Question

2. 10 23 x · [In(x)]2 Jg x+2 In Problems 1-26, evaluate each improper integral, or show why it diverges. po 1 5 1. dx dx 2 3. dx 4. dx 13 1 + x2 5 X 5. dx 6. dx x. In(x) Jo 1 + x2 1 7 dx 8. dx -2 J3 (x - 2)2 1 1 9. dx 10. dx (x - 2) 1 11. dx 12. dx J3 (x+2)3 14 1 1 13. dx 14. dx Vx so fo 1 00 1 (x + 2)2 JIS T24 so 16 1 1 15. dx 16. dx To √x √2-x lot lot 17 1 dx 14- x² 18. 3x2 dx V8-73

Answer 1

solution > 4 I dx TX Here I 9=0, b=4 Since 4 f(x) is infinite at lower limit, da is an improper integral of So va Type - and In type-and, If f(x) is unbounded at lower limit then, so f(x) dx lim s f(x) da - E-yo ate 4 0 00 in de Z line Tv dx E-yo ore lim [282 4 & to E va) live [254-25€ = {ro 2x 2 - 2x0 4 Therefore, Given Integral de tot de da is convergent. Answer

6 9 2 In this question, we will check that - u-3n+ 5 is a solution or not of x-y" - y = 3 equation y = 423475 differentiation with respect to a y = 2x-3 Again differentiation with respect to a 3 g" 다. 2 L.H.S. of equation s. y - y = r. (2) - (2x - 3) [by usind a = 22-(2x-3) = 22-2x+ 3 = 3 x.yu - y = 3 which is equation @ Therefore equation is solution of given differential equation (eq. ). a Answer