Question

How do you evaluate the definite integral by the limit definition given `int x/2dx` from [0,4]?

Answer 1

4

Explanation:

`int_0^4x/2dx`

`=[x^2/2xx1/2]_0^4`

`=[x^2/4]_0^4`

`=[x^2/4]^4-cancel([x^2/4]_0`

`=4^2/4=16/4=4`

Answer 2

Please see the explanation section below.

Explanation:

Here is a limit definition of the definite integral. (I hope it's the one you are using.) I will use what I think is somewhat standard notation in US textbooks.

.`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.)

I prefer to do this type of problem one small step at a time.

`int_0^4 x/2 dx`.

Find `Delta x`

For each `n`, we get

`Deltax = (b-a)/n = (4-0)/n = 4/n`

Find `x_i`

And `x_i = a+iDeltax = 0+i4/n = (4i)/n`

Find `f(x_i)`

`f(x_i) = x_i/2 =1/2 * x_i`

`= 1/2 * (4i)/n`

`= (2i)/n`

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 ( (2i)/n) 4/n`

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

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

Evaluate the sums

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

(We used a summation formula 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)`

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

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

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

To finish the calculation, we have

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

`= 4(1) = 4`