Calculus II/Test 1

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

Prob.

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

x=\sin t

and

y=\cos ^{2}t

Limits Sec 4.5 p290

Prob.

Evaluate the limit:

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

Prob.

Evaluate the limit: (see page 297 problem 47)

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

solution

Let $y=e^{x}$

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

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

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

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.

Integration by parts Sec 5.6 p283

Prob.

Evaluate

\int _{0}^{\pi /2}e^{x}\sin xdx

sample alternative: :

\int _{0}^{\pi /3}e^{x}\cos xdx

this was hard. we do integration by parts and solve 2 equations in two unknowns.

http://www.petervis.com/mathematics/integration_by_parts/integrate_e_x_sinx.html

Prob.

Evaluate

\int _{0}^{1}\arccos xdx

hint: https://www.youtube.com/watch?v=htTerwAGeLY

hint

This one does $\int \arcsin x\;dx$ :

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

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

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

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

Prob.

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

hint- see also p.389 in textbook

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

Do the second term with the substitution:

$u=\sin(x)\Rightarrow du=\cos(x)$

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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