Solução_Calculo_Stewart_6e

F.
TX.10
6 APPLICATIONS OF INTEGRATION
6.1 Areas Between Curves
1. A =
3. A =
x=4
x=0
y=1
y=−1
(y T − y B) dx =
4
1
(x R − x L ) dy =
0
−1
(5x − x 2 ) − x 4
dx = (4x − x 2 ) dx = 2x 2 − 1 x3 4
=
3
32 − 64 0 3
e y − (y 2 − 2) 1
dy = e y − y 2 +2 dy
= e y − 1 3 y3 +2y 1
−1 = e 1 − 1 3 +2 − e −1 + 1 3 − 2 = e − 1 e + 10 3
0
−1
− (0) =
32
3
2
5. A = (9 − x 2 ) − (x +1) dx
−1
=
2
−1
(8 − x − x 2 ) dx
2
=
8x − x2
2 − x3
3
−1
=
16 − 2 − 8 3 − −8 −
1
+
1
2 3
=22− 3+ 1 2 = 39 2
7. The curves intersect when x = x 2 ⇔ x 2 − x =0 ⇔
x(x − 1) = 0 ⇔ x =0or 1.
A =
1
0
(x − x 2 ) dx = 1
2 x2 − 1 3 x3 1
0 = 1 2 − 1 3 = 1 6
9. A =
2
1
1
x − 1
dx = ln x + 1 2
x 2 x
1
= ln 2 + 1 2
− (ln 1 + 1)
=ln2− 1 2 ≈ 0.19
11. A =
=
1
0
√ x − x
2 dx
2
3 x3/2 − 1 3 x3 1
0
= 2 3 − 1 3 = 1 3
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