Definite Integral as a limit of a sum
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Definite Integral as a limit of a sum
Last updated at April 16, 2024 by Teachoo
Question 1 β«1_π^πβγπ₯ ππ₯γ β«1_π^πβγπ₯ ππ₯γ Putting π =π π =π π=(π β π)/π π(π)=π₯ We know that β«1_π^πβγπ(π₯) ππ₯γ =(πβπ) (πππ)β¬(πββ) 1/π (π(π)+π(π+β)+π(π+2β)β¦+π(π+(πβ1)β)) Hence we can write β«1_π^πβγπ₯ ππ₯γ =(πβπ) limβ¬(nββ) 1/π (π(π)+π(π+β)+π(π+2β)+β¦ +π(π+(πβ1)β) Here, π(π)=π₯ π(π)=π π(π+π)=π+β π (π+ππ)=π+2β β¦ π(π+(πβπ)π)=π+(πβ1)β Hence, our equation becomes β΄ β«_π^πβπ π π = (πβπ) (πππ)β¬(πββ) 1/π (π(π)+π(π+β)+π(π+2β)β¦+π(π+(πβ1)β)) = (πβπ) (πππ)β¬(πββ) 1/π (π+(π+β)+(π+2β)+ β¦+(π+(πβ1)β)) = (πβπ) (πππ)β¬(πββ) 1/π ( π+π+ β¦+π +β+2β+ β¦β¦+(πβ1)β) = (πβπ) (πππ)β¬(πββ) 1/π ( ππ +β+2β+ β¦β¦+(πβ1)β) = (πβπ) (πππ)β¬(πββ) 1/π ( ππ+β (π+π+ β¦β¦β¦+(πβπ))) π πππππ We know that 1+2+3+ β¦β¦+π= (π (π + 1))/2 1+2+3+ β¦β¦+πβ1= ((π β 1) (π β 1 + 1))/2 = (π (π β π) )/π = (πβπ) (πππ)β¬(πββ) 1/π ( ππ+(π . π(π β π))/π) = (πβπ) (πππ)β¬(πββ) ( ππ/π+π(π β 1)β/2π) = (πβπ) (πππ)β¬(πββ) ( π+(π β 1)π/2) = (πβπ) (πππ)β¬(πββ) ( π+(π β 1)(π βπ)/(2 . π)) = (πβπ) (πππ)β¬(πββ) ( π+(π/π β π/π) ((π β π) )/2) [ππ πππ β=(π β π)/π] = (πβπ) (πππ)β¬(πββ) ( π+(πβ π/π) ((π β π) )/2) = (πβπ)( π+(1β π/β) ((π β π) )/2) = (πβπ)( π+(1βπ) ((π β π) )/2) = (πβπ)( π+ (π β π )/2) = (πβπ)((2π + π β π )/2) = (π β π)(π + π)/2 = (π^π β π^π)/π