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:

$\underset{x\to 0}{\lim }\frac{1-\cos (2x)}{x^2}$

Prob.

Evaluate the limit: (see page 297 problem 47)

$ℒ=\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{e^x})}^{e^x}$

Template:Cot Let $y=e^x$

$ℒ=\underset{y\to \mathrm{\infty }}{\lim }{(1+\frac{1}{y})}^y$

$\ln ℒ=\underset{y\to \mathrm{\infty }}{\lim }y\cdot \ln (1/y+1)=\underset{y\to \mathrm{\infty }}{\lim }\frac{\ln (1/y+1)}{1/y}{=}^H\underset{z\to 0}{\lim }\frac{\ln (1+z)}{z}=\underset{z\to 0}{\lim }\frac{\frac{1}{1+z}}{1}=1$

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

• 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:

$\underset{\to 0^{+}}{\lim }{(1+\sin 4x)}^{\cot x}$ See Example 8 Sec 4.5

Integration by parts Sec 5.6 p283

Prob.

Evaluate

$\underset{0}{\overset{\pi /2}{\int }}{e}^x\sin xdx$

sample alternative: :$\underset{0}{\overset{\pi /3}{\int }}{e}^x\cos xdx$ 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

$\underset{0}{\overset{1}{\int }}\arccos xdx$

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

Template:Cot This one does $\int \arcsin xdx$:

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

$\Rightarrow du=\frac{1}{\sqrt{1-x^2}}dx$. 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: $x=\sin u\to dx=\cos udu=\sqrt{1-{\sin }^{2}u}du=\sqrt{1-x^2}du$ Template:Cob

Prob.

$\underset{0}{\overset{\pi /4}{\int }}{\cos }^{3}xdx$

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

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 Template:Cob

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.