Let R be the region bounded by y = 1/x, the lime x = 1, the line x = 3 and the x-axis. The line x = k divides R into two regions of equal area. Determine k.
so
[integral] 1/x dx form 1 to k = [integral] 1/x dx from k to 3
lnx from 1 to k = lnx from k to 3
ln k - ln1 = ln3 - lnk
2lnk = ln3 + ln1
ln (k^2) = ln3 + 0
k^2 = 3
k = √3
Thank you!!
To determine the value of k that divides the region R into two equal areas, we can set up the integral to find the area under the curve.
First, let's find the points of intersection between y = 1/x and the x-axis:
1/x = 0
x = 0
So, the point of intersection is (0, 0).
To find the other point of intersection, we can set y = 0:
1/x = 0
x = infinity
Since x cannot be equal to infinity, there is no other point of intersection.
Next, we need to determine the bounds of integration. We are given that the region is bounded by x = 1 and x = 3, so the bounds of integration are from x = 1 to x = 3.
Now, let's set up the integral to find the area under the curve:
A = ∫[1 to 3] (1/x) dx
A = ∫(1/x) dx from 1 to 3
Evaluating the integral:
A = ln|x| from 1 to 3
A = ln(3) - ln(1)
A = ln(3)
We want to find the value of k such that the regions on either side of x = k have equal areas. This means the area from x = 1 to k is equal to the area from k to 3.
So, we can set up the following equation:
ln(k) = ln(3) - ln(k)
Applying the laws of logarithms:
ln(k) + ln(k) = ln(3)
2 * ln(k) = ln(3)
ln(k^2) = ln(3)
Taking the exponent of both sides:
k^2 = 3
Solving for k:
k = √3
Therefore, k = √3 divides the region R into two equal areas.
To determine the value of k, we need to find the area of the region R and then find where it is divided into two equal halves. Here's how we can do that step by step:
Step 1: Find the area of region R
Region R is bounded by the curves y = 1/x, the line x = 1, the line x = 3, and the x-axis. To find the area of R, we need to calculate the definite integral of the curve y = 1/x between x = 1 and x = 3.
∫[1 to 3] (1/x) dx
To calculate this integral, we can use the natural logarithm:
= ln|x| [1 to 3]
= ln|3| - ln|1|
= ln(3)
So, the area of region R is ln(3).
Step 2: Find the value of k
Now, we know that the line x = k divides region R into two equal areas. Let's call the left region R1 and the right region R2.
The area of R1 can be obtained by integrating the curve from 1 to k:
∫[1 to k] (1/x) dx
And the area of R2 can be obtained by integrating the curve from k to 3:
∫[k to 3] (1/x) dx
To find the value of k, we need to set the two areas equal to each other and solve for k:
∫[1 to k] (1/x) dx = ∫[k to 3] (1/x) dx
Using the fundamental theorem of calculus, we can evaluate these integrals:
ln|k| - ln|1| = ln|3| - ln|k|
Simplifying:
ln|k| = ln|3| - ln|k|
Combining logarithms:
ln|k^2| = ln|3|
Since the natural logarithm is a one-to-one function, we can drop the logarithms:
k^2 = 3
Taking the square root:
k = ±√3
However, since we are looking for a positive x-value, we can conclude that k = √3.
Therefore, the value of k that divides region R into two equal areas is √3.