Solução_Calculo_Stewart_6e

F.
238 ¤ CHAPTER 5 INTEGRALS
TX.10
3. f(x) =e x − 2, 0 ≤ x ≤ 2. ∆x = b − a
n = 2 − 0 = 1 4 2 .
Since we are using midpoints, x ∗ i = x i = 1 2
(xi−1 + xi).
M 4 = 4 f(x i) ∆x =(∆x) [f(x 1)+f(x 2)+f(x 3)+f(x 4)]
i=1
= 1 2 f 1
4 + f 3
4 + f 5
4 + f 7
4
(e 1/4 − 2) + (e 3/4 − 2) + (e 5/4 − 2) + (e 7/4 − 2)
= 1 2
≈ 2.322986
The Riemann sum represents the sum of the areas of the three rectangles above the x-axis minus the area of the rectangle
below the x-axis; that is, the net area of the rectangles with respect to the x-axis.
5. ∆x =(b − a)/n =(8− 0)/4 =8/4 =2.
(a) Using the right endpoints to approximate 8
f(x) dx, wehave
0
4
f(x i) ∆x =2[f(2) + f(4) + f(6) + f(8)] ≈ 2[1+2+(−2) + 1] = 4.
i=1
(b) Using the left endpoints to approximate 8
f(x) dx, wehave
0
4
f(x i−1 ) ∆x =2[f(0) + f(2) + f(4) + f(6)] ≈ 2[2+1+2+(−2)] = 6.
i=1
(c) Using the midpoint of each subinterval to approximate 8
f(x) dx, wehave
0
4
f(x i ) ∆x =2[f(1) + f(3) + f(5) + f(7)] ≈ 2[3+2+1+(−1)] = 10.
i=1
7. Since f is increasing, L 5 ≤ 25
0
f(x) dx ≤ R 5 .
Lower estimate = L 5 = 5 f(x i−1 ) ∆x =5[f(0) + f(5) + f(10) + f(15) + f(20)]
i=1
=5(−42 − 37 − 25 − 6 + 15) = 5(−95) = −475
Upper estimate = R 5 = 5 f(x i) ∆x =5[f(5) + f(10) + f(15) + f(20) + f(25)]
i=1
=5(−37 − 25 − 6 + 15 + 36) = 5(−17) = −85
9. ∆x =(10− 2)/4 =2, so the endpoints are 2, 4, 6, 8,and10, and the midpoints are 3, 5, 7,and9. The Midpoint Rule
gives 10
√
x3 +1dx ≈ 4 f(x
2
i ) ∆x =2 √ 3 3 +1+ √ 5 3 +1+ √ 7 3 +1+ √ 9 3 +1 ≈ 124.1644.
i=1
11. ∆x =(1− 0)/5 =0.2, so the endpoints are 0, 0.2, 0.4, 0.6, 0.8, and 1, and the midpoints are 0.1, 0.3, 0.5, 0.7, and 0.9.
The Midpoint Rule gives
1
0 sin(x2 ) dx ≈ 5 f(x i ) ∆x =0.2 sin(0.1) 2 +sin(0.3) 2 +sin(0.5) 2 +sin(0.7) 2 +sin(0.9) 2 ≈ 0.3084.
i=1