LEt f and g be continous functions with the following properties
i. g(x) = A-f(x) where A is a constant
ii. for the integral of 1 to 2 f(x)dx= the integral of 2 to 3 of g(x)dx
iii. for the integral from 2 to 3 f(x)dx = -3A
a find the integral from 1 to 3 of f(x)dx in terms of A.
b. find the average value of g(x) in terms of A, over the interval [1,3]
c. Find the value of k if the integral from 0 to 1 f(x+1)dx=kA
need the steps for these problems worked out plz
Sure! Let's go through each problem step by step:
a) To find the integral from 1 to 3 of f(x)dx in terms of A, we need to use the given information. We know that g(x) = A - f(x), so we can substitute this into the integral from 2 to 3 in property ii:
∫[2 to 3] g(x)dx = ∫[2 to 3] (A - f(x))dx
Now, we are given that this integral is equal to the integral from 1 to 2 of f(x)dx:
∫[2 to 3] (A - f(x))dx = ∫[1 to 2] f(x)dx
So, let's write the integral from 1 to 3 in terms of A:
∫[1 to 3] f(x)dx = ∫[1 to 2] f(x)dx + ∫[2 to 3] f(x)dx
Since we know that ∫[1 to 2] f(x)dx = ∫[2 to 3] (A - f(x))dx, we can substitute it in:
∫[1 to 3] f(x)dx = ∫[2 to 3] (A - f(x))dx + ∫[2 to 3] f(x)dx
Simplifying this expression, we get:
∫[1 to 3] f(x)dx = A∫[2 to 3] dx
∫[1 to 3] f(x)dx = A(x)|[2 to 3]
∫[1 to 3] f(x)dx = A(3 - 2)
∫[1 to 3] f(x)dx = A
Therefore, the integral from 1 to 3 of f(x)dx in terms of A is simply A.
b) To find the average value of g(x) in terms of A over the interval [1,3], we need to use the formula for the average value of a function:
avg(g) = (1/(b-a)) * ∫[a to b] g(x)dx
In this case, a = 1, b = 3, and g(x) = A - f(x). Substituting these values, we have:
avg(g) = (1/(3-1)) * ∫[1 to 3] (A - f(x))dx
avg(g) = (1/2) * ∫[1 to 3] (A - f(x))dx
avg(g) = (1/2) * [A(x) - ∫[1 to 3] f(x)dx]
Since we found in part (a) that ∫[1 to 3] f(x)dx = A, we can substitute it in:
avg(g) = (1/2) * [A(x) - A]
avg(g) = (1/2) * [A(x - 1)]
avg(g) = (A/2) * (x - 1)
Therefore, the average value of g(x) in terms of A over the interval [1,3] is (A/2) * (x - 1).
c) To find the value of k if ∫[0 to 1] f(x+1)dx = kA, we need to make a substitution. Let's substitute u = x + 1, which means du = dx. The limits of integration also need to be adjusted.
When x = 0, u = 0 + 1 = 1
When x = 1, u = 1 + 1 = 2
Therefore, the new limits of integration become [1 to 2]. Let's substitute the variables into the integral:
∫[0 to 1] f(x+1)dx = ∫[1 to 2] f(u)du
We are given that this integral is equal to kA. So, we have:
∫[1 to 2] f(u)du = kA
Now, let's use property ii to write the integral from 1 to 2 in terms of g(x):
∫[1 to 2] f(u)du = ∫[2 to 3] g(x)dx
We know that ∫[2 to 3] g(x)dx = ∫[2 to 3] (A - f(x))dx. Substituting this in, we get:
∫[1 to 2] f(u)du = ∫[2 to 3] (A - f(x))dx
Now, let's change the variables in the right-hand side integral to match the left-hand side:
∫[1 to 2] f(u)du = ∫[2 to 3] (A - f(x))dx = ∫[1 to 2] (A - f(x-1))dx
Since the integrals are over the same interval, we can equate the integrands:
f(u) = A - f(x-1)
Since this equation holds for all x in the interval [1, 2], we can equate the values of u and x-1:
u = x - 1
Now, let's solve this equation to express u in terms of x:
u + 1 = x
Therefore, we can express f(u) in terms of f(x):
f(x) = A - f(u)
f(x) = A - f(x-1)
Now, let's substitute this expression into the left-hand side integral:
∫[1 to 2] f(u)du = ∫[1 to 2] (A - f(x-1))dx
Now, we can equate the integrals:
∫[1 to 2] f(u)du = ∫[1 to 2] (A - f(x-1))dx
Now, let's take the integral of both sides with respect to x:
∫[1 to 2] f(u)du = ∫[1 to 2] (A - f(x-1))dx
∫[1 to 2] f(u)du = A(x - 1) - ∫[1 to 2] f(x-1)dx
We're left with an integral of f(u) on the left side, but we want an integral of f(x) in terms of u. To convert the integral, we can use the substitution u = x - 1, which means du = dx. The limits of integration also need to be adjusted.
When x = 1, u = 1 - 1 = 0
When x = 2, u = 2 - 1 = 1
Therefore, the new limits of integration become [0 to 1]. Let's substitute the variables into the integral:
∫[1 to 2] f(u)du = A(x - 1) - ∫[1 to 2] f(x-1)dx
∫[0 to 1] f(u)du = A((u + 1) - 1) - ∫[0 to 1] f(u)du
Now, let's bring the integral of f(u) to the left side of the equation:
2∫[0 to 1] f(u)du = A(u + 1) - A
Now, let's evaluate the integral on the left side:
2∫[0 to 1] f(u)du = A(u + 1) - A
2∫[0 to 1] f(u)du = A(u + 1) - A
2(kA) = A(u + 1) - A
2k = u + 1 - 1
2k = u
Finally, we substitute u = 2k back into the equation:
f(x) = A - f(x-1)
f(x) = A - f(u)
f(x) = A - f(2k)
Therefore, the value of k is 2k.