MATH MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS Calculus, Fall 2017 Professor: Jared Speck. Problem 1. Approximate the integral

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1 MATH 8. - MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS 8. Calculus, Fall 7 Professor: Jared Speck Problem. Approimate the integral 4 d using first Simpson s rule with two equal intervals and then the trapezoid rule with two equal intervals. Compare our approimations to the eact value. Problem. Evaluate the integral e sin d. Problem 3. Evaluate the integral d ( ) 5/. Problem 4. Let F () that F () G() sin Problem 5. Evaluate the integral cos d and G() + C. What is the value of C? sin d (ln ) 4. Problem 6. Consider the following curve in parametric form: t + ln t, t ln t. Find the arc length of the portion of the curve corresponding to t. d. Show that there is a constant C such

2 MATH 8. - MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS Problem. Approimate the integral Solutions 4 d using first Simpson s rule with two equal intervals and then the trapezoid rule with two equal intervals. Compare our approimations to the eact value. Solution: We set f() 4 and. Then the Simpson approimation is 4 d 3 [f() + 4f() + f()] ( ) /3. 3 The trapezoid approimation is [ 4 d f() + f() + ] f() ( + + 8) 9. The eact value of the integral is 4 d 5 5 3/5. Problem. Evaluate the integral e sin d. Solution: We integrate b parts with u e, du e, dv sin d, v cos, which leads to e sin d u dv uv v du e cos + e cos d. Similarl, we compute that e cos d e sin e sin d. Inserting this identit for e cos d back into the previous formula, we deduce that e sin d e cos + e cos d e cos + e sin e sin d. We then add e sin d to both sides of the previous equation, divide b two, and add the integration constant C, which leads to e sin d {e sin e cos } + C.

3 MATH 8. - MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS 3 Problem 3. Evaluate the integral d ( ) 5/. Solution: We use the inverse trig substitution sin u, d cos u du, the identit (cos u) (sin u), and the identit (sec u) + (tan u), which leads to cos u (cos u) du (sec u) 4 du 5 [ + (tan u) ](sec u) du. We then make the substitution v tan u, dv (sec u) du, which leads to [ + (tan u) ](sec u) du + v dv v + v3 3 + C tan u + tan3 u + C 3 ( ) C. / 3 ( ) 3/ In the last step above, we used a right triangle to deduce that tan u. ( ) / Problem 4. Let F () cos d and G() sin d. Show that there is a constant C such that F () G() sin + C. What is the value of C? Solution: We use integration b parts in the form u, du d, dv cos d, v sin, u dv uv v du to deduce that Hence, and C sin. cos d } {{ } F () sin + sin sin sin + F () G() sin d sin d. } {{ } G() sin, Problem 5. Evaluate the integral d (ln ) 4.

4 4 MATH 8. - MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS Solution: We first make the substitution u ln, du d, which leads to the integral du (u 4). / We then make the inverse trigonometric substitution u sec θ, du sec θ tan θ dθ and use the identit (sec θ) (tan θ), which leads to du sec θ tan θ (u 4) dθ / tan θ sec θ dθ ln sec θ + tan θ + C, u u ln C ln (ln ) ln C. In the net-to-last line above, we used a right triangle to deduce that tan θ u 4 u. 4 Problem 6. Consider the following curve in parametric form: t + ln t, t ln t. Find the arc length of the portion of the curve corresponding to t. Solution: We first compute that d + t, d t, (d ds The arc length is therefore Arc Length ) + + t t +. t ( ) d ds t +. t

5 MATH 8. - MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS 5 To evaluate the integral, we first make the inverse trig substitution t tan u, sec u, use tan u + sec u, multipl the top and bottom b sin u, and use sin cos u, which leads to t + t sec u tan u sec u du sin u cos u du sin u sin u cos u du sin u ( cos u) cos u du. We then make the substitution v cos u, dv sin u du, which leads to sin u ( cos u) cos u du ( v )v dv v ( v)( + v) dv. To evaluate the v integral, we first make a partial fraction decomposition: v ( v)( + v) A v + B v + C v + D + v. The cover up method ields B, C, D. Then plugging the value v into both sides of the decomposition ields that A. Hence, v ( v)( + v) v + ( ) + ( ), v + v and v ( v)( + v) dv v + ln v + v + c cos u + ln cos u + cos u + c + t + ln ( + t ) / + ( + t ) / + c + t + + ln t + t + + c. In our calculations above, we used a right triangle to deduce that cos u ( + t ) /.

6 6 MATH 8. - MIDTERM 4 - SOME REVIEW PROBLEMS WITH SOLUTIONS Finall, we reinsert the factor of and the bounds of integration to conclude that Arc Length t + t { + t + } ln + t t + t + t { 5 + ln 5 } 5 + ln + { 5 + ln + } 5 + ln. 5