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Question Number 122885 by CanovasCamiseros last updated on 20/Nov/20

Commented by CanovasCamiseros last updated on 20/Nov/20

Please help

Pleasehelp

Answered by ebi last updated on 20/Nov/20

I=∫_0 ^2 e^(1−(x/2)) dx let u=1−(x/2) (du/dx)=−(1/2) ⇒dx=−2 du I=∫_0 ^2 −2 e^u du I=−2∫_0 ^2 e^u du I=−2e^u ∣_0 ^2 I=−2e^(1−(x/2)) ∣_0 ^2 I=−2(e^0 −e^1 ) I=2e−2

I=∫20e1−x2dxletu=1−x2dudx=−12⇒dx=−2duI=∫20−2euduI=−2∫20euduI=−2eu∣02I=−2e1−x2∣02I=−2(e0−e1)I=2e−2

Answered by MJS_new last updated on 20/Nov/20

∫e^(1−(x/2)) dx=e∫e^(−(x/2)) dx=e×(−2)e^(−(x/2)) =−2e^(1−(x/2)) +C ⇒ answer is 2e−2

∫e1−x2dx=e∫e−x2dx=e×(−2)e−x2=−2e1−x2+C⇒answeris2e−2

Answered by Bird last updated on 20/Nov/20

I=∫_0 ^2 e^(1−(x/2)) dx ⇒I=_((x/2)=t) ∫_0 ^1 e^(1−t) 2dt =2e ∫_0 ^(1 ) e^(−t) dt =2e[−e^(−t) ]_0 ^1 =2e{1−e^(−1) } =2e−2

I=∫02e1−x2dx⇒I=x2=t∫01e1−t2dt=2e∫01e−tdt=2e[−e−t]01=2e{1−e−1}=2e−2

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