Okay, you've been whining about this for days. Now it's time for some thought.
You know that the area between the curves f(x) and g(x), where f(x) > g(x) over the interval [a,b] is ∫[a,b] f(x) - g(x) dx, right?
Well, here you have set up an unusual situation, where x is the variable of integration, as well as being used to form the boundary (the line between (0,y) and (x,y).
So, how about we change things and use
y = f(t) = y = (1/2*t-2)^6+5
Now consider the graph of y=t^6. It is symmetric about the vertical line t=0.
Our function is just a scaled, shifted version of that, symmetric about the line t=4. So, if we use an arbitrary value of x, then the area under the curve is ∫(1/2*t-2)^6+5 dt and integrate on the interval [4,x]
Now the area is
2∫[4,x] (1/2*t-2)^6+5 dt
Now, since y = (1/2*t-2)^6+5 = [(1/2)(t-4)]^6 + 5, at some value x,
f(x) = [(1/2)(x-4)]^6 + 5 = 1/64 (x-4)^6 + 5
Due to symmetry, the area under the curve is 2∫[4,x] (1/2*t-2)^6+5 dt
if we assume x > 4. (If x < 4, then just use -x; symmetry makes that unimportant)
So now the area between our line f(t) = 1/64 (x-4)^6 + 5 and the curve g(t) = 1/64 (t-4)^6 + 5 is just
2∫[4,x] f(x) - g(t) dt
(remember, x is just some number here)
But the area under the line is just width*height = 2(x-4)(1/64 (x-4)^6 + 5)
So the area between the graphs is
2(x-4)(1/64 (x-4)^6 + 5) - 2∫[4,x] g(t) dt
= 2(x-4)(1/64 (x-4)^6 + 5) - 2(1/64 * 1/7 (t-4)^7 + 5t) [4,x]
Now, evaluating ∫g(t) dt at t=4, we get (0 - 5*4)
and at t=x, we get 1/64 (x-4)^7 + 5(x-4)
So, our final expression for the desired area is
1/32 (x-4)^7 + 10(x-4) + 5*4 - 1/64 (x-4)^7 + 5(x-4)
= 1/64 (x-4)^7 + 5(x-4) + 20
If I've done my arithmetic right. At this point you should be able to check to see whether I've missed something.
Find the area above curve y=(1/2*x-2)^6+5 enclosed by a line cutting at (0,y) and (x,y).
Note.It is a straight line.
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asked by Raj
Dec 22, 2019
Huh? It seems that you want the area under some arbitrary horizontal line.
So, let's say the line is through (0,6) and (2,6)
The area would then be (using symmetry of the region)
A = 2∫[0,2] 6 - ((1/2*x-2)^6+5) dx
That's just a simple substitution, letting
u = x/2 - 2
2du = dx
A = 2∫[-2,-1] 1-u^6 du
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oobleck
Dec 22, 2019
Answer is 438+6/7
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posted by Raj
Dec 22, 2019
I guess if you want an arbitrary x value, you could use this.
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oobleck
Dec 22, 2019
Oops. That's the area under the curve, not the area above the curve. You can adjust it as needed, I guess.
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oobleck
Dec 22, 2019
Oops. I forgot to note that the axis of symmetry is x=4, not x=0. So the area would be, for some arbitrary x>=4, since the area desired is a rectangle of area 2(x-4)y, you'd get
2((x-4)*((x/2-2)^6+5)-(∫[2,x] ((t/2-2)^6 + 5) dt))
= 53/1952 (x-4)^7 - 72/7
Better double-check my math ...
How did length become 2(x-4) and area of unwanted region (∫[2,x] ((t/2-2)^6 + 5) dt)) with bounds 2 and x?
8 answers
How width became 2(x-4)
x=4
(x-4);
But how times 2(x-4)
(x-4);
But how times 2(x-4)
You used symmetry for times 2.Am I right?
Why only 2(x-4) no 2x?
How to know what value of x to take?
Finally i got the answer 438+(6/7)
I do not know which value of x to take. You have not specified it.
0 answers