Class 13 - Review (10/3/2017, Tuesday)

2nd Gateway

You must pass the second Gateway by Oct 13, Friday. Otherwise, your grade will be damaged a lot!

Webwork

• 8.5 due 10/4 Wed
• 4.8 due 10/7 Sat

Quiz 3

Tomorrow

Team HW 3 Hints

Team HW 3 will appear on Quiz 3. Here are some guidelines to initiate solving them.

1a. Don't look at the erroneous solution, but just solve it on your own. Take $t = x^{2}$ so that you get $dt = 2xdx$ which gives $xdx = (1/2)dt.$ Make sure you change bounds in $x$ to bounds in $t.$ Then look at the erroneous solution to figure out the nonsense.

1b. Don't look at the erroneous solution, but just solve it on your own. Take $u' = x$ and $v = \ln(x).$ Then $u = x^{2}/2$ and $v' = 1/x,$ so use integration by parts. Then look at the erroneous solution to figure out the nonsense.

1c. Don't look at the erroneous solution, but just solve it on your own. Take $t = (x - 1)/2$ so that $dt = (1/2)dx$ and thus $dx = 2dt.$ Change the bounds carefully. Then look at the erroneous solution to figure out the nonsense.

1d. Don't look at the erroneous solution, but just solve it on your own. The antiderivative must be

\[Q(x) = 3 + \int_{1}^{x}q(t)dt.\]

A quick way to check whether the above is correct is to check $Q(1) = 3.$ Now, compute the rest and look at the erroneous solution to figure out the nonsense.

2a. The "line segment" here means the part of the line on the $xy$-plane, which has slope $-2.5$ passing through $(1, 0).$

2b - i. You can do this on your own. Just read the problem carefully.

2b - ii. I apologize that the wording of the problem is extremely confusing. Assume that the shortest side of such a triangle has length $x$. Now, compute the lengths of the other two sides of the triangle in terms of $x.$

2b - iii. Use Part ii.

2c. Find the volume of each slice. Thickness will be $dy$ and the area will be given by 2b - iii. Multiply them and write everything in $y.$ Look at where $y$ varies. Units will be observed when you compute the volume of a slice.

3a. Recall how midpoint and trapezoid approximations are related to concave down functions. Draw a picture to remember what the relationship was.

3b. Remember, this means $8$ subintervals and taking heights of the rectangles in the middle.

3c. Don't listen to the suggestion in the document. Compute the left-hand side and the right-hand side separately and see that the two computations match. This is much faster than simplification.

3d. Recall the arc length formula. Indeed, 3c will be helpful.

3e. Use your intuition. This is what people call isometry. A rigorous explanation should involve the fact the such a transformation (only involving reflection and shifting) does not affect the integral. This problem is too theoretically phrased, but it does contain some good lessons.

4a. Have fun!

4b. Ask me this in person if you get stuck.

4c. I think it means circumference of the region $\mathcal{R},$ but don't count on me.