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Question Number 59659 by aliesam last updated on 13/May/19

Commented by Mr X pcx last updated on 13/May/19

yes sir.

yessir.

Commented by Mr X pcx last updated on 13/May/19

I =∫_0 ^2 (1−2it −t^2 )dt =2−2i∫_0 ^2 tdt−∫_0 ^2 t^2 dt =2−2i[(t^2 /2)]_0 ^2 −[(t^3 /3)]_0 ^2 =2+2i(2)−(8/3) =−(2/3) +4i .

I=∫02(1−2it−t2)dt=2−2i∫02tdt−∫02t2dt=2−2i[t22]02−[t33]02=2+2i(2)−83=−23+4i.

Commented by aliesam last updated on 13/May/19

thank you sir great solution

thankyousirgreatsolution

Commented by MJS last updated on 13/May/19

(1−it)^2 =1−2it−t^2

(1−it)2=1−2it−t2

Commented by aliesam last updated on 13/May/19

Answered by meme last updated on 13/May/19

=∫_0 ^2 (1−2it−t^2 )dt(because (a−b)^2 =a^2 −2ab+b^2 or a=1 and b=it) =∫_0 ^2 dt−∫_0 ^2 (2it)dt−∫_0 ^2 t^2 dt =[t]_0 ^2 −[it^2 ]_0 ^2 −3[(1/3)t^3 ]_0 ^2 =2−4i−8 =−7−4i

=∫02(1−2it−t2)dt(because(a−b)2=a2−2ab+b2ora=1andb=it)=∫02dt−∫02(2it)dt−∫02t2dt=[t]02−[it2]02−3[13t3]02=2−4i−8=−7−4i

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