IntegrationInduction (1 Viewer)

[kami]

Hi, I know I probably should know how to do this but unfortunately I don't really understand the question - would one of you be so helpful enough as to explain to me whatever I've missed?

(i)Explain why, for α>0,
∫(0 to 1) x^α. e^x.dx < 3/(α + 1)

(You may assume that e<3)

(ii) Show, via induction, that for n= 0, 1, 2,....there exist integers a_n and b _n such that
∫(0 to 1)xⁿ. e^x.dx = a_n + b_n.

Thank you to anyone who can help.

 

[KFunk]

Hi, I know I probably should know how to do this but unfortunately I don't really understand the question - would one of you be so helpful enough as to explain to me whatever I've missed?

(i)Explain why, for α>0,
∫(0 to 1) x^α. e^x.dx < 3/(α + 1)

(You may assume that e<3)

For that bit you know that 1 &le; e^x &le; e < 3 (for 0 < x < 1) hence:

e^x < 3

x^&alpha;e^x &le; 3x^&alpha; (since x &ge; 0)

&there4; &int; (0 to 1) x^&alpha;e^x dx < 3&int; (0 to 1) x^&alpha; dx = 3/(&alpha; + 1)

&there4; &int; (0 to 1) x^&alpha;e^x dx < 3/(&alpha; + 1)

 

[KFunk]

(ii) Show, via induction, that for n= 0, 1, 2,....there exist integers a_n and b _n such that
∫(0 to 1)xⁿ. e^x.dx = a_n + b_n.

Skipping over the n=1 case (and letting S_n represent the proposition), assume true for n= k so:

&int; (0 to 1) x^ke^x dx = a_k + b_ke ...(1)

... then when n= k+1

&int; (0 to 1) x^k+1e^x dx

= [x^k+1e^x](0 to 1) - (k+1)&int; (0 to 1) x^ke^x dx ...(using int. by parts)

= [e - 0] - (k+1)&int; (0 to 1) x^ke^x dx

= e - (k+1)(a_k + b_ke) ... using (1)

= -(k+1)a_k + (1 - (k+1)b_k)e

= a_k+1 + b_k+1e (since all the terms above [except e] are integers]

hence S_k ==> S_k+1 etc...

 

[kami]

I think I get it now, thanks KFunk!

 

[KFunk]

No probs 10 chars