Download Math 242 Midterm #1 Review

Math 242 Midterm #1 Review
Dr. Achilles A. Beros
Formulas: The following formulas will be provided on the front of the exam. These are not all
the formulas you need to know in order to solve the problems on the exam.
1. sin(2θ) = 2 sin(θ) cos(θ)
2. cos(2θ) = cos2 (θ) − sin2 (θ)
3. cos2 (θ) =
1+cos(2θ)
2
4. sin2 (θ) = 1−cos(2θ)
2
R
5. sec(x)dx = ln | sec(x) + tan(x)| + C
R
6. csc(x)dx = − ln | csc(x) + cot(x)| + C
Rb
7. Trapezoidal Method: a f (x)dx ≈ ∆x
(y0 + 2y1 + · · · + 2yn−1 + yn )
2
(b−a)3
00
If f is continuous and M is an upper bound for |f 00 | on [a, b], then |ET | ≤ M12n
.
2
Rb
8. Simpson’s Method: a f (x)dx ≈ ∆x
(y0 + 4y1 + 2y2 + · · · + 2yn−2 + 4yn−1 + yn )
3
(b−a)5
(4)
If f is continuous and M is an upper bound for |f (4) | on [a, b], then |ES | ≤ M180n
4 .
Problem 1 Find the following derivatives.
d 2
1.
x tan−1 (x)
dx
√
d (2x + 2)2 x − 1 2.
dx
(x2 + 2)2
d
3.
g(x)f −1 (x)
dx
d x sin(x) 4.
2
dx
Solution
d 2
x2
1.
x tan−1 (x) = 2x tan−1 (x) +
dx
1 + x2
1
√
√
(2x + 2)2 x − 1
d (2x + 2)2 x − 1 . Let y =
2.
.
2 + 2)2
dx
(x2 + 2)2
(x
(2x + 2)2 √x − 1 1
ln(y) = ln
= 2 ln(x + 1) + 2 ln(2) + ln(x − 1) − 2 ln(x2 + 2).
2
2
(x + 2)
2
y0
2
1
4x
= (ln(y))0 =
+
−
. Thus,
y
x + 1 2(x − 1) x2 + 2 √
2
1
4x (2x + 2)2 x − 1
y0 =
+
− 2
x + 1 2(x − 1) x + 2
(x2 + 2)2
g(x)
d
−1
g(x)f (x) = g 0 (x)f −1 (x) + 0 −1
3.
dx
f (f (x))
d x sin(x) 4.
2
= (ex sin(x) ln(2) )0 = ln(2)(sin(x) + x cos(x))ex sin(x) ln(2)
dx
= ln(2)(sin(x) + x cos(x))2x sin(x) .
Problem 2 Solve the differential equation
dy
dx
=
sec2 (x) tan3 (x)
.
y2
Solution First, separating the variables, we get
Z
Z
2
y dy = sec2 (x) tan3 (x)dx.
Thus,
y3
=
3
Z
Z
=
=
sec2 (x) tan3 (x)dx
u3 du, where u = tan(x) and du = sec2 (x)dx
u4
tan4 (x)
+C =
+ C.
4
4
The general solution to the equation is
y=
3 tan4 (x)
4
Problem 3 Evaluate the limits.
x2 − 1
x→1 sin(πx)
1. lim
2
+C
1/3
.
x2
x→0 ln(sec(x))
2. lim
Solution
2x
2
x2 − 1
= lim
=−
1. By l’Hôpital’s rule, lim
x→1 π cos(πx)
x→1 sin(πx)
π
x2
2x
2
= lim
= lim
=2
x→0 ln(sec(x))
x→0 tan(x)
x→0 sec2 (x)
2. By l’Hôpital’s rule, lim
Z
Problem 4 Evaluate the integral
e2x cos(3x)dx.
Solution
e2x Z 2x e
e cos(3x)dx =
cos(3x) −
(−3 sin(x))dx
2
2
Z 2x e2x i
3 h e2x e
=
cos(3x) +
sin(3x) −
(3 cos(x))dx
2
2
2
2
Z
1
3
9
= e2x cos(3x) + e2x sin(3x) −
e2x cos(x)dx
2
4
4
Z
1
3
9
e2x cos(3x)dx = e2x cos(3x) + e2x sin(3x) + C
1+
4
2
4
Z
2
3
e2x cos(3x)dx = e2x cos(3x) + e2x sin(3x) + C
13
13
Z
2x
3
Z
Problem 5 Evaluate the integral
√
x2 1 − x2 dx.
Solution
Z
Z
q
√
2
2
2
x 1 − x dx = sin (θ) 1 − sin2 (θ) cos(θ)dθ, where x = sin(θ) and dx = cos(θ)dθ
Z
= sin2 (θ) cos2 (θ)dθ
Z
1
=
(1 − cos(2θ))(1 + cos(2θ))dθ
4
Z
1
(1 − cos2 (2θ))dθ
=
4
Z
1 + cos(4θ) 1 1−
dθ
=
4
2
1 θ sin(4θ) =
−
+C
4 2
8
1
2 sin(2θ) cos(2θ) =
θ−
+C
8
4
1
4 sin(θ) cos(θ)(cos2 (θ) − sin2 (θ)) =
θ−
+C
8
4
√
1
= (sin−1 (x) − x 1 − x2 (1 − x2 − x2 )) + C
8
√
1
= (sin−1 (x) − x 1 − x2 (1 − 2x2 )) + C
8
1
x
θ
√
1 − x2
4
Z
Problem 6 Evaluate the integral
2x2 − x + 1
dx.
x(x − 1)2
Solution
B
C
A
2x2 − x + 1
+
= +
2
x(x − 1)
x x − 1 (x − 1)2
2x2 − x + 1 = A(x − 1)2 + Bx(x − 1) + Cx
x = 0 =⇒ 1 = A
x = 1 =⇒ 2 = C
A + B = 2 =⇒ B = 1
Z
Z 2x2 − x + 1
1
1
2
+
+
dx
dx
=
x(x − 1)2
x x − 1 (x − 1)2
2
= ln |x| + ln |x − 1| −
+C
x−1
∞
Z
Problem 7 Evaluate the improper integral
0
1
dx.
1 + x2
Solution
Z
0
∞
1
dx = lim
b→∞
1 + x2
Z
b
1
dx
2
0 1+x
b
= lim tan−1 (x)
b→∞
0
= lim tan−1 (b)
b→∞
π
=
2
There will also be an extra credit problem on numerical integration (Simpson’s Rule and Trapezoidal rule).
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