# Calculus II/Test 1

• 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

### Prob.

Write this parametric equation in the form y=f(x), and sketch the graph:

${\displaystyle x=\sin t}$ and ${\displaystyle y=\cos ^{2}t}$

## Limits Sec 4.5 p290

### Prob.

Evaluate the limit:

${\displaystyle \lim _{x\rightarrow 0}{\frac {1-\cos(2x)}{x^{2}}}}$

### Prob.

Evaluate the limit: (see page 297 problem 47)

${\displaystyle {\mathcal {L}}=\lim _{x\rightarrow \infty }\left(1+{\frac {1}{e^{x}}}\right)^{e^{x}}}$

solution

Let ${\displaystyle y=e^{x}}$

${\displaystyle {\mathcal {L}}=\lim _{y\rightarrow \infty }\left(1+{\frac {1}{y}}\right)^{y}}$

${\displaystyle \ln {\mathcal {L}}=\lim _{y\rightarrow \infty }y\cdot \ln \left(1/y+1\right)=\lim _{y\rightarrow \infty }{\frac {\ln \left(1/y+1\right)}{1/y}}=^{H}\lim _{z\rightarrow 0}{\frac {\ln(1+z)}{z}}=\lim _{z\rightarrow 0}{\frac {\frac {1}{1+z}}{1}}=1}$

This is correct, but we need ${\displaystyle {\mathcal {L}}=e^{\ln({\mathcal {L}})}=e^{1}=e}$

• 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:

${\displaystyle \lim _{\rightarrow 0^{+}}\left(1+\sin 4x\right)^{\cot x}}$ See Example 8 Sec 4.5

## Integration by parts Sec 5.6 p283

### Prob.

Evaluate

${\displaystyle \int _{0}^{\pi /2}e^{x}\sin xdx}$
sample alternative: :${\displaystyle \int _{0}^{\pi /3}e^{x}\cos xdx}$ this was hard. we do integration by parts and solve 2 equations in two unknowns.

### Prob.

Evaluate

${\displaystyle \int _{0}^{1}\arccos xdx}$
hint

This one does ${\displaystyle \int \arcsin x\;dx}$:

${\displaystyle u=\arcsin x{\text{ and }}dv=dx}$

${\displaystyle \Rightarrow du={\frac {1}{\sqrt {1-x^{2}}}}\;dx}$. Now let ${\displaystyle {\tilde {u}}=1-x^{2}}$

I think ${\displaystyle \int \arcsin xdx=x\arcsin x+{\sqrt {1-x^{2}}}+c}$

Also, the derivative of the arcsin should be obtained using this trick: ${\displaystyle x=\sin u\rightarrow dx=\cos u\;du={\sqrt {1-\sin ^{2}u}}\;du={\sqrt {1-x^{2}}}\;du}$

### Prob.

${\displaystyle \int _{0}^{\pi /4}\cos ^{3}xdx}$

${\displaystyle \cos ^{3}x=(\cos x)^{3}=\cos x\left(\cos ^{2}x\right)=\cos x\left(1-\sin ^{2}x\right)}$ ${\displaystyle =\cos x-\cos x\sin ^{2}x}$

Do the second term with the substitution:

${\displaystyle u=\sin(x)\Rightarrow du=\cos(x)}$

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

### Probs.

• 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

### Probs.

• 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.

### Probs

• 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.
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