MATH 101: PRACTICE MIDTERM 2

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Gloria Dawson
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1 MATH : PRACTICE MIDTERM INSTRUCTOR: PROF. DRAGOS GHIOCA March 7, Duration of examination: 7 minutes This examination includes pages and 6 questions. You are responsible for ensuring that your copy of the paper is complete. Bring any discrepancy to the attention of your invigilator. Special instructions: Use the exam sheet to write all your answers. For each problem write your solution on the page that contains the text of the problem. If you need more space to write your solution to any of the problems, use the two blank pages from the end of your exam. In that case, indicate exactly what problem you are solving. No calculator is allowed. You are not allowed to use any notes or other assistance (electronic, etc.). In case you attempt to use a calculator, or any notes, or other assistance, your exam will be taken and you will receive points. For rough work you can use the two extra blank pages at the end of your exam. Make sure that you indicate what is your rough work and what is your actual solution to the problem because otherwise I will grade everything that you wrote and in case that you have mistakes in your rough work, then you will lose points for them. So, indicate exactly what you want to be graded. Name: Student ID: Mark:

2 You can use this page for rough work

3 Problem. (6 points) Compute arctan(6x) dx Solution. We use integration by parts arctan(6x) dx arctan(6x) dx 6 arctan(6x) x + 36x x dx 6x x arctan(6x) 36x + dx. For the last integral we use the substitution So, arctan(6x) dx u 36x + and thus du 7x dx. x arctan(6x) x arctan(6x) 6x 36x + dx u du x arctan(6x) ln u + C x arctan(6x) ln(36x + ) + C. 3

4 Problem. (8 points) Compute 3 x x dx Solution. We first complete the square inside the square-root and obtain 3 x x 4 (x + x + ) 4 (x + ). This suggests the following substitution x + sin(t) and so, dx cos(t) dt. We compute 3 x x dx 4 4 sin (t) cos(t) dt cos(t) cos(t) dt 4 cos (t) dt. For the last integral we use first the trigonometric identity and so, 4 cos (t) dt cos (t) cos(t) +, cos(t) + dt sin(t) + t + C, where for the last integral, if you wish, you could use the very simple substitution u t and so, du dt. Now, in order to express our answer back into the variable x, we note that ( ) x + sin(t) sin(t) cos(t) (x + ) (x + ) 3 x x. So, 3 x x dx (x + ) ( ) 3 x x x + + arcsin + C. 4

5 Problem 3. ( points) Compute sec 3 (x) dx Solution. We rewrite the integral as sec 3 (x) dx cos 3 (x) dx cos(x) (cos (x)) dx We make the substitution u sin(x) which gives du cos(x) dx. So, we obtain sec 3 (x) dx We look to find the partial fractions for the rational map (u ) (u + ) A u + du ( u ) (u ) du (u ) (u + ) du. B (u ) + C u + + We bring everything to a common denominator and therefore obtain cos(x) dx ( sin (x)). D (u + ). A (u )(u + ) + B (u + ) + C (u ) (u + ) + D (u ). We plug in u and then u in the above identity of polynomials and obtain: B 4; therefore B 4 and D 4; therefore D 4. We substitute B D 4 in the above identity of polynomials and obtain: A (u )(u + ) + B (u + ) + C (u ) (u + ) + D (u ) A (u )(u + ) + 4 (u + ) + C (u ) (u + ) + (u ) 4 A(u )(u + u + ) + 4 (u + u + ) + C (u u + )(u + ) + 4 (u u + ) Au 3 + Au Au A + u 4 + u Cu3 Cu Cu + C + u 4 u + ( 4 (A + C)u 3 + A C + ) u (A + C)u A + C +. Equating each term from each side of the above identity of polynomials, i.e. noting that the coefficients of u 3, u, u and the constant term must be the same in both the left and the right hand side above, we obtain The first and the third equations yield A + C A C + A C A + C +. A C, 5

6 while the second and the fourth equations yield Overall, we obtain that C C, and so, A C. C 4 and thus, A 4. We conclude that Therefore sec 3 (x) dx (u ) (u + ) 4 u + 4 (u ) + 4 u (u + ). ( 4 ( 4 ( 4 ( 4 ln du u + du (u ) + du u + + ) du (u + ) ) + C ln u + ln u + u u + ln sin(x) sin(x) + ln sin(x) + sin(x) + sin(x) + sin(x) sin(x) ) sin + C (x) 4 ln (sin(x) + ) sin (x) 4 sin(x) cos (x) + C 4 ln (sin(x) + ) cos (x) + sin(x) cos (x) + C 4 ln ( sin(x) + cos(x) ) + tan(x) sec(x) + C ln tan(x) + sec(x) + tan(x) sec(x) + C. ) + C 6

7 Problem 4. (6 points) Determine whether the following integral is convergent or divergent x x e x dx Solution. First we note that the function e x is greater than or equal to on the entire interval [, ). This allows us to apply the comparative test for integrals. The exponential function is much larger than any polynomial; this suggests that our integral is convergent. We look for a larger function to compare our original function; since e x for x, we observe that e x e x and thus x e x x e x xe x. Now, noting that e x dx e x + C and integrating by parts, we obtain x e x dx x ( e x ) e x dx xe x + e x dx Therefore x e x dx xe x e x + C. lim where in the above limit we used the fact that lim t t t xe x dx xe x dx lim t ( xe x e x) t lim t t e t e t, t e. Therefore, the integral is indeed convergent. t 7

8 Problem 5. (6 points) Find the centroid of the region bounded by the curves y cos(x), y, x, and x π. Solution. The area of the region is So the coordinates of the centroid are: x A π A π x cos(x) dx x sin(x) π π + cos(x) π π cos(x) dx sin(x) π. π sin(x) dx using integration by parts and ȳ A π π 8. π cos (x) dx + cos(x) dx 4 ) ( x 4 + sin(x) 8 π 8

9 Problem 6. (4 points) Solve the differential equation with initial condition y(). y x 3 y Solution. We divide by y both sides and then integrate; we get and then and so, We exponentiate both sides and get y y x3 dy y x 3 dx ln y x4 4 + C. y(x) K e x4 4, where K : e C is a positive constant. We solve for K by using that y(), and therefore K e and so, K. We obtain that y(x) e x4 4. Note that we are allowed to drop the absolute values in the above formula for y(x) since the right-hand side is guaranteed to be positive due to the exponential function. 9

10 You may use this page for rough work.

11 You may use this page for rough work

Math 106: Review for Exam II - SOLUTIONS

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