Question

How do you evaluate the definite integral by the limit definition given `int (2x+5)dx` from [0,2]?

Answer 1

Here is a limit definition of the definite integral. (I hope it's the one you are using.)

.`int_a^b f(x) dx = lim_(nrarroo) sum_(i=1)^n f(x_i)Deltax`.

Where, for each positive integer `n`, we let `Deltax = (b-a)/n`

And for `i=1,2,3, . . . ,n`, we let `x_i = a+iDeltax`. (These `x_i` are the right endpoints of the subintervals.)

Let's do one small step at a time.

`int_0^2 (2x+5) dx`.

Find `Delta x`

For each `n`, we get

`Deltax = (b-a)/n = (2-0)/n = 2/n`

Find `x_i`

And `x_i = a+iDeltax = 0+i2/n = (2i)/n`

Find `f(x_i)`

`f(x_i) = 2(x_i) +5`

`= 2((2i)/n) +5`

`= (4i)/n+5`

Find and simplify `sum_(i=1)^n f(x_i)Deltax` in order to evaluate the sums.

`sum_(i=1)^n f(x_i)Deltat = sum_(i=1)^n ( (4i)/n+5) 2/n`

`= sum_(i=1)^n( (8i)/n^2 + 10/n)`

`= 8/n^2 sum_(i=1)^n(i) + 10/n sum_(i=1)^n 1`

Evaluate the sums

`= 8/n^2((n(n+1))/2) + 10/n(n)`

(We used summation formulas for the sums in the previous step.)

Rewrite before finding the limit

`sum_(i=1)^n f(x_i)Deltax = 8/n^2((n(n+1))/2) +10/n(n)`

`= 4((n(n+1))/n^2)) + 10`

Now we need to evaluate the limit as `nrarroo`.

`lim_(nrarroo) ((n(n+1))/n^2) = lim_(nrarroo) (n/n*(n+1)/n) = 1*1=1`

To finish the calculation, we have

`int_0^2 (2x+5) dx = lim_(nrarroo) ( 4((n(n+1))/n^2))+10`

`= 4(1) +10 = 14`