Dear alwin86,
Detailed solutions follow. Remember, these are for study purposes
only. You must not attempt to pass them off as your own work.
1.
The distance between a point (x, y) and the point (4, 0) is the square
root of
(x - 4)^2 + y^2 .
Substituting y = f(x) = sqrt(x) in this expression, we see that
g(x) = x^2 - 8x + 16 + x
= x^2 - 7x + 16
is the square of the distance.
To minimize the distance, we find the first derivative
g'(x) = 2x - 7
which tells us that when
2x - 7 = 0
x = 3.5
the parabola g(x) reaches its lowest point. Thus, the point
(3.5, sqrt(3.5))
is the point on f(x) closest to (4, 0).
2.
The average value of f(x) = x^3 over [0, 2] is the same as its average
height, which is the area under the curve divided by 2, the length of
the interval.
The first integral of f(x) is
x^4 / 4
which tells us that the area under f(x) from 0 to 2 is
2^4 / 4 = 16 / 4
= 4 .
Dividing this by 2, we obtain
4 / 2 = 2
as the average value of f(x) over the interval [0, 2]. The function
assumes this value when
x^3 = 2
x = 2^(1/3)
= 1.26
with two decimal digits of precision, so
(1.26, 2)
is approximately where the curve passes through its average value over
the interval [0, 2].
3.
The graph of f(x) lies below the x-axis in the interval [0, 1] and over
the x-axis in [1, 2]. We must therefore calculate each area separately
before summing.
The first integral of f(x) is
x^3 / 3 - x .
Evaluated at x = 1, this gives us
1^3 / 3 - 1 = 1/3 - 1
= -2/3
so the area between the curve and the x-axis is 2/3 in the interval
[0, 1].
Evaluating the integral at x = 2, we have
2^3 / 3 - 2 = 8/3 - 2
= 2/3
and subtracting the earlier obtained -2/3 at x = 1, we obtain
2/3 - (-2/3) = 2/3 + 2/3
= 4/3
as the area under the curve in the interval [1, 2].
To find the total area between the curve and the x-axis in the interval
[0, 2], we sum the areas to obtain
2/3 + 4/3 = 2 .
4.
According to the second fundamental theorem of calculus --
Mathworld: Second Fundamental Theorem of Calculus
http://mathworld.wolfram.com/SecondFundamentalTheoremofCalculus.html
-- the first derivative of F(x) is simply the function under the integral
evaluated at t = x.
F'(x) = 1/x^2
5.
The first integral of the function y = 1 - x^4 is
x - x^5 / 5 .
In the interval [0, 1], the area under the curve is therefore
1 - 1^5 / 5 = 1 - 1/5
= 4/5 .
Because the curve is symmetrical, the area covered in [-1, 0] will be
the same.
Thus, the total area covered in the interval [-1, 1] is
4/5 + 4/5 = 8/5 .
If you have any concerns about the accuracy of these solutions, please
advise me through a Clarification Request and allow me the opportunity
to fully meet your needs before you rate this answer.
Regards,
leapinglizard
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