Ex 7.9, 4
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex7.9, 4 β«_0^(π₯/4)βsinβ‘2π₯ππ₯ Step 1 :- F(π₯)=β«1βγπ ππ 2π₯ ππ₯γ =1/2 (βπππ 2π₯) =(β1)/( 2) πππ 2π₯ Hence F(π₯)=(β1)/( 2) πππ 2π₯ Step 2 :- β«_0^(π/4)βγπ ππ2π₯ ππ₯γ=πΉ(π/4)βπΉ(0) =(β1)/( 2) πππ (2 Γ π/4)β((β1)/( 2) πππ (2 Γ0)) =(β1)/( 2) πππ π/2+1/2 πππ 0 =(β1)/( 2) Γ0+1/2 Γ1 = π/π
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