Question

How do you use limits to evaluate `int (x+2)dx` from [1,4]?

Answer 1

Perform the integration as if the integral were indefinite but without a constant of integration.
Subtract the expression evaluated at the lower limit from the expression evaluated at the upper limit.

Explanation:

Given: `int_1^4 (x+2)dx`

Perform the integration as if the integral were indefinite but without a constant of integration:

`int_1^4 (x+2)dx= {:x^2/2+2x]_1^4`

Subtract the resulting expression evaluated at the lower limit from the expression evaluated at the upper limit:

`int_1^4 (x+2)dx= (4^2/2+2(4))- (1^2/2+2(1))`

`int_1^4 (x+2)dx= 8+8- 1/2-2`

`int_1^4 (x+2)dx= 13.5`

Answer 2

`int_1^4 \ (x+2) \ dx = 13.5`

Explanation:

By definition of an integral, then

`int_a^b \ f(x) \ dx`

represents the area under the curve `y=f(x)` between `x=a` and `x=b`. We can estimate this area under the curve using thin rectangles. The more rectangles we use, the better the approximation gets, and calculus deals with the infinite limit of a finite series of infinitesimally thin rectangles.

That is

`int_a^b \ f(x) \ dx = lim_(n rarr oo) (b-a)/n sum_(i=1)^n \ f(a + i(b-a)/n)`

Here we have `f(x)=x+2` and we partition the interval `[1,4]` using `Delta = {1, 1+3*1/n, 1+3*2/n, ..., 1+3*n/n }`

And so:

`I = int_1^4 \ (x+2) \ dx`
`\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ f(1+3*i/n)`
`\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ ((1+3*i/n)+2)`
`\ \ = lim_(n rarr oo) 3/n sum_(i=1)^n \ (3+(3i)/n)`
`\ \ = lim_(n rarr oo) 3/n {sum_(i=1)^n 3+ sum_(i=1)^n (3i)/n }`
`\ \ = lim_(n rarr oo) 3/n {3n+ 3/nsum_(i=1)^n i }`

Using the standard summation formula:

`sum_(r=1)^n r = 1/2n(n+1)`

we have:

`I = lim_(n rarr oo) 3/n {3n+ 3/n*1/2n(n+1) }`

`\ \ = lim_(n rarr oo) 3/n {3n+ 3/2(n+1) }`

`\ \ = lim_(n rarr oo) 3/n*3/2 {2n+ (n+1) }`

`\ \ = lim_(n rarr oo) 9/(2n) (3n+1)`

`\ \ = lim_(n rarr oo) (27/2+9/(2n))`
`\ \ = 27/2`

`\ \ = 13.5`