Take a Diagnostic Exam 33
simplifies to
=x^2 sinx+ 2 xcosx− 2
∫
cosxdx,
and the final integration gives you
=x^2 sinx+ 2 xcosx−2 sinx+C.
Chapter 11
33.
∫ 4
1
1
√
x
dx=
∫ 4
1
x−^1 /^2 dx=
x^1 /^2
1 / 2
] 4
1
= 2 x^1 /^2
] 4
1
=2(4)^1 /^2 −2(1)^1 /^2 = 4 − 2 = 2
34.
∫k
− 1
(2x−3)dx=x^2 − 3 x
]k
− 1
=k^2 − 3 k−(1+3)
=k^2 − 3 k− 4
Setk^2 − 3 k− 4 = 6 ⇒k^2 − 3 k− 10 = 0
⇒(k−5)(k+2)= 0 ⇒k=5ork=− 2.
KEY IDEA
You can check your answer by
evaluating
∫− 2
− 1
(2x−3)dxand
∫ 5
− 1
(2x−3)dx.
- h(x)=
∫π
π/ 2
√
sintdt⇒h′(x)=
√
sinx
h′(π)=
√
sinπ=
√
0 = 0
- Letu= 3 x;du= 3 dxor
du
3
=dx.
∫
g(3x)dx=
∫
g(u)
du
3
=
1
3
∫
g(u)du
=
1
3
f(u)+c=
1
3
f(3x)+c
∫ 2
0
g(3x)dx=
1
3
[f(3x)]^20
=
1
3
f(6)−
1
3
f(0)
Thus, the correct choice is (A).
37.
∫x
π
sin(2t)dt=
[
−cos(2t)
2
]x
π
=
−cos(2x)
2
−
(
−
cos(2π)
2
)
=−
1
2
cos(2x)+
1
2
38. I.
∫c
a
f(x)dx=
∫b
a
f(x)dx+
∫c
b
f(x)dx
The statement is true, since the upper and
lower limits of the integrals are in sequence,
i.e.a→c=a→b→c.
II.
∫b
a
f(x)dx=
∫c
a
f(x)dx−
∫b
c
f(x)dx
=
∫c
a
f(x)dx+
∫c
b
f(x)dx
The statement is not always true.
III.
∫c
b
f(x)dx=
∫a
b
f(x)dx−
∫a
c
f(x)dx
=
∫a
b
f(x)dx+
∫c
a
f(x)dx
The statement is true.
Thus, only statements I and III are true.
- Sinceg(x)=
∫ x
π/ 2
2 sintdt, then
g′(x)=2 sinx.
Setg′(x)= 0 ⇒2 sinx= 0 ⇒x=πor 2π
g′′(x)=2 cosxandg′′(π)=2 cosπ=
−2 andg′′(2π)= 1.
Thusg has a local minimum atx= 2 π. You
can also approach the problem geometrically
by looking at the area under the curve. See
Figure DS-11.
+
+
y= 2 sint
2 π
2
5 π
2 2
0 ππ^3 π
–2
2
y
t
Figure DS-11