Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 5.3 THE FUNDAMENTAL THEOREM OF CALCULUS ¤ 243
3. (a) g(x) = x
f(t) dt.
0
g(0) = 0
f(t) dt =0
0
g(1) = 1
f(t) dt =1· 2=2 [rectangle],
0
g(2) = 2
f(t) dt = 1
f(t) dt + 2
f(t) dt = g(1) + 2
f(t) dt
0 0 1 1
=2+1· 2+ 1 · 1 · 2=5 [rectangle plus triangle],
2
(d)
g(3) = 3
0 f(t) dt = g(2) + 3
2 f(t) dt =5+1 2 · 1 · 4=7,
g(6) = g(3) + 6
f(t) dt [the integral is negative since f lies under the x-axis]
3
=7+ − 1
2 · 2 · 2+1· 2 =7− 4=3
(b) g is increasing on (0, 3) because as x increases from 0 to 3,wekeepaddingmorearea.
(c) g has a maximum value when we start subtracting area; that is, at x =3.
5. (a)ByFTC1withf(t) =t 2 and a =1, g(x) = x
1 t2 dt ⇒
g 0 (x) =f(x) =x 2 .
(b) Using FTC2, g(x) = x
1 t2 dt = 1
3 t3 x
1 = 1 3 x3 − 1 3
⇒ g 0 (x) =x 2 .
7. f(t) = 1
x
t 3 +1 and g(x) = 1
t 3 +1 dt,sobyFTC1, g0 (x) =f(x) = 1 . Note that the lower limit, 1, could be any
x 3 +1
real number greater than −1 and not affect this answer.
1
9. f(t) =t 2 sin t and g(y) = y
2 t2 sin tdt,sobyFTC1,g 0 (y) =f(y) =y 2 sin y.
11. F (x) =
π
x
√
1+sectdt= −
x
π
√
1+sectdt ⇒ F 0 (x) =− d
dx
13. Let u = 1 x .Thendu dx = − 1 x .Also,dh
2 dx = dh du
du dx ,so
h 0 (x) = d
dx
0
1/x
2
arctan tdt= d
du
u
2
x
π
√
1+sectdt= −
√
1+secx
arctan tdt· du
du
=arctanu
dx dx = − arctan(1/x) .
x 2
15. Let u =tanx. Then du
dx =sec2 x.Also, dy
dx = dy du
du dx ,so
y 0 = d tan x
t + √ tdt= d u
t + √ tdt· du
dx
du
dx = u + √ u du
dx = tan x + √ tan x sec 2 x.
17. Let w =1− 3x. Then dw = −3. Also,dy
dx dx = dy
dw
y 0 = d 1
u 3
dx 1−3x 1+u du = d 1
2 dw w
19.
2
−1
0
dw
dx ,so
u 3 dw
du ·
1+u2 dx = − d
dw
w
1
u 3 dw
du ·
1+u2 dx = −
w3
3(1 − 3x)3
(−3) =
1+w2 1+(1− 3x) 2
x 3 − 2x x
4
2 2
4
(−1)
4
dx =
4 − x2 =
−1
4 − 22 − − (−1) 2 =(4− 4) − 1
4
− 1 =0−
− 3 4 4 =
3
4