Concept of Definite Integral

Related Calculators: Definite and Improper Integral Calculator , Riemann Sum Calculator

In Area Problem note we saw that limit of the form `lim_(n->oo)f(x_i^(**))Delta x` arises when we compute an area.

It turns out that this same type of limit occurs in a wide variety of situations even when `f` is not necessarily a positive function.

Thus, this limit has special name and notation.

Definition of a Definite Integral. If `f` is a continuous function defined for `a<=x<=b`, we divide the interval `[a,b]` into `n` subintervals of equal width `Delta x=(b-a)/n`. We let `x_0(=a),x_1,x_2,...,x_n(=b)` be the endpoints of these subintervals and we choose sample points `x_1^(**),x_2^(**),...,x_n^(**)` in these subintervals, so `x_i^(**)` lies in the i-th subinterval `[x_(i-1),x_i]`. Then the definite integral of `f` from `a` to `b` is `int_a^bf(x)dx=lim_(n->oo)sum_(i=1)^nf(x_i^(**))Delta x`.

As with indefinite integral case `f(x)` is called integrand.

`a` is lower limit and `b` is upper limit.

Note, that definite integral is a number, it doesn't depend on `x`. We can choose another variable: `int_a^bf(x)dx=int_a^bf(t)dt=int_a^bf(u)du`.

The sum `sum_(i=1)^nf(x_i^(**))Delta x` is called Riemann sum.

If sample points `x_i^**` are left endpoints then sum is called Left Riemann Sum, if right endpoints then sum is called Right Riemann Sum.

Now let's see what will be if `f` can take both positive and negative values.net area

If `f` takes on both positive and negative values, then the Riemann sum is the sum of the areas of the rectangles that lie above the x-axis minus sum of areas of the rectangles that lie below the x-axis.

Thus, definite integral is net area: area above x-axis minus area below x-axis:

`int_a^bf(x)dx=S_1-S_2+S_3+S_4-S_5+S_6`.

Note, that although we defined `int_a^bf(x)dx` by dividing interval `[a,b]` into `n` intervals of equal width, it is sometimes advantageous to work with intervals of unequal width. If the subinterval widths are `Delta x_1,Delta x_2,...,Delta x_n` then we have to ensure that all these widths approach 0 in the limiting process. This happens if the largest width, `max Delta x_i` , approaches 0. So in this case definition of integral becomes `int_a^bf(x)dx=lim_(max Delta x_i->0) sum_(i=1)^nf(x_i^(**)) Delta x_i`.

Now let's go through a couple of examples.

Example 1. Express `lim_(n->oo)sum_(i=1)^n[3x_i^2-cos(x_i^5)]Delta x` as an integral on the interval `[0,pi]` .

Comparing the given limit with the limit in definition of integral, we see that they will be identical if we choose `f(x)=3x^2-cos(x^5)` and `x_i^(**)=x_i` (so the sample points are right endpoints). We are given that `a=0` and `b=pi`, therefore `lim_(n->oo)sum_(i=1)^n[3x_i^2-cos(x_i^5)]Delta x=int_0^pi(3x^2-cos(x^5))dx`.

It is very important to recognize limits of sums as integrals. In general, if `lim_(n->oo)sum_(i=1)^nf(x_i^(**))Delta x=int_a^bf(x)dx` then we replace `lim sum` by `int`, `x_i^(**)` by `x` and `Delta x` by `dx`.

Example 2. Evaluate `int_0^2(3x^2-x^3)dx`.

Divide interval `[0,2]` into `n` subintervals of width `Delta x=(2-0)/n=2/n`.

Since we can use any point within subinterval, let's use right endpoints.

Right endpoint of i-th subinterval is `(2i)/n`, so

`int_0^2(3x^2-x^3)dx=lim_(n->oo)sum_(i=1)^nf((i)/n)2/n=lim_(n->oo)sum_(i=1)^n (3((2i)/n)^2-((2i)/n)^3)2/n=lim_(n->oo)sum_(i=1)^n(24 i^2/n^2-16i^3/n^3)1/n=`

`=lim_(n->oo)sum_(i=1)^n(24/n^3 i^2-16/n^4 i^3)=lim_(n->oo)(sum_(i=1)^n(24/n^3 i^2)-sum_(i=1)^n(16/n^4 i^3))=lim_(n->oo)(24/n^3 sum_(i=1)^n i^2-16/n^4 sum_(i=1)^n i^3)=`

On this stage we need the following two formulas:

`sum_(i=1)^n i^2=(n(n+1)(2n+1))/6`.

`sum_(i=1)^ni^3=((n(n+1))/2)^2`.

So, `int_0^2(3x^2-x^3)dx=lim_(n->oo)(24/n^3 (n(n+1)(2n+1))/6-16/n^4 ((n(n+1))/2)^2)=`

`=lim_(n->oo)((4(n+1)(2n+1))/n^2-(4(n+1)^2)/n^2)=`

`=lim_(n->oo)(4(1+1/n)(2+1/n)-4(1+2/n+1/n^2))=4(1+0)(2+0)-4(1+2*0+0)=4`

So, `int_0^2(3x^2-x^3)dx=4`.

Example 3. Evaluate the following integral by interpreting it in terms of areas: `int_(-2)^2sqrt(4-x^2)dx`.

Since `f(x)=sqrt(4-x^2)>=0`, we can interpret this integral as the area under the curve `y=sqrt(4-x^2)` from -2 to 2.

Squaring both sides give `y^2=4-x^2` then `x^2+y^2=4` and the required area is area of semicircle with radius 2.

Therefore, `int_(-2)^2sqrt(4-x^2)dx=1/2 pi*(2)^2=2pi` .

Example 4. Evaluate the following integral by interpreting it in terms of areas: `int_-1^2(2|x|-1)dx`.example of net area

First we draw graph of the function `y=2|x|-1` on interval `[-1,2]`.

Recall that definite integral is net area: `int_(-1)^2(2|x|-1)dx=S_1-S_2-S_3+S_4`.

Now area `S_1` is area of right-angled triangle with legs `1/2` and 1, so `S_1=1/2*1/2*1=1/4`.

Similarly, `S_2=1/2*1/2*1=1/4`, `S_3=1/2*1/2*1=1/4` and `S_4=1/2*3/2*3=9/4`.

So, `int_(-1)^2 (2|x|-1)dx=1/4-1/4-1/4+9/4=2`.