1. Class 12
2. Important Question for exams Class 12

Transcript

Ex7.9, 20 β«_0^1β(π₯ π^π₯+sinβ‘γππ₯/4γ ) ππ₯ Step 1 :- Let F(π₯)=β«1β(π₯π^π₯+π ππ π/4 π₯)ππ₯ =β«1βγπ₯π^π₯ ππ₯+β«1βγπ ππ π/4 π₯ ππ₯γγ Solving π°π β«1βγπ₯π^π₯ ππ₯γ Integration by parts β«1βγπ₯π^π₯ ππ₯=π₯β«1βγπ^π₯ ππ₯ββ«1β[(ππ₯/ππ₯) β«1βγπ^π₯ ππ₯γ]ππ₯γγ =π₯π^π₯ββ«1β(+1.π^π₯ ππ₯)ππ₯ =π₯π^π₯ββ«1βγπ^π₯ ππ₯γ =π₯π^π₯βπ^π₯ =π^π₯ (π₯β1) Therefore, F(π₯)=β«1βγπ₯π^π₯ ππ₯+β«1βγπ ππ π/4 π₯ ππ₯γγ =π^π₯ (π₯β1)+1/(π/4) (βπππ  ππ₯/4) =π^π₯ (π₯β1)β4/π πππ  ππ₯/4 Hence, F(π₯)=π^π₯ (π₯β1)β4/π πππ  ππ₯/4 Step 2 :- β«_0^1β(π₯π^π₯+π ππ ππ₯/4) ππ₯=πΉ(1)βπΉ(0) =π^1 (1β1)β4/π πππ  ((πΓ1))/4βπ^0 (0β1)+4/π πππ  ((πΓ0))/4 =πΓ0β4/π πππ  ((π))/4β1(β1)+4/π cosβ‘0 =(β4)/( π) πππ  ((π))/4+1+4/π =(β4)/( π) 1/β2+1+4/π =π+π/πβ(πβπ)/π

Class 12
Important Question for exams Class 12