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Question
Evaluate `int_(-1)^2 (7x - 5)"d"x` as a limit of sums
Solution
Here a = –1
b = 2
And h = `(2 + 1)/"n"`
i.e, nh = 3 and f(x) = 7x – 5.
Now, we have
`int_(-1)^2 (7x - 5)"d"x = lim_("k" -> 0) "h"["f"(-1) + "f"(-1 + "h") + "f"(-1 + 2"h") + ... + (-1 + ("n" - 1)"h")]`
Note that
f(–1) = –7 – 5 = –12
f(–1 + h) = –7 + 7h – 5 = –12 + 7h
f(–1 + (n –1)h) = 7 (n – 1)h – 12.
Therefore, `int_(-1)^2 (7x - 5)"d"x = lim_("h" -> 0) "h"[(-12) + (7"h" - 12) + (14"h" - 12) + ... + (7("n" - 1)"h" - 12)]`
= `lim_("h" -> 0) "h"[7"h"[1 + 2 + ... +("n" - 1)] - 12"n"]`
= `lim_("h" -> 0) "h"[7"h" (("n" - 1)"n")/2 - 12 "n"]`
= `lim_("h" -> 0) [7/2("nh")("nh" - "h") - 12"nh"]`
= `7/2(3 - 0) - 12 xx 3`
= `(7 xx 9)/2 - 36`
= `(-9)/2`.
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