Find the area of the region between the curves y=lnx and y=ln2x from x=1 and x=5.
∫[1,5] (ln2x - lnx) dx
= ∫[1,5] ln2+lnx-lnx dx
= ∫[1,5] ln2 dx
= 4ln2
thanks bro
To find the area between two curves, we need to integrate the difference between the two functions over the desired interval. In this case, we want to find the area between the curves y = ln(x) and y = ln(2x) from x = 1 to x = 5.
First, let's find the points of intersection between the two curves by setting the two equations equal to each other:
ln(x) = ln(2x)
We can simplify this equation by taking the exponential of both sides:
x = 2x
Now, we can solve for x:
x - 2x = 0
-x = 0
x = 0
However, x = 0 is not within the desired interval of x = 1 to x = 5. Therefore, there are no points of intersection between the two curves in this interval.
To find the area, we need to integrate the difference between the two functions from x = 1 to x = 5:
A = ∫(ln(2x) - ln(x)) dx from 1 to 5
Simplifying the integrand:
A = ∫ln(2x) dx - ∫ln(x) dx from 1 to 5
Using the properties of logarithms, we can write this as:
A = ∫ln(2) dx - ∫ln(x) dx from 1 to 5
Since ln(2) is a constant factor, we can take it out of the integral:
A = ln(2) ∫dx - ∫ln(x) dx from 1 to 5
Evaluating the definite integrals, we get:
A = ln(2)(5 - 1) - [(5ln(5) - 5) - (ln(1) - 1)]
Simplifying further:
A = ln(2)(4) - (5ln(5) - 5 + 1)
A = 4ln(2) - 5ln(5) + 6
Thus, the area of the region between the curves y = ln(x) and y = ln(2x) from x = 1 to x = 5 is 4ln(2) - 5ln(5) + 6.
To find the area of the region between two curves, we need to integrate the difference between the upper curve and the lower curve with respect to x over the given interval. In this case, the upper curve is y = ln2x and the lower curve is y = lnx.
Step 1: Find the points of intersection
To determine the interval of integration, we need to find the x-values where these two curves intersect. Setting the equations equal to each other:
lnx = ln2x
Using the properties of logarithms, we can simplify this to:
x = 2x
Simplifying further:
x - 2x = 0
-x = 0
x = 0
Since we are only interested in the interval between x = 1 and x = 5, we can ignore the x = 0 solution.
Step 2: Set up the integral
The integral to find the area of the region between the curves is given by:
∫ [upper curve - lower curve] dx
In this case, the upper curve is ln2x and the lower curve is lnx, so the integral becomes:
∫ [ln2x - lnx] dx
Step 3: Evaluate the integral
Now we can integrate the expression:
∫ [ln2x - lnx] dx = ∫ ln2x dx - ∫ lnx dx
The integral of ln2x can be calculated using u-substitution. Let u = 2x, then du/dx = 2, and rearranging, dx = du/2.
Plugging this into the first integral:
∫ ln2x dx = ∫ ln u du/2
Using integration by parts, letting dv = ln(u) du/2 and u = 1/2, we can calculate dv as:
dv = ln(u) du/2 = (1/2) 1/u du = (1/2) ln(u) du
Now we can integrate by parts, using v = (u/2) ln(u) - (1/4)u as the antiderivative of dv:
∫ ln2x dx = [(u/2) ln(u) - (1/4)u] - ∫ [(u/2) ln(u) - (1/4)u] du
= (u/2) ln(u) - (1/4)u - [(u/2) ln(u) - (1/4)u]
= -(1/4)u
Now we can integrate the second expression:
∫ lnx dx = x ln(x) - x
Step 4: Evaluate the definite integral
To find the area, we evaluate the definite integral from x = 1 to x = 5:
Area = ∫[ln2x - lnx] dx = ∫ ln2x dx - ∫ lnx dx = -(1/4)u - (x ln(x) - x) from 1 to 5
Plugging in the values:
Area = [-(1/4)(2x)] - [x ln(x) - x] evaluated from 1 to 5
= -(1/2)x - x ln(x) + x - (-(1/4)(2(1))) - (1 ln(1) - 1)
Simplifying further:
Area = -(1/2)(5) - 5 ln(5) + 5 - (-1/4) - (0 - 1)
= -5/2 - 5 ln(5) + 5 + 1/4 + 1
Thus, the area of the region between the curves y = lnx and y = ln2x from x = 1 to x = 5 is approximately:
Area ≈ -5/2 - 5 ln(5) + 5 + 1/4 + 1