Integration Full Chapter Explained - Integration Class 12 - Everything you need



Last updated at Dec. 20, 2019 by Teachoo
Transcript
Ex 7.9, 20 β«_0^1β(π₯ π^π₯+sinβ‘γππ₯/4γ ) ππ₯ Let F(π₯)=β«1β(π₯π^π₯+π ππ ππ₯/4)ππ₯ =β«1βγπ₯π^π₯ ππ₯+β«1βγsinβ‘(ππ₯/4) ππ₯γγ Solving I1 and I2 separately Solving π°π β«1βγπ₯π^π₯ ππ₯γ =π₯β«1βγπ^π₯ ππ₯ββ«1β[(ππ₯/ππ₯) β«1βγπ^π₯ ππ₯γ]ππ₯γ =π₯π^π₯ββ«1β(1.π^π₯ ππ₯)ππ₯ =π₯π^π₯ββ«1βγπ^π₯ ππ₯γ =π₯π^π₯βπ^π₯ =π^π₯ (π₯β1) Solving by parts β«1βγπ(π₯)π(π₯)ππ₯=π(π₯)β«1βγπ(π₯)ππ₯ββ«1β(π^β² (π₯) β«1βπ(π₯)ππ₯)ππ₯γγ Let π(π₯)=π₯ and π(π₯)=π^π₯ Solving I2 β«1βγsinβ‘(ππ₯/4) ππ₯γ = 1/(π/4) (βcosβ‘(ππ₯/4) ) = (β4)/π cosβ‘(ππ₯/4) Therefore, F(π₯)=β«1βγπ₯π^π₯ ππ₯+β«1βγπ ππ π/4 π₯ ππ₯γγ =π^π₯ (π₯β1)β4/π cosβ‘(ππ₯/4) Now, β«_0^1β(π₯π^π₯+π ππ ππ₯/4) ππ₯=πΉ(1)βπΉ(0) =(π^1 (1β1)β4/π cosβ‘((π Γ 1)/4) )β(π^0 (0β1)+4/π πππ ((π Γ 0)/4)) =πΓ0β4/π πππ π/4β1(β1)+4/π cosβ‘0 =(β4)/( π) πππ π/4+1+4/π =(β4)/( π) 1/β2+1+4/π =(β2β2)/( π) +1+4/π =π+π/π β(πβπ)/π
Ex 7.9
Ex 7.9, 2
Ex 7.9, 3
Ex 7.9, 4
Ex 7.9, 5
Ex 7.9, 6
Ex 7.9, 7
Ex 7.9, 8 Important
Ex 7.9, 9
Ex 7.9, 10
Ex 7.9, 11
Ex 7.9, 12
Ex 7.9, 13
Ex 7.9, 14
Ex 7.9, 15
Ex 7.9, 16 Important
Ex 7.9, 17 Important
Ex 7.9, 18
Ex 7.9, 19 Important
Ex 7.9, 20 Important You are here
Ex 7.9, 21 Important
Ex 7.9, 22
About the Author