Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 5.2 THEDEFINITEINTEGRAL ¤ 237
21. lim
n
n→∞ i=1
π iπ
tan
4n 4n can be interpreted as the area of the region lying under the graph of y =tanx on the interval
0, π 4 ,
since for y =tanx on
0, π π/4 − 0
4 with ∆x =
n
n
n iπ
A = lim f (x ∗ i ) ∆x = lim tan
n→∞
n→∞ 4n
i=1
i=1
= π iπ
, xi =0+i ∆x =
4n 4n ,andx∗ i = x i, the expression for the area is
π
. Note that this answer is not unique, since the expression for the area is
4n
the same for the function y =tan(x − kπ) on the interval kπ, kπ + π 4
,wherek is any integer.
23. (a) y = f(x) =x 5 . ∆x = 2 − 0
n
(b)
(c)
A = lim Rn = lim
n→∞ n→∞
n
i=1
= 2 n and x i =0+i ∆x = 2i
n .
n
f(x i) ∆x = lim
i=1
i 5 CAS
= n2 (n +1) 2 2n 2 +2n − 1
12
n→∞ i=1
n 2i
n
5
· 2
n = lim
n→∞
n
i=1
32i 5
n · 2
5 n = lim 64
n→∞ n 6
64
lim
n→∞ n · n2 (n +1) 2 2n 2 +2n − 1
= 64
n 2
6 12
12 lim +2n +1 2n 2 +2n − 1
n→∞
n 2 · n 2
= 16 1+
3 lim 2 n→∞ n + 1 2+ 2 n 2 n − 1
n 2
25. y = f(x) =cosx. ∆x = b − 0
n
A = lim R n = lim
n→∞ n→∞
= b n
n
f(x i ) ∆x = lim
i=1
If b = π 2 ,thenA =sinπ 2 =1.
and xi =0+i ∆x =
bi
n .
n→∞ i=1
n bi
cos
n
· b CAS
= lim
n n→∞
⎡
⎢
⎣
= 16 3
1
b sin b
n
i 5 .
i=1
· 1 · 2=
32
3
⎤
2n +1 − b
⎥
b 2n ⎦ CAS
= sinb
2n sin
2n
5.2 The Definite Integral
1. f(x) =3− 1 b − a
x, 2 ≤ x ≤ 14. ∆x =
2
n = 14 − 2 =2.
6
Since we are using left endpoints, x ∗ i = x i−1 .
L 6 = 6 f(x i−1 ) ∆x
i=1
=(∆x)[f(x 0)+f(x 1)+f(x 2)+f(x 3)+f(x 4)+f(x 5)]
=2[f(2) + f(4) + f(6) + f(8) + f(10) + f(12)]
=2[2+1+0+(−1) + (−2) + (−3)] = 2(−3) = −6
The Riemann sum represents the sum of the areas of the two rectangles above the x-axis minus the sum of the areas of the
three rectangles below the x-axis; that is, the net area of the rectangles with respect to the x-axis.