Sketch the region R defined by 1 ≤ x ≤ 2 and 0 ≤ y ≤ 1/x3.
a) Find (exactly) the number a such that the line x = a divides R into two
parts of equal area
b) Then find (to 3 places) the number b such that the line y = b divides R
into two parts of equal area
What's the trouble? I assume you know how to integrate, so you just want to solve
∫[1,a] 1/x^3 dx = ∫[a,2] 1/x^3 dx
-1/2x^2 [1,a] = -1/2x^2 [a,2]
-1/2a^2 + 1/2 = -1/8 + 1/2a^2
1/a^2 = 5/8
a^2 = 8/5
a = √(8/5)
Now do a similar integral using x = 1/∛y to find b.
To sketch the region R, we need to understand the boundaries and shape of the region. The given inequalities define the following limits:
1 ≤ x ≤ 2: This means that the values of x are between 1 and 2, inclusive.
0 ≤ y ≤ 1/x^3: The values of y are limited by the inverse cube of x. As x approaches 1, y approaches infinity (since 1/1^3 = 1), and as x approaches 2, y approaches 1/8. So, y is bounded above by the graph of y = 1/x^3.
To sketch the region R, we draw the lines x = 1 and x = 2, and the curve y = 1/x^3. The region R is the area between these boundaries.
a) To find the number a such that the line x = a divides R into two parts of equal area, we can calculate the area of region R and find the value that splits it equally.
The area of region R can be found by integrating the function y = 1/x^3 between x = 1 and x = 2.
Area(R) = ∫[1, 2] (1/x^3) dx.
To find the value of a that divides R into two equal parts, we need to calculate the integrals:
∫[1, a] (1/x^3) dx = ∫[a, 2] (1/x^3) dx.
The value of a can be determined by solving these two integrals equal to each other. This process involves solving a definite integral and usually requires numerical methods or calculator software.
b) To find the number b such that the line y = b divides R into two parts of equal area, we can calculate the area of region R and find the value that splits it equally.
In this case, we calculate the integral of (1/x^3) with respect to x from 1 to 2, which gives us the area of region R. Then we need to find a value of b such that the integral of (b) with respect to y from 0 to b equals half of the area of region R.
We can use numerical methods or calculator software to find the value of b that splits the area of region R equally.