University:
San Diego State University
Course:
MATH 151 | Calculus II
Academic year:

2023
Views:

405

Pages:

1
Author:

skarvurdld6

Workshop #11 √ 1. A right triangle has a side of length 4 − x2 and a side of length x, neither of which are the hypotenuse. What is the length of the other side? Draw a labeled right triangle. √ 2. A right triangle has a side of length 4 + x2 and a side of length 2, one of which is the hypotenuse. What is the length of the other side? Draw a labeled right triangle. √ 3. A right triangle has a side of length x2 − 4 and a side of length 2, neither of which is the hypotenuse. What is the length of the other side? Draw a labeled right triangle. 4. Consider the integral below. Z √ 1 dx 4 − x2 (a) Look at your triangle from Problem 1. There are two non-right angles, for which of the angles will sin u = x2 ? Label that angle u. (b) If sin u = x2 , then x = 2 sin u. What does dx equal? (c) Rewrite the above integral in terms of u. Z 1 √ dx. (d) Use the above substitution to compute 4 − x2 5. In the above problem we used the fact that sin u = Z (a) If the integral was x 2 Z to help us figure out √ 1 dx. 4 − x2 1 √ dx instead, we would use the triangle in problem 2. 4 + x2 In your triangle for Problem 2 label the angle u for which tan u = x2 . (b) Make this substitution to get the integral in terms of u. Why is this substitution useful? Z (c) If the integral was 1 √ dx instead, we would use the triangle in problem 3. 2 x −4 In your triangle for Problem 3 label the angle u for which sec u = x2 . (d) Make this substitution to get the integral in terms of u. Why is this substitution useful? Compute the following integrals. Z x3 √ 6. dx x2 + 1 Z √ 7. x2 − 1 dx x3 Z 8. ex √ 1 dx 1 − e2x