Prove-e-1-e-t-2-e-t-2-dt-1-1-e-2-e-the-base-of-natural-logarithm- Tinku Tara April 13, 2025 Integration 0 Comments FacebookTweetPin Question Number 218624 by Nicholas666 last updated on 13/Apr/25 Prove;∫1eet2et2dt⩽1−1e2e−thebaseofnaturallogarithm Answered by Nicholas666 last updated on 13/Apr/25 ∫1/eet2et2dt⩽1−1e2⇒f′t=ddt(t2e−t2)=2te−t2+t2(−2te−t2)=2te−t2(1−t2)⇒f′(0.5)=2(0.5)e−(0.5)2=e−0.25(1−0.25)=0.75e−25>0⇒f′(2)=2(2)e−22(1−22)=4e−4(1−4)=−12e−4<0⇒f(1)=12e12=1e⇒∫1/eet2et2dt⩽∫1/ee1edt⇒∫1/ee1edt=1e∫1/ee1dt=1e[t]1/ee=1e(e−1e)=e2−1e2=1−1e2⇒∫1/eet2et2dt⩽1−1e2Q.E.D Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Prove-that-for-all-real-numbers-a-and-b-with-a-lt-b-the-following-inequality-holds-a-b-1-dx-3-b-a-a-b-x-a-1-2-dx-a-b-1-a-x-1-3-dx-Next Next post: Question-218629 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![∫_(1/e) ^e (t^2 /e^t^2 ) dt ≤1−(1/e^2 ) ⇒f′t=(d/dt)(t^2 e^(−t^2 ) )=2te^(−t^2 ) +t^2 (−2te^(−t^2 ) )= 2te^(−t^2 ) (1−t^2 ) ⇒f′(0.5)=2(0.5)e^(−(0.5)^2 ) = e^(−0.25) (1−0.25)=0.75e^(−25) >0 ⇒f′(2)=2(2)e^(−2^2 ) (1−2^2 )= 4e^(−4) (1−4)=−12e^(−4) <0 ⇒f(1)=(1^2 /e^(12) )=(1/e) ⇒∫_(1/e) ^e (t^2 /e^t^(2 ) )dt≤∫_(1/e) ^e (1/e)dt ⇒∫_(1/e) ^e (1/e)dt=(1/e)∫_(1/e) ^e 1 dt=(1/e)[t]_(1/e) ^e = (1/e)(e−(1/e))=(e^(2−1) /e^2 )=1−(1/e^2 ) ⇒∫_(1/e) ^e (t^2 /e^t^2 )dt ≤1−(1/e^2 ) Q.E.D](https://www.tinkutara.com/question/Q218625.png)