392 STEP 5. Build Your Test-Taking Confidence- The correct answer is (C).
∫− 2
− 35 x
(x+2)(x−3)
dxis an improper integralsincef(x)=
5 x
(x+2)(x−3)
has an infinite
discontinuity atx=−2, one of the limits of
integration. Therefore,
∫− 2− 35 x
(x+2)(x−3)
dxis equal ton→lim− 2 −∫n− 35 x
(x+2)(x−3)
dx.- The correct answer is (C).
Use a partial fraction decomposition with∫
A
2 x− 1
dx+∫
B
x+ 5
dx. Then
A(x+ 5 )+B( 2 x− 1 )= 1 ⇒
Ax+ 2 Bx= 0 ⇒A=− 2 B. Substituting and
solving, 5A−B= 1 ⇒ 5 (− 2 B)−B=1, so
B=−1
11
andA=2
11
. Then
∫
dx
( 2 x− 1 )(x + 5 )=
1
11
∫
2 dx
( 2 x− 1 )−1
11
∫
dx
(x + 5 )=1
11
ln∣∣
2 x− 1∣∣
−1
11
ln∣∣
x + 5∣∣=
1
11
ln∣∣
∣
∣2 x− 1
x+ 5∣∣
∣
∣+C.- The correct answer is (C).
Average value=1
(π/6)− 0∫π/ 602 sin(2x)dx=
6
π[
−cos(2x)]π/ 6
0=6
π[
−cos(
π
3)
−(−cos 0)]=
6
π[
−1
2
+ 1
]
=3
π.
- The correct answer is (C).
x=sin^2 t⇒
dx
dt
=2 sintcost=sin( 2 t)andy=cos( 2 t)⇒
dy
dt
=−2 sin( 2 t). Then
(
dx
dt) 2
=sin^2 ( 2 t)and
(
dy
dt) 2
=(−2 sin( 2 t))^2 =4 sin^2 ( 2 t). For0 ≤t≤
π
2, the length of the arc the particle
traces out isL=∫ π 20√
sin^2 ( 2 t)+4 sin^2 ( 2 t)dt=
∫ π 20√
5 sin^2 ( 2 t)dt=√
5∫ π 20sin( 2 t)dt.Note that√
sin^2 (2t)=|sin(2t)|=sin(2t) for
0 ≤t≤
π
2- The correct answer is (A).
y=3 sin^2(
x
2)
;dy
dx
=6 sin(
x
2)[
cos(
x
2)]
1
2=3 sin(
x
2)
cos(
x
2)dy
dx∣∣
∣∣
x=π=3 sin(
π
2)
cos(
π
2)
=3(1)(0)= 0Atx=π,y=3 sin^2(
π
2)
=3(1)^2 = 3 .Thepoint of tangency is (π,3).Equation of
tangent atx=πisy− 3 =0(x−π)⇒y=3.