Question

Evaluate the following as limit of sum:

`int _0^2 (x^2 + 3) "d"x`

Answer 1

We know that `int_"a"^"b" "f"(x) "d"x = lim_("n" -> oo) "h" sum_("r" = 0)^("n" - 1) "f"("a" + "rh")`

For I = `int_0^2 (x^2 + 3) "d"x`

We have a = 0 and b = 2

I = `int_00^2 (x^2 + 3) "d"x`

Here, a = 0, b = 2 and h = `("b" - "a")/"n" = (2 - 0)/"n" = 2/"n"`

⇒ nh = 2

And f(x) = `(x^2 + 3)`

∴ I = `int_0^2 (x^2 + 3)"d"x = lim_("h" -> 0) "h" sum_("r" = 0)^("n" - 1) "f"("a" + "rh")`

= `lim_("h" -> 0) "h" sum_("r" = 0)^("n" - 1) "f"("rh")`

= `lim_("h" -> 0) "h" sum_("r" = 0)^("n" - 1) (3 + "r"^2"h"^2)`

= `lim_("h" -> 0) "h"[3"n" + "h"^2 ((("n" - 1)("n" - 1 + 1)(2"n" - 2 + 1))/6)]`

= `lim_("h" -> 0) "h"[3"n" + "h"^2 ((("n"^2 - "n")(2"n" - 1))/6)]`

= `lim_("h" -> 0) "h" [3"n" + "h"^2/6 (2"n"^3 - 3"n"^2 + "n")]`

= `lim_("h" -> 0) [3"nh" + (2"n"^3"h"^3 - 3"n"^2"h"^2 * "h" + "nh" * "h"^2)/6]`

= `lim_("h" -> 0) [3.2 + (2.2^3 - 3.2^2 * "h" + 2 * "h"^2)/6]`

= `6 + 16/6`

= `26/3`