find the area of the region under y=6 ln(3 x) and above y=8 for 4<x<7
Integrate for x=4 to 7 the vertical slices
from y2(x) to y1(x).
y1(x)=6ln(3x)
y2(x)=8
so
I=∫[6ln(x)-8]dx from 4 to 7
I get 6.46 approx.
I got 26.23
Did you take out the 3 from ln(3x)?
Yes, your answer is correct.
I missed out the three.
To find the area of the region under the curve y = 6 ln(3x) and above y = 8 for the interval 4 < x < 7, we need to calculate the definite integral of the function over that interval. Here's how you can do it step by step:
Step 1: Write down the given functions and interval.
- The upper function is y = 8.
- The lower function is y = 6 ln(3x).
- The interval is 4 < x < 7.
Step 2: Determine the limits of integration.
- Since we are integrating with respect to x, we need to find the x-values that correspond to the points where the upper and lower functions intersect.
- Set 8 equal to 6 ln(3x) and solve for x:
8 = 6 ln(3x)
ln(3x) = 8/6
ln(3x) = 4/3
3x = e^(4/3)
x = e^(4/3)/3
- The upper limit of integration is the x-value where the upper and lower functions intersect, which is e^(4/3)/3.
- The lower limit of integration is the given lower bound of the interval, which is x = 4.
Step 3: Set up the definite integral.
- The area of the region is given by the integral of the difference between the upper and lower functions over the interval from 4 to e^(4/3)/3:
Area = ∫[4, e^(4/3)/3] (8 - (6 ln(3x))) dx
Step 4: Evaluate the definite integral.
- Integrate the difference between the upper and lower functions with respect to x over the given interval.
- Evaluate the integral:
Area = [8x - 6x ln(3x)] evaluated from 4 to e^(4/3)/3
Step 5: Calculate the area.
- Substitute the upper and lower limits of integration into the evaluated integral and calculate the difference:
Area = [(8 * (e^(4/3)/3)) - (6 * (e^(4/3)/3) * ln(3 * (e^(4/3)/3)))] - [(8 * 4) - (6 * 4 * ln(3 * 4))]
Step 6: Simplify and approximate the area if needed.
- Calculate the values inside the square brackets and subtract the results.
Area ≈ [(8 * (e^(4/3)/3)) - (6 * (e^(4/3)/3) * ln(e^(4/3)))] - [(8 * 4) - (6 * 4 * ln(12))]
And that's how you can find the area of the region under y = 6 ln(3x) and above y = 8 for 4 < x < 7.