F(x) = int[a,x]f(t)dt = x^3-2x^2+x-a; find "a", range where F(x)>0, area of region

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robertvitoriano

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F(x) = int[a,x]f(t)dt = x^3-2x^2+x-a; find "a", range where F(x)>0, area of region

2. Consider the following function:

. . .\(\displaystyle \displaystyle F(x)\, =\, \int_a^x\, f(t)\, dt\, =\, x^3\, -\, 2x^2\, +\, x\, -\, a,\quad (a\, \neq\, 0)\)

(1) Find \(\displaystyle a\)

(2) Find the range of \(\displaystyle x\) where \(\displaystyle F(x)\, >\, 0\)

(3) Find the area of the region surrounded by the x-axis and the graph of \(\displaystyle f(x)\)


i think it has something to do with polynomial division, but I don't know how to proceed, I have never seen this kind of integration (constant to x), can someone help me? if you find, give some links of similar problems.

The answers are

(1)2

(2) X > 2

(3) 4/27
 

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robertvitoriano said:
2. Consider the following function:

. . .\(\displaystyle \displaystyle F(x)\, =\, \int_a^x\, f(t)\, dt\, =\, x^3\, -\, 2x^2\, +\, x\, -\, a,\quad (a\, \neq\, 0)\)

(1) Find \(\displaystyle a\)

(2) Find the range of \(\displaystyle x\) where \(\displaystyle F(x)\, >\, 0\)

(3) Find the area of the region surrounded by the x-axis and the graph of \(\displaystyle f(x)\)


i think it has something to do with polynomial division, but I don't know how to proceed, I have never seen this kind of integration (constant to x), can someone help me? if you find, give some links of similar problems.

The answers are

(1)2

(2) X > 2

(3) 4/27
Click to expand...
Hint:

F(a) = ?
 
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Still trying...

it turns out to be 0, since I integrated f(t) and replaced x with a to a, I still don't know how to get to the correct answer, what does this zero mean?is that a "a root of the integral"? I'm not sure about the needed concepts, I'd apreciate some help.
 
robertvitoriano said:
it turns out to be 0, since I integrated f(t) and replaced x with a to a, I still don't know how to get to the correct answer, what does this zero mean?is that a "a root of the integral"? I'm not sure about the needed concepts, I'd apreciate some help.
Click to expand...

first the limits are a to x, not a to a.
f(x)=x3-2x2+x-a, and f(x)= int{f(t)dt}, at a to x.
f(t) is the derivative of f(x). integrate, evaluate from a to x, and set equal to f(x) given.
this yields value for a.
 
robertvitoriano said:
i think it has something to do with polynomial division, but I don't know how to proceed, I have never seen this kind of integration (constant to x), can someone help me? if you find, give some links of similar problems. The answers ar 1)2 2) X > 2 3) 4/27View attachment 9254
Click to expand...
Subhotosh Khan gave you the correct hint. Now I'll spoon feed you what that hint is really saying.
You are given two ways to calculate F(a). One using the integral (now solve for F(a) ONLY using the integral). The 2nd way you can solve for F(a) is using the polynomial (so do that now). Now equate the two and you'll be able to solve for a.
 
robertvitoriano said:
it turns out to be 0, since I integrated f(t) and replaced x with a to a, I still don't know how to get to the correct answer, what does this zero mean?is that a "a root of the integral"? I'm not sure about the needed concepts, I'd apreciate some help.
Click to expand...
If the limits of an integral is a to a, then the integral = 0
 
sinx said:
first the limits are a to x, not a to a.
f(x)=x3-2x2+x-a, and f(x)= int{f(t)dt}, at a to x.
f(t) is the derivative of f(x). integrate, evaluate from a to x, and set equal to f(x) given.
this yields value for a.
Click to expand...

Well, I'm not really sure, how can I evaluate a definite integral that doesn't deal with numbers? this is my try, is it correct?2018-03-14 17.28.56.jpg
 
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robertvitoriano said:
2. Consider the following function:

. . .\(\displaystyle \displaystyle F(x)\, =\, \int_a^x\, f(t)\, dt\, =\, x^3\, -\, 2x^2\, +\, x\, -\, a,\quad (a\, \neq\, 0)\)

(1) Find \(\displaystyle a\)

(2) Find the range of \(\displaystyle x\) where \(\displaystyle F(x)\, >\, 0\)

(3) Find the area of the region surrounded by the x-axis and the graph of \(\displaystyle f(x)\)
Click to expand...
Since several days have gone by - I'll post a complete solution for part (1):

.\(\displaystyle \displaystyle F(a)\, =\, \int_a^a\, f(t)\, dt\, = 0\) ..........................................(i)

and

\(\displaystyle \displaystyle F(a)\, = \, a^3\, -\, 2a^2\, +\, a\, -\, a\)........................(ii)

From (i) and (ii)

a^3 - 2a^2 = 0

a^2 * (a - 2) = 0

a = 2 ( given \(\displaystyle a \ne 0\))
 
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Subhotosh Khan said:
Since several days have gone by - I'll post a complete solution for part (1):

.\(\displaystyle \displaystyle F(a)\, =\, \int_a^a\, f(t)\, dt\, = 0\) ..........................................(i)

and

\(\displaystyle \displaystyle F(a)\, = \, a^3\, -\, 2a^2\, +\, a\, -\, a\)........................(ii)

From (i) and (ii)

a^3 - 2a^2 = 0

a^2 * (a - 2) = 0

a = 2 ( given \(\displaystyle a \ne 0\))
Click to expand...
Thanks man, now I see what was proposed:

integral of F(x)= f(t) if F(x) = derivative of f(t),The fundamental theorem of calculus
 
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Subhotosh Khan said:
Since several days have gone by - I'll post a complete solution for part (1):

.\(\displaystyle \displaystyle F(a)\, =\, \int_a^a\, f(t)\, dt\, = 0\) ..........................................(i)

and

\(\displaystyle \displaystyle F(a)\, = \, a^3\, -\, 2a^2\, +\, a\, -\, a\)........................(ii)

From (i) and (ii)

a^3 - 2a^2 = 0

a^2 * (a - 2) = 0

a = 2 ( given \(\displaystyle a \ne 0\))
Click to expand...

you get the right answer this way, but i don't know how.
the limits are a to x, not a to a.
the soln to part one;
you integrate (the derivative) 3x2-4x+1 from a to x, set this equal to x3-2x2+x-a, and you get a.
 
sinx said:
you get the right answer this way, but i don't know how.
the limits are a to x, not a to a.
the soln to part one;
you integrate (the derivative) 3x2-4x+1 from a to x, set this equal to x3-2x2+x-a, and you get a.
Click to expand...
It is true that the limits are a to x. But this makes this integral a function of x. Khan suggested that you let x=a and let things work out from there.

On one hand F(a) = 0, since an integral from a to a = 0. On the other hand you get F(a) = a3 -2a2 + a -a = a3 -2a2

Now equate a3 -2a2 = 0 and get a=2
 
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Here is a nice way of doing this problem.

The integral is said to equal x3 - 2x2 + x -a. But it also should equal x3 - 2x2 + x - (a3 - 2a2 + a) so a3 - 2a2 = 0. Then a = 0 or a=2. Hence a=2

I would have thought Sir Denis would have said this.
 
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