How to prove that the
{The integral with limit a to b *(the integral with limit x to b f(t)dt}dx=integral with limit a to b (x-a)f(x)dx
Help..
doesn't matter. Just expand the expressions, and there will be lots of F(x). F(b), F(a)'s floating around. It will be the same on both sides of the equation
I'll get you started. Suppose
F(x) = ∫f(x) dx
∫(x-a)f(x) dx = ∫xf(x) dx - ∫af(x) dx = ∫xf(x) dx - aF(x)
Now, using integration by parts, let
u = x, du = dx
dv = f(x) dx, v = F(x)
∫xf(x) dx = xF(x) - ∫F(x) dx
∫[x,b] f(t) dt = F(b)-F(x)
see what you can do from here.
What is the function of f(b)?
To prove the given equation:
∫[a to b]{∫[x to b] f(t) dt} dx = ∫[a to b] (x - a) f(x) dx,
we need to use the properties of definite integration and consider the order of integration.
Let's start by expanding the left-hand side (LHS) of the equation:
LHS = ∫[a to b]{∫[x to b] f(t) dt} dx
Now, we can evaluate the inner integral with respect to t:
LHS = ∫[a to b]{∫[x to b] f(t) dt} dx
= ∫[a to b]{[F(t)] from x to b} dx
Here, F(t) represents the antiderivative (integral) of f(t) with respect to t.
Next, we apply the fundamental theorem of calculus to simplify the expression even further. According to the theorem, if F(t) is an antiderivative of f(t), then:
[F(t)] from x to b = F(b) - F(x)
Using this property, the equation becomes:
LHS = ∫[a to b]{[F(b) - F(x)]} dx
= ∫[a to b] F(b) dx - ∫[a to b] F(x) dx
Since F(b) is a constant, we can pull it out of the integral:
LHS = F(b) * ∫[a to b] dx - ∫[a to b] F(x) dx
Now, we evaluate the definite integral:
LHS = F(b)(b - a) - ∫[a to b] F(x) dx
Finally, we apply the second part of the fundamental theorem of calculus, which states that the integral of a function over its own interval is equal to its antiderivative function:
∫[a to b] F(x) dx = [F(x)] from a to b = F(b) - F(a)
Applying this, our equation becomes:
LHS = F(b)(b - a) - (F(b) - F(a))
Combining like terms:
LHS = F(b)(b - a) - F(b) + F(a)
Factoring out F(b):
LHS = F(b)[(b - a) - 1] + F(a)
Now, we can rewrite the equation as:
LHS = (b - a)F(b) - F(b) + F(a)
Further simplification gives:
LHS = (b - a)F(b) - F(b) + F(a)
= (b - a)F(b) - F(b) + F(a)
At this point, we can rearrange the terms and observe that (b - a)F(b) - F(b) is equivalent to (b - a - 1)F(b). Hence:
LHS = (b - a - 1)F(b) + F(a)
Now, let's compare the left-hand side (LHS) with the right-hand side (RHS) of the equation, which is:
RHS = ∫[a to b] (x - a)f(x) dx
If we take the antiderivative (integral) of (x - a)f(x) with respect to x, we will obtain F(x), which is the antiderivative of f(x).
Comparing LHS and RHS, we can see that:
LHS = (b - a - 1)F(b) + F(a)
= RHS
Thus, we have proved that the equation:
∫[a to b]{∫[x to b] f(t) dt} dx = ∫[a to b] (x - a) f(x) dx
is true.