Tutorial 8 - Eastern Mediterranean University Open CourseWares

Tutorial 8 - Eastern Mediterranean University Open CourseWares

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e
qs .r
y .ZZ_6I
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:
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Questions related to SECTION 5.3
suortrunJeeruo^Alaqry(t)J,r[
x
'"
(s).a
'p
:
(r)V
Bur,Lrollo; eql elenp,tg
.suotcunJ_e€Jp
Z
F (x) =
f (t) dt
2
(s)y
= (r).{ puerye)!,0[
1. The graph of f is shown in the figure. Let
Z x
A(x) =
f (t) dt
and
(o)v
."
./
rog
(E)v .l
(z)v .e
(s)a
(z)t 'z
.q
0
:t
Tutorial 8 - May 2, 2014
be two area functions of f . Evaluate the following area functions.
)l .tz
)l.at
aiq .1Z_EZ
uauDpunl
luatsrcuor
Butuollolaqtlap*uorrrffi:frf&:r::tr::rtr:#:i;:;:#
f (t) dt = 8
.r,
0
@)v
uop,uny(X)!o;;;);:,';;:#:t
f (t) dt = −5
:
(b) F (5) =
2
0
Z
(c) A(0) =
:l' :
5
'1 to{ tp
U)t
Z
rT
(a) A(2) =
4
2
Z
f (t) dt = 0
0
8
Z
f (t) dt = 8 − 5 − 11 = −8
(d) A(8) =
0
Z
8
f (t) dt = −5 − 11 = −16
(e) F (8) =
2
5
2. Evaluate the following integrals using the Fundamental Theorem of Calculus.
Z 2
(a)
(x2 − 4) dx
.gg_g7
ualanpunl
−2
2
8
8
16
32
− 16 = −
= −8+ −8=
3
3
3
3
Ilr'f .r,
c-
-
n
'7,:
(t)t 'vt
(rl
'9r
−2
:
x3
− 4x
3
'0r
C-=r'c : (t){ 'sI
0:D'g = (t){ '€I
−2
(x2 − 4) dx =
l:rt
2
t apolls put)
Z
c
2012
Department of Mathematics, Eastern Mediterranean University
nq
(x)y
0
:
f (t) dt = 8 − 5 = 3
(f) A(8) =
erre
Z
1
Tutorial 8 - May 2, 2014
1
Z
(x−3 − 8) dx
(b)
1
2
1
Z
(x−3 − 8) dx =
1
2
Z
π
4
(c)
1
x
5
1
− 8x = − − 8 + 2 + 4 = −
1
−2
2
2
−2
2
sec2 (θ) dθ
0
π
4
Z
0
Z
1
2
(d)
0
Z
2
(e)
1
dx
1 − x2
Z 1
2
21
π
dx
π
1
−1
−1
√
= sin x = sin
− sin−1 (0) = − 0 =
2
6
6
0
1 − x2
0
√
3
dt
t
Z
1
2
2
3
dt = (3 ln | t |) = 3 ln(2) − 3 ln(1) = 3 ln(2) = ln(8)
t
1
√
x− x
dx
x3
4
√
Z 9
Z 9
Z 9
x− x
1
1
−2
− 25
dx
=
−
dx
=
x
−
x
dx =
x3
x2 x 45
4
4
4
!9 9
3
1
2 1
2
1
2
13
x−1 x− 2 − 3 = − + 3 =− +
+ −
=
=
−1
x 3x 2 9 81 4 24
162
−2 Z
(f)
π
π 4
sec2 (θ) dθ = (tan θ) = tan
− tan(0) = 1 − 0 = 1
4
0
9
4
4
3. Simplify the following expressions.
Z x
d
(a)
(t2 + t + 1) dt
dx 3
Z x
d
(t2 + t + 1) dt = x2 + x + 1
dx 3
Z 10
d
dz
(b)
2
dx x2 z + 1
Z 10
Z x2
d
dz
d
dz
1
2x
=−
=− 4
2x = − 4
2
2
dx x2 z + 1
dx 10 z + 1
x +1
x +1
c
2012
Department of Mathematics, Eastern Mediterranean University
2
Tutorial 8 - May 2, 2014
d
(c)
dx
Z
d
dx
Z
tan−1 x
cos(ln t)tan t dt
a
tan−1 x
−1
cos(ln t)tan t dt = cos(ln(tan−1 x))tan(tan
a
x)
1
cos(ln(tan−1 x))x
=
1 + x2
1 + x2
Questions related to SECTION 5.4
1. Use symmetry to evaluate the following integrals.
Z π
4
(a)
cos x dx
− π4
Because cos(−x) = cos(x), cos(x) function is even. Therefore the integral
can be calculated as:
!
√
Z π
Z π
π
√
4
4
2
4
cos x dx = 2(sin(x)) = 2
cos x dx = 2
−0 = 2
2
0
− π4
0
Z
π
4
(b)
sin5 (x) dx
− π4
Because sin(−x) = − sin(x), the sin x function is odd. Therefore the integral can be calculated as:
Z
π
4
sin5 (x) dx = 0
− π4
Questions related to SECTION 5.5
1. Find the following integrals using substitution rule.
Z
(a)
2x(x2 + 1)4 dx
Let
du
du
u = x2 + 1 ⇒
= 2x ⇒ dx =
dx Z
Z
Z 2x
du
u5
(x2 + 1)5
2x(x2 + 1)4 dx = 2x u4
= u4 du =
+C =
+C
2x
5
5
c
2012
Department of Mathematics, Eastern Mediterranean University
3
Tutorial 8 - May 2, 2014
Z
sin3 (x) cos x dx
(b)
Let
du
du
= cos x ⇒ dx =
u = sin(x) ⇒
dxZ
Z
Z cos x 4
du
u
sin4 (x)
sin3 (x) cos x dx = u3 cos x
= u3 du =
+C =
+C
cos x
4
4
Z
(c)
(x2 + x)10 (2x + 1) dx
Let
du
du
= 2x + 1 ⇒ dx =
dx
2x + 1
Z
Z
Z
u11
(x2 + x)11
du
2
10
10
= u10 du =
+C =
+C
(x +x) (2x+1) dx = u (2x+1)
2x + 1
11
11
Z
1
√
(d)
dx
1 − 9x2
Z
Z
1
1
√
p
dx =
dx
2
1 − 9x
1 − (3x)2
u = x2 + x
⇒
Let
u = 3x
Z
(e)
⇒
du
du
= 3 ⇒ dx =
dx Z
3
1
1
1
sin−1 (3x)
p
du = sin−1 (u) + C =
+C
3
3
3
1 − (u)2
x9 sin x10 dx
Let
u = x10
Z
Z
(f)
⇒
x9 sin x10 dx =
du
= 10x9
dx
Z
du
dx =
9
Z 10x
du
1
1
1
x9 sin u
=
sin
u
du
=
−
cos
u+C
=
−
cos x10 +C
10x9
10
10
10
⇒
ex − e−x
dx
ex + e−x
Let
u = ex + e−x
Z
ex − e−x
dx =
ex + e−x
⇒
Z
du
du
= ex − e−x ⇒ dx = x
dx
e − e−x
Z
x
−x
e −e
du
1
dx
=
du = ln | u | +C = ln | ex +e−x | +C
u
ex − e−x
u
c
2012
Department of Mathematics, Eastern Mediterranean University
4
Tutorial 8 - May 2, 2014
2
Z
(g)
(x2
0
2x
dx
+ 1)2
Let
du
du
= 2x ⇒ dx =
dx
2x
−1 5
Z u=5
Z x=2
1
2x
2x du
u
4
=
u−2 du =
dx =
=− +1=
2
2
2
(x + 1)
−1 5
5
x=0 u 2x
u=1
u = x2 + 1
2
Z
0
π
2
Z
⇒
1
sin2 (θ) cos θ dθ
(h)
0
Let
du
du
= cos θ ⇒ d θ =
dθ
cos θ
3 1
Z u=1
Z θ= π
2
u 1
du
u2 du =
=
sin2 (θ) cos θ dθ =
u2 cos θ
=
cos θ
3 3
u=0
θ=0
⇒
u = sin θ
Z
0
π
4
Z
(i)
0
π
2
0
sin x
dx
cos2 (x)
Let
du
du
= − sin x ⇒ dx =
dx
− sin x
√
−1 22
Z x= π
Z u= √2
4 sin x
2
u
sin x
du
−2
u
du
=
−
dx
=
=
−
=
2
2
cos (x)
u − sin x
−1 x=0
u=1
1
√
√
2
2− 2
= √ −1= √
= 2−1
2
2
⇒
u = cos x
Z
0
Z
π
4
2
x2 ex
(j)
3 +1
dx
−1
Let
du
du
= 3x2 ⇒ dx = 2
dx
3x
9
Z 2
Z x=2
Z u=9
u
du
1
e
e9 − 1
3
x2 ex +1 dx =
x2 eu 2 =
eu du = =
3x
3 u=0
3
3
−1
x=−1
u = x3 + 1
⇒
0
c
2012
Department of Mathematics, Eastern Mediterranean University
5
Tutorial 8 - May 2, 2014
4
Z
(k)
0
p
p
dp
9 + p2
Let
u = 9 + p2
Z
0
4
du
= 2p
dp
⇒
p
p
dp =
9 + p2
Z
p=4
p=0
⇒
p du
1
√
=
2
u 2p
dp =
Z
u=25
u=9
du
2p
25
√
√
√
1
1
√ du =
2 u = 25− 9 = 2
2
u
9
c
2012
Department of Mathematics, Eastern Mediterranean University
6