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
To solve these problems step by step, let's break them down one by one:
a) Find the integral from 1 to 3 of f(x)dx in terms of A:
We know that g(x) = A - f(x). To find the integral of f(x) from 1 to 3, we can use the given information in property ii. According to property ii, the integral of 1 to 2 f(x)dx is equal to the integral of 2 to 3 of g(x)dx:
∫[1 to 2] f(x) dx = ∫[2 to 3] g(x) dx
Since g(x) = A - f(x), we can rewrite the equation as:
∫[1 to 2] f(x) dx = ∫[2 to 3] (A - f(x)) dx
Expanding the right side of the equation:
∫[1 to 2] f(x) dx = ∫[2 to 3] A dx - ∫[2 to 3] f(x) dx
Using property iii, we know that the integral from 2 to 3 of f(x) dx is equal to -3A, so the equation becomes:
∫[1 to 2] f(x) dx = A(x)∣[2 to 3] - (-3A)
Evaluating the definite integral, we get:
∫[1 to 2] f(x) dx = A(3) - 2A + 3A
Simplifying the expression, we have:
∫[1 to 2] f(x) dx = 3A - 2A + 3A = 4A
Therefore, the integral of f(x) from 1 to 3 is equal to 4A.
b) Find the average value of g(x) in terms of A over the interval [1, 3]:
The average value of a function g(x) over an interval [a, b] is given by:
Avg(g(x)) = (1/(b-a)) * ∫[a to b] g(x) dx
In this case, we need to find the average value of g(x) over the interval [1, 3].
Since g(x) = A - f(x), we can substitute this into the average value formula:
Avg(g(x)) = (1/(3-1)) * ∫[1 to 3] (A - f(x)) dx
Simplifying:
Avg(g(x)) = (1/2) * ∫[1 to 3] (A - f(x)) dx
Expanding the integral:
Avg(g(x)) = (1/2) * ∫[1 to 3] A dx - ∫[1 to 3] f(x) dx
Using property iii, we know that the integral from 2 to 3 of f(x) dx is equal to -3A, so the equation becomes:
Avg(g(x)) = (1/2) * (A(x)∣[1 to 3]) - (-3A)
Evaluating the definite integral, we get:
Avg(g(x)) = (1/2) * (A(3) - A)
Simplifying the expression, we have:
Avg(g(x)) = (1/2) * (3A - A) = (1/2) * (2A) = A
Therefore, the average value of g(x) in terms of A over the interval [1, 3] is simply equal to A.
c) Find the value of k if the integral from 0 to 1 of f(x+1)dx = kA:
To find the value of k, we need to evaluate the integral from 0 to 1 of f(x+1)dx.
Substituting f(x+1) into the integral expression:
∫[0 to 1] f(x+1) dx = kA
Let u = x + 1, and therefore, dx = du.
The new bounds of integration become u = 1 when x = 0, and u = 2 when x = 1.
The integral expression becomes:
∫[1 to 2] f(u) du = kA
Since ∫[1 to 2] f(u) du represents the integral from 1 to 2 of f(x) dx (with u as the new variable), we can use the result from property ii, where the integral of 1 to 2 of f(x) dx is equal to the integral of 2 to 3 of g(x) dx.
Thus, we have:
∫[1 to 2] f(u) du = ∫[2 to 3] g(x) dx
∫[1 to 2] f(u) du = ∫[2 to 3] (A - f(x)) dx
Using property iii, we know that the integral from 2 to 3 of f(x) dx is equal to -3A, so the equation becomes:
∫[1 to 2] f(u) du = ∫[2 to 3] (A - f(x)) dx = A(x)∣[2 to 3] - (-3A)
Evaluating the definite integrals, we get:
∫[1 to 2] f(u) du = A(3) - 2A + 3A
Simplifying the expression:
∫[1 to 2] f(u) du = -2A + 3A = A
From the previous step, we know:
∫[1 to 2] f(u) du = kA
Setting the two equations equal to each other:
A = kA
Dividing both sides by A:
1 = k
Therefore, the value of k is 1.
To solve these problems, we need to use the given properties of the functions f and g. Let's go step by step:
a. To find the integral from 1 to 3 of f(x)dx in terms of A, we can use the properties ii and iii. Let's start by using property iii and substituting it into the integral from 2 to 3:
∫2 to 3 f(x)dx = -3A
Now, using property ii, we can rewrite the integral from 2 to 3 in terms of g(x):
∫2 to 3 g(x)dx = ∫2 to 3 (A - f(x))dx
Note that ∫1 to 2 f(x)dx is equal to ∫2 to 3 g(x)dx according to property ii. Therefore, we can rewrite the integral as:
∫1 to 2 f(x)dx = ∫2 to 3 g(x)dx = ∫2 to 3 (A - f(x))dx
Combining the integrals:
∫1 to 3 f(x)dx = ∫1 to 2 f(x)dx + ∫2 to 3 f(x)dx
∫1 to 3 f(x)dx = ∫1 to 2 f(x)dx + ∫2 to 3 (A - f(x))dx
Now, we can substitute the values we have:
∫1 to 3 f(x)dx = ∫1 to 2 f(x)dx - 3A
b. To find the average value of g(x) in terms of A over the interval [1,3], we can use the definition of the average value of a function. The average value of a function f(x) over an interval [a,b] is given by:
Average value = (1 / (b-a)) * ∫a to b f(x)dx
In this case, we need to find the average value of g(x) over [1,3]. Since g(x) = A - f(x), we can substitute this into the above formula:
Average value of g(x) = (1 / (3-1)) * ∫1 to 3 (A - f(x))dx
Simplifying:
Average value of g(x) = (1 / 2) * ∫1 to 3 (A - f(x))dx
c. To find the value of k if ∫0 to 1 f(x+1)dx = kA, we can use the property i. According to property i, g(x) = A - f(x). Substituting this into the integral, we get:
∫0 to 1 (A - g(x+1))dx = kA
Expanding the integral:
∫0 to 1 (A - g(x+1))dx = ∫0 to 1 (A - (A - f(x+1)))dx
Simplifying:
∫0 to 1 (A - g(x+1))dx = ∫0 to 1 f(x+1)dx
Therefore, we have:
∫0 to 1 f(x+1)dx = kA
Now you have the steps to solve each problem using the given properties of f and g.