Definite and indefinite integrals

Key Questions

  • How do I evaluate definite integrals?

    A definite integral looks like this:

    #int_a^b f(x) dx#

    Definite integrals differ from indefinite integrals because of the #a# lower limit and #b# upper limits.

    According to the first fundamental theorem of calculus, a definite integral can be evaluated if #f(x)# is continuous on [#a,b#] by:

    #int_a^b f(x) dx =F(b)-F(a)#

    If this notation is confusing, you can think of it in words as:

    The integral of a function (#f(x)#) with limits #a# and #b# is the integral of that function evaluated at the upper limit (#F(b)#) minus the integral of that function evaluated at the lower limit (#F(a)#)
    #F(x)# just denotes the integral of the function.

    Note that you will get a number and not a function when evaluating definite integrals. Also, you have to check whether the integral is defined at the given interval.

    Let's look at an example .

    #int_-2^6 x^3+2 dx#

    #x^3+2# is defined for all real numbers, so the boundaries of #a# and #b# are defined. To evaluate this definite integral, we first find the integral function and then plug in the upper limit of 6 into the integral function, and subtract the integral function evaluated at the lower limit of -2.

    #int_-2^6 x^3+2 dx = [1/4x^4+2x ]_-2^6=(1/4(6)^4+2(6))-(1/4(-2)^4+2(-2))= (336)-(0)= 336#

    Sally · 1 · Oct 14 2014
  • How do I evaluate indefinite integrals?

    Indefinite integrals are antiderivatives in general form.

    #int f(x) dx=F(x)+C#,

    where #F'(x)=f(x)#.

    I hope that this was helpful.

    Wataru · · Oct 19 2014
  • What is the difference between definite and indefinite integrals?

    Indefinite integrals have no lower/upper limits of integration. They are general antiderivatives, so they yield functions.

    #int f(x) dx=F(x)+C#,

    where #F'(x)=f(x)# and #C# is any constant.

    Definite integrals have lower and upper limits of integration (#a# and #b#). They yield values.

    #int_a^b f(x) dx = F(b)-F(a)#,

    where #F'(x)=f(x)#.

    I hope that this was helpful.

    Wataru · 4 · Oct 10 2014

Questions

Introduction to Integration