Question

Evaluate `int_(-1)^2 (7x - 5)"d"x` as a limit of sums

Answer 1

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`.