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
a.
Let
I1=∫f(x)dx from 1 to 2
I2=∫f(x)dx from 2 to 3
J1=∫g(x)dx from 1 to 2
J2=∫g(x)dx from 2 to 3
∫f(x)dx from 1 to 3
= I1 + I2
= J2 + I2 [I1=J2 from ii]
= ∫(f(x)+g(x))dx from 2 to 3
= ∫Adx from 2 to 3
= A
b.
The average value of g(x) from 1 to 3 is the integral of g(x) over the same limits divided by (3-1)=2.
∫g(x)dx 1 to 3
=∫(A-f(x))dx from 1 to 3 [from i]
=[∫Adx - ∫f(x)dx] from 1 to 3
=2A - A
=A
Average value of g(x) = A/(3-1)=A/2.
c.
∫f(x+1)dx from 0 to 1
= ∫f(x)dx from 1 to 2
= I1
= J2
= ∫(A-f(x))dx from 2 to 3
= A - (-3A) [ from ii ]
= 4A
Therefore if 4A=kA, k=4
Sure! Let's break down each part of the problem and go through the steps to find the solutions.
a) To find the integral from 1 to 3 of f(x)dx in terms of A, we can use the properties given in the problem:
i. g(x) = A - f(x)
ii. ∫(1 to 2) f(x)dx = ∫(2 to 3) g(x)dx
iii. ∫(2 to 3) f(x)dx = -3A
To solve this, let's combine properties i and iii. We can substitute g(x) from property i into property iii:
∫(2 to 3) (A - g(x))dx = -3A
Expanding and integrating, we get:
[Ax - ∫(2 to 3) g(x)dx] + C = -3A
Now, let's substitute property ii into the equation:
[Ax - ∫(2 to 3) f(x)dx] + C = -3A
Since we know that ∫(1 to 2) f(x)dx = ∫(2 to 3) g(x)dx, we can replace the integral with ∫(1 to 2) f(x)dx:
[Ax - ∫(1 to 2) f(x)dx] + C = -3A
Rearranging the terms, we have:
Ax - ∫(1 to 2) f(x)dx = -3A - C
To find the integral from 1 to 3 of f(x)dx, we integrate f(x) from 1 to 2 and add the integral from 2 to 3:
∫(1 to 3) f(x)dx = ∫(1 to 2) f(x)dx + ∫(2 to 3) f(x)dx
Plugging in the equation we derived earlier, we get:
∫(1 to 3) f(x)dx = (Ax - ∫(1 to 2) f(x)dx) + (-3A - C)
Simplifying further, we have:
∫(1 to 3) f(x)dx = Ax - ∫(1 to 2) f(x)dx - 3A - C
Finally, we can rearrange the terms to get the integral in terms of A:
∫(1 to 3) f(x)dx = Ax - ∫(1 to 2) f(x)dx - 3A - C
= Ax - 3A - ∫(1 to 2) f(x)dx - C
= (Ax - 3A - C) - ∫(1 to 2) f(x)dx
Therefore, the integral from 1 to 3 of f(x)dx in terms of A is (Ax - 3A - C) - ∫(1 to 2) f(x)dx.
b) To find the average value of g(x) in terms of A over the interval [1,3], we can use the following formula:
Average value = 1/(b-a) * ∫(a to b) g(x)dx
In this case, a = 1 and b = 3. Plugging in the expression for g(x) from property i into the average value formula, we have:
Average value = 1/(3-1) * ∫(1 to 3) (A - f(x))dx
Simplifying:
Average value = 1/2 * [(Ax - 2A) - ∫(1 to 3) f(x)dx]
Using the result from part a, we can substitute the integral from 1 to 3 of f(x)dx:
Average value = 1/2 * [(Ax - 2A) - [(Ax - 3A - C) - ∫(1 to 2) f(x)dx]]
Simplifying further, we get:
Average value = 1/2 * [(Ax - 2A) - (Ax - 3A - C) + ∫(1 to 2) f(x)dx]
Canceling out the common terms, we have:
Average value = 1/2 * [Ax - 3A - C + ∫(1 to 2) f(x)dx]
Therefore, the average value of g(x) in terms of A over the interval [1,3] is 1/2 * [Ax - 3A - C + ∫(1 to 2) f(x)dx].
c) To find the value of k if ∫(0 to 1) f(x+1)dx = kA, we need to apply a substitution. Let u = x + 1, so du = dx.
Now, we can rewrite the integral:
∫(0 to 1) f(x+1)dx = ∫(0+1 to 1+1) f(u)du
= ∫(1 to 2) f(u)du
Using property ii from the problem, ∫(1 to 2) f(x)dx = ∫(2 to 3) g(x)dx, we can substitute:
∫(0 to 1) f(x+1)dx = ∫(1 to 2) f(u)du = ∫(2 to 3) g(x)dx
From property iii, we know that ∫(2 to 3) f(x)dx = -3A. Therefore, the equation becomes:
∫(0 to 1) f(x+1)dx = ∫(1 to 2) f(u)du = ∫(2 to 3) g(x)dx = -3A
We're given that the integral on the left-hand side is kA. So we have:
kA = -3A
Dividing both sides by A, we get:
k = -3
Therefore, the value of k is -3.