a)Find the number a such that the line x = a bisects the area under the curve y= 1/x^2, 1<x<3.
a=3/2
b)Find the number b such that the line y = b bisects the area in part (a)
b=?
To find the number b such that the line y = b bisects the area under the curve y = 1/x^2, 1 < x < 3, we need to divide the total area under the curve into two equal parts.
First, let's find the total area under the curve. The area, A, under the curve y = 1/x^2 from x = 1 to x = 3 can be calculated using definite integration:
A = ∫[1 to 3] (1/x^2) dx
Simplifying the integral:
A = ∫[1 to 3] x^(-2) dx
Using the power rule of integration, we have:
A = [-x^(-1)] [from 1 to 3]
Substituting the limits of integration:
A = [-(1/(3^1)) - (1/(1^1))]
A = [-(1/3) + 1]
A = [2/3]
So, the total area under the curve is 2/3.
To find the number b such that the line y = b bisects the area, we need to find the point where the line intersects with the curve. The line y = b intersects the curve y = 1/x^2 when 1/x^2 = b.
Solving for x, we have:
1/x^2 = b
x^2 = 1/b
x = sqrt(1/b)
To bisect the area, we need to find two points on the curve that have an equal area between the line y = b and the curve y = 1/x^2.
Let's integrate the region between x = sqrt(1/b) and x = 3:
Area between y = 1/x^2 and y = b = ∫[sqrt(1/b) to 3] (1/x^2 - b) dx
= ∫[sqrt(1/b) to 3] (1/x^2) - (b) dx
= [-x^(-1)] - (bx) [sqrt(1/b) to 3]
= [-(1/(3^1)) - 3b] - [-(1/(sqrt(1/b))^1) - sqrt(1/b)b]
= [-1/3 - 3b] + [sqrt(b) - sqrt(1/b)b]
To bisect the area, this expression should be equal to half of the total area, which is 2/3 divided by 2:
[-1/3 - 3b] + [sqrt(b) - sqrt(1/b)b] = (2/3)/2
Simplifying the equation:
[-1/3 - 3b] + [sqrt(b) - sqrt(1/b)b] = 1/3
Now, we can solve this equation for b.
To find the number b such that the line y = b bisects the area under the curve y= 1/x^2, 1<x<3, we can begin by finding the total area under the curve and then dividing it by 2.
The total area under the curve y= 1/x^2, 1<x<3, can be found by evaluating the definite integral:
A = ∫[1,3] (1/x^2) dx
To evaluate this integral, we can use the power rule for integration. Assuming you are familiar with the power rule, we have:
A = [-1/x] evaluated from 1 to 3
= (-1/3) - (-1/1)
= -1/3 + 1
= 2/3
So, the total area under the curve is 2/3.
To find the number b such that the line y = b bisects this area, we need to find the value of y = b such that the area under the curve y = 1/x^2, 1<x<3, from the x-axis up to the line is equal to half of the total area.
Therefore, we need to solve the following equation:
∫[1,a] (1/x^2) dx + ∫[a,3] (1/b) dx = (1/2) * (2/3)
Let's evaluate each integral separately:
∫[1,a] (1/x^2) dx = [-1/x] evaluated from 1 to a
= (-1/a) - (-1/1)
= -1/a + 1
∫[a,3] (1/b) dx = [x/b] evaluated from a to 3
= (3/b) - (a/b)
Substituting these values back into the equation:
(-1/a + 1) + ((3/b) - (a/b)) = 1/3
Combining like terms:
-1/a + 1 + 3/b - a/b = 1/3
Multiplying through by ab:
-b + a + 3a - ab = ab/3
Rearranging terms:
4a - b - ab = ab/3
Now, we can solve for b:
4a - b - ab = ab/3
Bringing all terms with b to one side:
4a - ab = b + ab/3
Combining like-terms:
4a - ab = (3b + ab)/3
Multiplying through by 3:
12a - 3ab = 3b + ab
Rearranging terms:
12a - 3b - 4ab = 0
Factoring out b:
b(12 - 4a) - 3b = 0
Factorizing the equation further:
b(12 - 4a - 3) = 0
Setting each part equal to zero:
b = 0 or 12 - 4a - 3 = 0
Simplifying the second equation:
12 - 4a - 3 = 0
-4a - 3 = -12
-4a = -12 + 3
-4a = -9
a = -9/-4
a = 9/4
Therefore, the value of b such that the line y = b bisects the area in part (a) is 0 or when a = 9/4.
int dx/x^2 = -1/x + c
from 1 to 3 = -1/3 - (-1/1) = 2/3
half of that is 1/3
from 1 to a = -1/a - (-1/1) = -1/a + 1
so
-1/a + 3/3 = 1/3
-1/a = -2/3
2 a = 3
a = 3/2