th-webwork2.unl.edu k2/106-Fall-2018/Hw.14-7.1a/6/?effectiveUser 04978436&user-04978436&key K2A8EyleZWsAEc6agojAqbE /we HW 14-7.1a: Problem 6 Previous Problem List Noxt (1 point) Consider the integral 52(1)dx. In the following, we will evaluate the integral using two methods. A. First, rewrite the integral by multiplying out the integrand: 5x2( 1)dx t dx Then evaluate the resulting integral term-by-term: 5x ( +1)dr B. Next, rewrite the integral using the substitution wx31: 5x (r1)d- dw Evaluate this integral (and back-substitute for w) to find the value of the original integral: С. How are your expressions from parts A and B different? What is the difference between the two? ignore the constant of integration. (answer from B)-(answer from A) Are both of the answers correct? (Be sure you can explain why they are)
Show transcribed image textth-webwork2.unl.edu k2/106-Fall-2018/Hw.14-7.1a/6/?effectiveUser 04978436&user-04978436&key K2A8EyleZWsAEc6agojAqbE /we HW 14-7.1a: Problem 6 Previous Problem List Noxt (1 point) Consider the integral 52(1)dx. In the following, we will evaluate the integral using two methods. A. First, rewrite the integral by multiplying out the integrand: 5x2( 1)dx t dx Then evaluate the resulting integral term-by-term: 5x ( +1)dr B. Next, rewrite the integral using the substitution wx31: 5x (r1)d- dw Evaluate this integral (and back-substitute for w) to find the value of the original integral: С. How are your expressions from parts A and B different? What is the difference between the two? ignore the constant of integration. (answer from B)-(answer from A) Are both of the answers correct? (Be sure you can explain why they are)