Find the exact area between y=lnx and y=ln(x^ 2 ) for 4≤x≤8 .
just plug and chug. The area is
∫[4,8] ln(x^2)-lnx dx = ∫[4,8] 2lnx-lnx dx = ∫[4,8] lnx dx
Now, using integration by parts,
u = lnx, du = 1/x dx
dv = dx, v = x
∫lnx dx = x lnx - ∫ dx = x lnx - x
So, the area is
(8ln8-8)-(4ln4-4)
= 24ln2 - 8 - 8ln2 + 4
= 16ln2 - 4
Sure, let's calculate the area between the curves y = ln(x) and y = ln(x^2) from x = 4 to x = 8.
First, let's find the points of intersection between these two curves.
ln(x) = ln(x^2)
x = x^2
x^2 - x = 0
x(x - 1) = 0
x = 0 or x = 1
Since 4 ≤ x ≤ 8, we know that x = 0 is not included in this range.
Now let's find the area between the curves. We'll integrate the function that gives the difference between the two curves with respect to x.
Area = ∫[4,8] (ln(x) - ln(x^2)) dx
To simplify this expression, we can use the properties of logarithms:
Area = ∫[4,8] (ln(x) - 2ln(x)) dx
Area = ∫[4,8] (-ln(x)) dx
Next, we integrate:
Area = [-xln(x) + x] |[4,8]
Substituting the limits of integration:
Area = (-8ln(8) + 8) - (-4ln(4) + 4)
Finally, we calculate the result:
Area = 8 - 8ln(8) + 4ln(4) - 4
So the exact area between y = ln(x) and y = ln(x^2) for 4 ≤ x ≤ 8 is 8 - 8ln(8) + 4ln(4) - 4.
To find the exact area between the curves y = ln(x) and y = ln(x^2) for 4 ≤ x ≤ 8, follow these steps:
Step 1: Determine the points of intersection:
To find the points of intersection, set the two equations equal to each other and solve for x:
ln(x) = ln(x^2)
Since the natural logarithm function is one-to-one, we can drop the logarithm and solve for x directly:
x = x^2
Rearranging the equation, we have:
x^2 - x = 0
Using the zero-product property, we can factor out x:
x(x - 1) = 0
Setting each factor equal to zero:
x = 0 and x - 1 = 0
Solving for x, we find two points of intersection: x = 0 and x = 1.
Step 2: Determine the area bound by the curves:
To find the area, integrate the difference between the upper curve (ln(x^2)) and the lower curve (ln(x)) between x = 4 and x = 8.
Since we have two parts for the integral (from 4 to 1 and from 1 to 8), we will evaluate them separately.
Area 1 (from x = 4 to x = 1):
∫ [ln(x^2) - ln(x)] dx
Breaking it down further:
∫ ln(x^2) dx - ∫ ln(x) dx
Using the properties of logarithms, we can simplify the integrals:
2∫ ln(x) dx - ∫ ln(x) dx
Combining like terms:
∫ ln(x) dx
Using the integration formula for natural logarithm:
x ln(x) - x + C
Evaluating the integral from 4 to 1:
[1 ln(1) - 1] - [4 ln(4) - 4]
Simplifying:
-3 - [4 ln(4) - 4]
Area 2 (from x = 1 to x = 8):
∫ [ln(x^2) - ln(x)] dx
Using the same simplification as before:
∫ ln(x) dx
Evaluating the integral from 1 to 8:
[8 ln(8) - 8] - [1 ln(1) - 1]
Simplifying:
8 ln(8) - 7
Step 3: Calculate the total area between the curves:
To find the total area, add the areas from step 2:
Total area = Area 1 + Area 2
= [-3 - (4 ln(4) - 4)] + (8 ln(8) - 7)
Simplifying and calculating the values using a calculator:
Total area ≈ -3 - (4 ln(4) - 4) + (8 ln(8) - 7)
≈ -19.956
To find the exact area between two curves, we need to set up an integral. The area between two curves can be found by taking the integral of the difference between the two functions.
In this case, we need to find the area between the curves y = ln(x) and y = ln(x^2), where x ranges from 4 to 8.
First, let's find the points where the two curves intersect. Setting the two equations equal to each other, we have:
ln(x) = ln(x^2)
To simplify the equation, we can rewrite ln(x^2) as 2ln(x):
ln(x) = 2ln(x)
Next, to solve for x, we can take the natural logarithm of both sides:
ln(x) = ln(e^(2ln(x)))
Since ln(e^a) = a, we can simplify the equation further:
ln(x) = 2ln(x) => x = e^(2ln(x))
Now, we need to find the values of x where the two curves intersect. To do this, we can solve for x:
x = e^(2ln(x))
Taking the exponential of both sides gives us:
x = x^2
Rearranging the equation, we have:
x^2 - x = 0
Factoring out an x from the equation, we get:
x(x - 1) = 0
Therefore, x = 0 or x - 1 = 0. Since the natural logarithm is only defined for positive values of x, we can disregard x = 0. Therefore, x = 1 is the only point where the two curves intersect.
Next, let's find the integral of the difference between the two functions over the range of x from 4 to 8:
∫[4, 8] (ln(x) - ln(x^2)) dx
We can simplify this by using the properties of logarithms:
∫[4, 8] (ln(x) - 2ln(x)) dx
Combining like terms, we have:
∫[4, 8] (ln(x) - ln(x^2)) dx = ∫[4, 8] (ln(x) - 2ln(x)) dx
Now, we can use the properties of logarithms to simplify the integral further. The difference of logarithms is equal to the logarithm of the quotient:
∫[4, 8] (ln(x) - 2ln(x)) dx = ∫[4, 8] ln(x/x^2) dx
Simplifying the logarithm again, we have:
∫[4, 8] ln(1/x) dx
Now, we can evaluate this integral:
∫[4, 8] ln(1/x) dx = ∫[4, 8] -ln(x) dx
Since we have a negative sign in front of the logarithm, we can flip the limits of integration:
∫[4, 8] -ln(x) dx = ∫[8, 4] ln(x) dx
Using the properties of logarithms once again, we can simplify the integral further:
∫[8, 4] ln(x) dx = [xln(x) - x] [8, 4]
Evaluating the limits of integration, we get:
[xln(x) - x] [8, 4] = (4ln(4) - 4) - (8ln(8) - 8)
Calculating the values, we finally get:
(4ln(4) - 4) - (8ln(8) - 8) ≈ -6.08
Therefore, the exact area between the curves y = ln(x) and y = ln(x^2) for 4 ≤ x ≤ 8 is approximately -6.08 square units. Note that the negative sign indicates that the area is below the x-axis.