1 to b is equal to the area from b to 7. Let's first solve for the integral from 1 to b.

Integral (1 to b, 1/x) dx
= ln|x| {evaluated from 1 to b}
= ln(b) - ln(1) = ln(b) - 0 = ln(b)

Now, let's solve for the integral from b to 7 of 1/x.
Integral (b to 7, 1/x) dx
= ln|x| {evaluated from b to 7}
= ln(7) - ln(b)

Note that these two areas must be the same, AND x = b is the value that we really want. Now, we equate these two areas.

ln(b) = ln(7) - ln(b)

Bring the -ln(b) to the other side, to get
2ln(b) = ln(7)

Bring the 2 outside of the ln to the inside of the log (by the log property that c*log[base b](a) = log[base b](a^c) ), to get

ln(b^2) = ln(7). Take the antilog of both sides (i.e. get rid of the logs) to get
b^2 = 7. Therefore, b = +/- sqrt(7), and your values for b are
b = {-sqrt(7), sqrt(7)}

HOWEVER, -sqrt(7) falls outside of our interval [1, 7], so we reject the negative solution, and
b = sqrt(7)

The value of x that divides the area under y = 1/x is
x = sqrt(7).

Answer 2

I can't write in math on here, so I'll have to do it in English.

1. Set the integral from 1 to t of 1/x equal to the integral from t to 7 of 1/x. T is the number where the line has to be drawn to yield equal area on both sides under the curve.

2. Integrate both sides to get ln(x) on both sides. You now have ln(x) from 1 to t = ln(x) from t to 7.

3. Use the integral theorem to get ln(t) - ln(1) = ln(7) - ln(t).

4. ln(1) = 0, so your equation now reads ln(t) = ln(7) - ln(t).

5. Add ln(t) to both sides: 2*ln(t) = ln(7).

6. In the rule of logs, a coefficiant can be moved to be the exponent of the argument: ln(t squared) = ln(7).

7. Raise both sides to the power of e: t squared = 7.

Therefore t = the square root of 7.