Integration by parts

Chapter 7 Class 12 Integrals
Concept wise

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### Transcript

Ex 7.6, 3 Integrate the function π₯^2 ππ₯ β«1βγπ₯^2 π^π₯ ππ₯γ = π₯^2 β«1βγππ₯ ππ₯γββ«1β(π(π₯^2 )/ππ₯ β«1βγππ₯ ππ₯γ) ππ₯ = π₯^2. ππ₯ ββ«1βγ2π₯ . ππ₯γ ππ₯ = π₯^2. ππ₯ β2β«1βγπ . ππγ ππ Now we know that β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(πβ²(π₯)β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = x2 and g(x) = ex β¦(1) Solving I1 β«1βγπ₯ π^π₯ ππ₯γ = π₯β«1βππ₯ ππ₯ββ«1β(ππ₯/ππ₯ β«1βπ^π₯ ππ₯) ππ₯ = π₯ππ₯ ββ«1βππ₯ ππ₯ = π₯ππ₯ βππ₯ Now we know that β«1βγπ(π₯) πβ‘(π₯) γ ππ₯=π(π₯) β«1βπ(π₯) ππ₯ββ«1β(πβ²(π₯)β«1βπ(π₯) ππ₯) ππ₯ Putting f(x) = x and g(x) = ex Putting value of I1 in our equation β΄ β«1βγπ₯^2 ππ₯" " γ ππ₯" = " π₯^2. ππ₯ β2β«1βγπ . ππγ ππ =π₯^2. ππ₯ β2(πππβπ^π )+πΆ =π₯^2. ππ₯ β2π₯ππ₯+γ2πγ^π₯+πΆ =ππ (π^πβππ+π)+πͺ