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.

Prob.
Write this parametric equation in the form y=f(x), and sketch the graph:
 * $$x = \sin t$$ and $$y= \cos^2 t$$

Prob.
Evaluate the limit:
 * $$\lim_{x\rightarrow 0}

\frac {1-\cos(2x)} {x^2}$$

Prob.
Evaluate the limit: (see page 297 problem 47)

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

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.

Similar problem:

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

Prob.
Evaluate
 * $$\int_0^{\pi/2} e^x \sin x dx$$


 * sample alternative: :$$\int_0^{\pi/3} e^x \cos x dx$$ 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
 * $$\int_0^1 \arccos x dx$$


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

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 $$\int \arcsin x dx = x\arcsin x+ \sqrt{1-x^2} + c$$

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

Prob.

 * $$\int_0^{\pi /4} \cos^3 x dx$$

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

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

Probs.

 * Section 5.7 (pp. 389-392): Examples 1, 2, and 4. You need not memorize the half-angle formulas.

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.