Question

9.3 Integral Test & Seric Use the Integral Test to determine the convergence or divergence of the series. 2 3n + 6 n = 1 Part 1 of 5 Recall the Integral Test. Iff is positive positive, continuous, and decreasing decreasing for x 2 1 and an = f(n), then an and f(x) dx either both converge or both diverge. n=1 Part 2 of 5 Let f(x) 2 3x + 6 Note that f(x) is positive, continuous, and decreasing for x 2 1. So, the Integral Test can can be applied to the given series 2 3 + 6 Part 3 of 5 2 3n + 6 n=1 2 From the Integral Test, Σ and dx either both converge or both diverge. So, evaluate 3x + 6 dx to determine whether the series 2 2 converges or diverges. First, find the antiderivative. 3x + 6 3n + 6 dx = 3x + 6 + C Submit Skip (you cannot come back)

Answer 1

oy Problem to use the the Integral tut to determine the Convergence divergence of 2 3h+6 the series 2 -ha- Solution Integral test lit fous be positive Continuos monoto notonically in lla) then the siris an and so fou dze Ceilisgent or digent, together decreasing n21 2 • 2_du hal So by integral tist ohnen and I= 3h+t 30th both Convergent or divergent as it integral I is finite I is finit then I is Convergent.. it I is infinite then I is divergent so we have to evaluated dn. And 2 Tinto S 2 first find 2 dna da 3n+69 304+2) substitute t=x+2 dt = da 2 so ( dt > lagt 3. t 3 Put the value of t henu 2 -dr = teg (x+2)te 3 3 +38767

Now take limet GO 2 S 2 (leg (242) ] din = Sentos C 3. T 3 [oo by z hence (36) 2 da = ∞ Integral so sirius is divergent sh 2 is also 3 nito het ditergest
first of all I filled your blank space by evaluating given integral then by integral test sereis is convergent or divergent according to integral convergent or divergent since value of integral is infinite hence integral is divergent so series is also divergent this is the full explanation as I mention everything in my hand written solution