Wright State University Lake Campus/2017-1/MTH2310
The four midterm tests are on 2/2/17, 3/9/17, 3/30/17, and 4/20/17 (all Thursdays) from 11:30 - 12:50 pm.
The final exam is on Monday 4/24/17, at 3:15-5:15 pm. T1=Test 1 (sections): 1.7, 4.5, 5.6, G,, 5.7, 5.10
For each equation, solve the problem in your private wiki and generate a variation using the Prob. provided.
Parametric equation Sec 1.7 p72
edit
Write this parametric equation in the form y=f(x) , and sketch the graph:
x
=
sin
t
{\displaystyle x=\sin t}
y
=
cos
2
t
{\displaystyle y=\cos ^{2}t}
Evaluate the limit:
lim
x
→
0
1
−
cos
(
2
x
)
x
2
{\displaystyle \lim _{x\rightarrow 0}{\frac {1-\cos(2x)}{x^{2}}}}
Evaluate the limit: (see page 297 problem 47)
L
=
lim
x
→
∞
(
1
+
1
e
x
)
e
x
{\displaystyle {\mathcal {L}}=\lim _{x\rightarrow \infty }\left(1+{\frac {1}{e^{x}}}\right)^{e^{x}}}
solution
Let
y
=
e
x
{\displaystyle y=e^{x}}
L
=
lim
y
→
∞
(
1
+
1
y
)
y
{\displaystyle {\mathcal {L}}=\lim _{y\rightarrow \infty }\left(1+{\frac {1}{y}}\right)^{y}}
ln
L
=
lim
y
→
∞
y
⋅
ln
(
1
/
y
+
1
)
=
lim
y
→
∞
ln
(
1
/
y
+
1
)
1
/
y
=
H
lim
z
→
0
ln
(
1
+
z
)
z
=
lim
z
→
0
1
1
+
z
1
=
1
{\displaystyle \ln {\mathcal {L}}=\lim _{y\rightarrow \infty }y\cdot \ln \left(1/y+1\right)=\lim _{y\rightarrow \infty }{\frac {\ln \left(1/y+1\right)}{1/y}}=^{H}\lim _{z\rightarrow 0}{\frac {\ln(1+z)}{z}}=\lim _{z\rightarrow 0}{\frac {\frac {1}{1+z}}{1}}=1}
This is correct, but we need
L
=
e
ln
(
L
)
=
e
1
=
e
{\displaystyle {\mathcal {L}}=e^{\ln({\mathcal {L}})}=e^{1}=e}
Because a reasonable person might forget to take antilog in the last step, the tests will not be multiple choice, but instead graded for partial credit. Similar problem:
lim
→
0
+
(
1
+
sin
4
x
)
cot
x
{\displaystyle \lim _{\rightarrow 0^{+}}\left(1+\sin 4x\right)^{\cot x}}
See Example 8 Sec 4.5
Integration by parts Sec 5.6 p283
edit
Evaluate
∫
0
π
/
2
e
x
sin
x
d
x
{\displaystyle \int _{0}^{\pi /2}e^{x}\sin xdx}
sample alternative: :
∫
0
π
/
3
e
x
cos
x
d
x
{\displaystyle \int _{0}^{\pi /3}e^{x}\cos xdx}
http://www.petervis.com/mathematics/integration_by_parts/integrate_e_x_sinx.html
Evaluate
∫
0
1
arccos
x
d
x
{\displaystyle \int _{0}^{1}\arccos xdx}
hint: https://www.youtube.com/watch?v=htTerwAGeLY
hint
This one does
∫
arcsin
x
d
x
{\displaystyle \int \arcsin x\;dx}
u
=
arcsin
x
and
d
v
=
d
x
{\displaystyle u=\arcsin x{\text{ and }}dv=dx}
⇒
d
u
=
1
1
−
x
2
d
x
{\displaystyle \Rightarrow du={\frac {1}{\sqrt {1-x^{2}}}}\;dx}
u
~
=
1
−
x
2
{\displaystyle {\tilde {u}}=1-x^{2}}
I think
∫
arcsin
x
d
x
=
x
arcsin
x
+
1
−
x
2
+
c
{\displaystyle \int \arcsin xdx=x\arcsin x+{\sqrt {1-x^{2}}}+c}
Also, the derivative of the arcsin should be obtained using this trick:
x
=
sin
u
→
d
x
=
cos
u
d
u
=
1
−
sin
2
u
d
u
=
1
−
x
2
d
u
{\displaystyle x=\sin u\rightarrow dx=\cos u\;du={\sqrt {1-\sin ^{2}u}}\;du={\sqrt {1-x^{2}}}\;du}
∫
0
π
/
4
cos
3
x
d
x
{\displaystyle \int _{0}^{\pi /4}\cos ^{3}xdx}
hint- see also p.389 in textbook
cos
3
x
=
(
cos
x
)
3
=
cos
x
(
cos
2
x
)
=
cos
x
(
1
−
sin
2
x
)
{\displaystyle \cos ^{3}x=(\cos x)^{3}=\cos x\left(\cos ^{2}x\right)=\cos x\left(1-\sin ^{2}x\right)}
=
cos
x
−
cos
x
sin
2
x
{\displaystyle =\cos x-\cos x\sin ^{2}x}
Do the second term with the substitution:
u
=
sin
(
x
)
⇒
d
u
=
cos
(
x
)
{\displaystyle u=\sin(x)\Rightarrow du=\cos(x)}
This should lead to: ∫ [cos(x) - cos(x)*sin(x)^2] dx = sin(x) - (1/3)sin(x)^3 + c
Partial Fractions Sec 5.7 p398
edit
Section 5.7 (pp. 389-392): Examples 1, 2, and 4. You need not memorize the half-angle formulas. Improper Integrals Sec 5.10 p413
edit
I think I did examples 2 and 4 pp416-417
Example 3 involves the arctan, which is the integral of 1/(1+x2 ), which I consider a low priority integral to memorize. Good project, if you show why. Time permitting, we will look at a "type 2" case: Example 9 is fun, because it uses the Comparison Theorem. But you need to be certain that you understand this theorem. 
