Çankaya University
Department of Mathematics
2016 - 2017 Fall Semester
MATH 155 - Calculus for Engineering I
Final Examination
1) Evaluate the following limits if they exist.
a)(7 pts)
lim √
x→0
tan(3x)
√
x sin(2 x)
b)(8 pts)
ex − 1 − x −
lim
x→0
x3 + x4
x2
2
2) a)(9 pts) Find the derivative of the function:
y = f (x) = (x + ex )ln x
b)(6 pts) Find the slope of the tangent line to the curve x8 + 4x2 y 2 + y 8 = 6 at the point
(1, 1).
3) Evaluate the integrals:
a)(7 pts)
Z
(x3 + 6x2 )7 (x2 + 4x)dx
b)(8 pts)
Z 2
1
3e3x + 2
dx
e3x + 2x
4) a)(10 pts) Find the area between the curve y = x2 − 5x + 4 and the line y = 18.
b)(10 pts) Evaluate the integral
Z
(3x + 1)e−2x dx
5) Evaluate the integrals.
a)(10 pts)
Z 2π
sin3 x cos12 x dx
π
2
b)(10 pts)
Z
dx
√
7 + x2
6) a)(9 pts) Evaluate the integral
Z
4x2 + 2x − 2
dx
(x2 + 1)(2x − 3)
b)(6 pts) Evaluate the integral if exists
Z ∞
e−2x dx
0
Answers
1) a) Using the change of variable u =
lim
u→0
√
x we obtain:
tan(3u2 )
u sin(2u)
Now multiply and divide by 3u and rearrange to obtain:
= lim
u→0
tan(3u2 )
3u
·
2
3u
sin(2u)
Similarly, multiply and divide by 2 and rearrange:
= lim
u→0
=
3 tan(3u2 )
2u
·
·
2
3u2
sin(2u)
3
3
·1·1=
2
2
b)This limit is in the form
ex − 1 − x −
lim
x→0
x3 + x 4
0
, so using L’Hôpital’s rule we obtain:
0
x2
2
= lim
ex − 1 − x
3x2 + 4x3
= lim
ex − 1
6x + 12x2
= lim
ex
6 + 24x
x→0
x→0
x→0
At this point, the limit is NOT in the form
Just insert x = 0 to obtain:
=
1
6
0
, so we can NOT use L’Hôpital.
0
2) a)
ln y = ln x ln(x + ex )
(ln y)
0
1 + ex
1
x
ln(x + e ) +
=
ln x
x
x + ex
ln(x + ex ) 1 + ex
y0
=
+
ln x
y
x
x + ex
ln(x + ex ) 1 + ex
0
x ln x
y = (x + e )
+
ln x
x
x + ex
b) Using implicit differentiation we obtain:
8x7 + 8xy 2 + 8x2 yy 0 + 8y 7 y 0 = 0
y0 =
−x7 − xy 2
x2 y + y 7
At (1, 1) the slope is:
y0 =
−2
= −1
2
3) a) The substitution u = x3 + 6x2 gives
du = (3x2 + 12x) dx
1
du = (x2 + 4x) dx
3
Rewriting the integral in terms of u, we obtain:
Z
1
(x + 6x ) (x + 4x) dx =
3
3
2 7
2
Z
u7 du
=
u8
+c
24
=
(x3 + 6x2 )8
+c
24
b) Use the substitution
u = e3x + 2x
du = (3e3x + 2) dx
⇒
The new integral limits are:
x=1
⇒
u = e3 + 2
x=2
⇒
u = e6 + 4
Rewriting the integral in terms of u, we obtain:
Z
1
2
3e3x + 2
dx =
e3x + 2x
Z
e6 +4
e3 +2
du
u
e6 +4
= ln |u| 3
e +2
= ln(e6 + 4) − ln(e3 + 2)
= ln
e6 + 4
e3 + 2
4) a)
y
20
y = 18
16
12
y = x2 − 5x + 4
8
4
x
−2
−1
1
2
3
4
5
−4
x2 − 5x + 4 = 18
x2 − 5x − 14 = 0
(x − 7)(x + 2) = 0
⇒
x = 7,
x = −2
The area is:
Z
7
A =
18 − (x2 − 5x + 4) dx
14 + 5x − x2 dx
−2
Z
7
=
−2
7
5 2 x3 = 14x + x −
2
3 −2
245 343
8
=
98 +
−
− −28 + 10 +
2
3
3
=
243
2
6
7
b) We have to use integration by parts.
u = 3x + 1
⇒
dv = e−2x dx
⇒
Z
(3x + 1)e
−2x
du = 3 dx
e−2x
−2
v =
e−2x
dx = (3x + 1)
−
−2
Z
e−2x
3 dx
−2
(3x + 1) e−2x 3
= −
+
2
2
Z
5) a)
e−2x dx
= −
3
(3x + 1) e−2x
− e−2x + c
2
4
= −
(6x + 5) e−2x
+c
4
2π
3
Z
2π
Z
12
sin2 x cos12 x sin x dx
sin x cos x dx =
π
2
π
2
2π
Z
(1 − cos2 x) cos12 x sin x dx
=
π
2
Using the substitution: u = cos x, du = − sin x dx
Z
1
−(1 − u2 )u12 du
=
0
Z
=
1
(u14 − u12 ) du
0
u15 u13
=
−
15
13
= −
2
195
1
0
b) We need the trigonometric substitution
√
√
x = 7 tan θ, dx = 7 sec2 θ dθ
Z
Z √
dx
7 sec2 θ dθ
√
√
=
7 + x2
7 + 7 tan2 θ
Z
sec2 θ dθ
√
1 + tan2 θ
Z
sec2 θ dθ
√
sec2 θ
=
=
Z
=
sec θ dθ
= ln | sec θ + tan θ| + c
x
x + √ +c
= ln sec arctan √
7
7
OR:
r
x2
x = ln 1 +
+ √ +c
7
7
3
√
√
= ln 7 + x2 + x − ln 7 + c
OR:
√
= ln 7 + x2 + x + k
3
3
6) a)Using partial fractions expansion, we find:
4x2 + 2x − 2
Ax + B
C
= 2
+
2
(x + 1)(2x − 3)
x +1
(2x − 3)
4x2 + 2x − 2 = (Ax + B)(2x − 3) + C(x2 + 1)
4x2 + 2x − 2 = (2A + C)x2 + (−3A + 2B)x + (−3B + C)
2A + C = 4
⇒
−3A + 2B = 2
⇒
A=
22
40
6
, B= , C=
13
13
13
−3B + C = −2
Z
4x2 + 2x − 2
6
dx =
2
(x + 1)(2x − 3)
13
Z
x dx
22
+
2
x + 1 13
Z
dx
40
+
2
x + 1 13
Z
3
22
20 2
=
ln |x + 1| +
arctan x +
ln x −
13
13
13 Z
b)
∞
−2x
e
Z
dx = lim
t→∞
0
t
e−2x dx
0
e−2x
= lim
t→∞ −2
= lim −
t→∞
=
1
2
The integral is convergent.
t
0
e−2t 1
+
2
2
dx
2x − 3
3 +c
2