Definite Integrals 253
Note thatf(x)=cosx−x^2 is an even
function. Thus, you could have written∫π
−π
(
cosx−x^2
)
dx= 2
∫π
0
(
cosx−x^2
)
dx
and obtained the same result.
Part B Calculators are allowed.
13.
∫ 2
0
(
x^3 +k
)
dx=
x^4
4
+kx
] 2
0
=
(
24
4
+k(2)
)
− 0
= 4 + 2 k
Set 4+ 2 k=10 andk=3.
- Enter
∫
( 2 x∗cos(x),x,− 1 .2, 3. 1 )and
obtain− 4. 70208 ≈− 4 .702.
15.
d
dx
(∫ x 3
1
√
t^2 + 1 dt
)
=
√
(x^3 )^2 + 1
d
dx
(
x^3
)
= 3 x^2
√
x^6 + 1
- Δx=
8 − 0
4
= 2
Midpoints arex=1, 3, 5, and 7.
∫ 8
0
(
x^3 + 1
)
dx=
(
13 + 1
)
(2)+
(
33 + 1
)
(2)
+
(
53 + 1
)
(2)+
(
73 + 1
)
(2)
=(2)(2)+(28)(2)+(126)(2)
+(344)(2)= 1000
- (a)
∫ 0
− 2
g(x)dx+
∫ 2
0
g(x)dx
=
∫ 2
− 2
g(x)dx
∫ 0
− 2
g(x)dx+ 3
= 8 .Thus,
∫ 0
− 2
g(x)dx= 5.
(b)
∫− 2
2
g(x)dx=−
∫ 2
− 2
g(x)dx=− 8
(c)
∫− 2
0
5 g(x)dx= 5
∫− 2
0
g(x)dx
= 5
(
−
∫ 0
− 2
g(x)dx
)
=5(−5)=− 25
(d)
∫ 2
− 2
2 g(x)dx= 2
∫ 2
− 2
g(x)dx
=2(8)= 16
18.
∫ 1 / 2
0
dx
√
1 −x^2
=sin−^1 (x)
] 1 / 2
0
=sin−^1
(
1
2
)
−sin−^1 (0)
=
π
6
− 0 =
π
6
19.
∫sinx
cosx
(2t+1)dt=
∫sinx
0
(2t+1)dt
−
∫cosx
0
(2t+1)dt
dy
dx
=
d
dx
(∫sinx
cosx
(2t+1)dt
)
=(2 sinx+1)
d
dx
(sinx)−(2 cosx+ 1 )
d
dx
(cosx)
=(2 sinx+1) cosx−(2 cosx+1)(−sinx)
=2 sinxcosx+cosx+2 sinxcosx+sinx
=4 sinxcosx+cosx+sinx
- Δx=
30 − 0
3
= 10
Midpoints arex=5, 15, and 25.
∫ 30
0
f(x)dx=[f(5)]10+[f(15)]10+[f(25)]10
=(2.6)(10)+(4.1)(10)+(5.2)10
= 119
21.
∫∞
0
e−xdx=limk→∞
∫k
0
e−xdx
=limk→∞
(
−e−x|k 0
)
=limk→∞
(
−e−k+ 1
)
= 1
