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प्रश्न
Evaluate the following integrals as the limits of sum:
`int_1^2 (4x^2 - 1)"d"x`
उत्तर
Here f(x) = 4x2 – 1
a = 1
b = 2
We use the formula
`int_"a"^"b" f(x)"d"x = lim_("n" -> oo) ("b" - "a")/"n" sum_("r" = 1)^"n" f("a" + ("b" - "a") "r"/"n")`
So we get,
= `lim_("n" -> oo) (2 - 1)/"n" sum_("r" = 1)^"n" (1 + "r"/"n")`
= `lim_("n" -> oo) 1/"n" sum_("r" = 1)^"n" [4(1 + "r"/"n")^2 - 1]`
= `lim_("n" -> oo) 1/"n" sum_("r" = 1)^"n" [4(1 + (2"r")/"n" + "r"^2/"n"^2) - 1]`
= `lim_("n" -> oo) 1/"n" sum_("r" = 1)^"n" [4 + (8"r")/"n" + (4"r"^2)/"n"^2 - 1]`
= `lim_("n" -> oo) 1/"n" [sum_("r" = 1)^"n" (4"r"^2)/"n"^2 + sum_("r" = 1)^"n" (8"r")/"n" + sum_("r" = 1)^"n" 3]`
= `lim_("n"-> oo) 1/"n" [4 xx ("n"("n" + 1)(2"n" + 1))/(6"n"^2) + 8/"n" xx ("n"("n" + 1))/2 + 3"n"]`
= `lim_("n"> oo) 1/"n" [4 xx ("n"^3 (1 + 1/"n")(2 + 1/"n"))/(6"n"^2) + (8"n")/2 (1 + 1/"n") + 3"n"]`
= `lim_("n" -> oo) [2/3 xx (1 + 1/"n")(2 + 1/"n") + 4(1 + 1/"n") + 3]`
= `4/3 + 4 + 3`
= `4/3 + 7`
= `(4 + 21)/3`
= `25/3`
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