Solução_Calculo_Stewart_6e

F.
234 ¤ CHAPTER 5 INTEGRALS
3. (a) R 4 = 4 f(x i ) ∆x ∆x = π/2 − 0 = π
4 8
i=1
=[f(x 1 )+f(x 2 )+f(x 3 )+f(x 4 )] ∆x
TX.10
4
= f(x i ) ∆x
i=1
= cos π 8 +cos 2π 8 +cos3π 8 +cos4π 8
π
8
≈ (0.9239 + 0.7071 + 0.3827 + 0) π 8 ≈ 0.7908
Since f is decreasing on [0,π/2],anunderestimate is obtained by using the right endpoint approximation, R 4 .
(b) L 4 = 4 4
f(x i−1) ∆x = f (x i−1) ∆x
i=1
i=1
=[f(x 0 )+f(x 1 )+f(x 2 )+f(x 3 )] ∆x
= cos 0 + cos π 8 +cos 2π 8 +cos3π 8
π
8
≈ (1 + 0.9239 + 0.7071 + 0.3827) π 8 ≈ 1.1835
L 4 is an overestimate. Alternatively, we could just add the area of the leftmost upper rectangle and subtract the area of the
rightmost lower rectangle; that is, L 4 = R 4 + f(0) · π − f
π
8 2 · π
. 8
5. (a) f(x) =1+x 2 and ∆x = 2 − (−1) =1 ⇒
3
R 3 =1· f(0) + 1 · f(1) + 1 · f(2) = 1 · 1+1· 2+1· 5=8.
∆x = 2 − (−1)
6
=0.5 ⇒
R 6 =0.5[f(−0.5) + f(0) + f(0.5) + f(1) + f(1.5) + f(2)]
=0.5(1.25 + 1 + 1.25 + 2 + 3.25 + 5)
=0.5(13.75) = 6.875
(b) L 3 =1· f(−1) + 1 · f(0) + 1 · f(1) = 1 · 2+1· 1+1· 2=5
L 6 =0.5[f(−1) + f(−0.5) + f(0) + f(0.5) + f(1) + f(1.5)]
=0.5(2 + 1.25 + 1 + 1.25 + 2 + 3.25)
=0.5(10.75) = 5.375
(c) M 3 =1· f(−0.5) + 1 · f(0.5) + 1 · f(1.5)
=1· 1.25 + 1 · 1.25 + 1 · 3.25 = 5.75
M 6 =0.5[f(−0.75) + f(−0.25) + f(0.25)
+ f(0.75) + f(1.25) + f(1.75)]
=0.5(1.5625 + 1.0625 + 1.0625 + 1.5625 + 2.5625 + 4.0625)
=0.5(11.875) = 5.9375
(d) M 6 appears to be the best estimate.