Question

How do you evaluate the definite integral by the limit definition given `int 4dx` from [-a,a]?

Answer 1

`int \ 4 \ dx = 8a`

Explanation:

We seek:

`A = int_(-a)^(a) \ 4 \ dx`

We assume wlog that a `gt 0`

Let us use our common sense, we know that a definite:

`A = int_alpha^beta \ f(x) \ dx`

represents the area under the curve `y=f(x)` between `x=alpha` and `x=beta`, thus the integral we seek represents the area under the fixed curve `y=4` between `x=-a` and `x=a`, which is a rectangle of height `4` and width `2a`.

Hence,

`int \ 4 \ dx = 8a`

We can establish the same result by using the definition of the integral in term of a riemann sum. 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)`

And we partition the interval `[a,b]` equally spaced using:

`Delta = {a+0((b-a)/n), a+1((b-a)/n), ..., a+n((b-a)/n) }`
`\ \ \ = {a, a+1((b-a)/n),a+2((b-a)/n), ... ,b }`

Here we have `f(x)=4` and so we partition the interval `[-a,a]` using:

`Delta = {-a, -a+2(2a)/n, -a+2 (2a)/n, -a+3 (2a)/n, ..., a }`

And so:

`I = int_(-a)^(a) \ 4 \ dx`
`\ \ = lim_(n rarr oo) (2a)/n sum_(i=1)^n \ f(-a+i*(2a)/n)`
`\ \ = lim_(n rarr oo) (2a)/n sum_(i=1)^n \ 4`
`\ \ = lim_(n rarr oo) (2a)/n (4n)`
`\ \ = lim_(n rarr oo) 8a`
`\ \ = 8a`

Using Calculus

If we use Calculus and our knowledge of integration to establish the answer, for comparison, we get:

`int_(-a)^(a) \ 4 \ dx = [ 4x ]_(-a)^(a)`

`" " = 4[ x ]_(-a)^(a)`
`" " = 4{ (a) - (-a) }`
`" " = 4(2a)`
`" " = 8a`