21.19
y = x
2 '
3
AREA AND ARC LENGTH 0 167
21.21
and
Fig. 21-17
Fig. 21-18
21.18
y = 3x-2 from * = 0 to jc = l.
So,
from x = 1 to x = 8.
Hence,
1 + (y')
2 = 1 + (4/9x
2 '
3 ) = (9x
213 + 4)/9x
2 '
3 .
Thus,
Then,
Let
21.20
jc
2 '
3 + y
2 '
3 = 4 from x = 1
By implicit differentiation,
to x = 8.
So,
Hence,
Therefore,
from x = 1 to x = 2.
So,
Hence,
21.22
Let &t consist of all points in the plane that are above the x-axis and below the curve whose equation is
y ~ -x
2 + 2* + 8. Find the area of $.
(see Fig. 21-17). To find where it cuts the*-axis, let -x
2 +2* + 8 = 0, x-2x-8 = 0, (x - 4)(* + 2) = 0,
or x = —2. Hence, the area of &i is J_ 2
21.23
Find the area bounded by the curves y = 2x
2 - 2 and y = ::
2 + x.
See Fig. 21-18. y = 2x
2 - 2 is a parabola with vertex at (0, -2). On the other hand, y = x
2 + x =
y' = 3.
L =
M = 9 x
2/3 + 4, du = 6x~
in dx.
y = -(X-8) = -[(X-l)2-9]=-(X-lY + 9.The parabola's vertex is (1,9) and it opens downward
2
x = 4
32)-(§+4-16) = 36.
(-x* + 2x + 8)dx = (- ;U + x* + 8x) f_ 2 = (- f + 16 +
(x + j)
2 - I is a parabola with vertex (-5, - j). To find the points of intersection, set 2x
2 - 2 = x
2 + x,
x
2 -x-2 = 0, (x - 2)(x + 1) = 0, x = 2 or *=-!. Thus, the points are (2,6) and (-1,0). Hence,
the area is J2, [(x2 + x) - (2x2 -2)] dx = /!, (2 + x - x2) dx = (2x + \x2 - ^3) ]2_, = (4 + 2- f) - (-2 +
J + J) = -§.