Integrals of Polynomial functions

Key Questions

  • How do you evaluate the integral #intx^3+4x^2+5 dx#?

    Because this equation only consists of terms added together, you can integrate them separately and add the results, giving us:

    #int x^3 + 4x^2 + 5dx = intx^3dx + int4x^2dx + int5dx#

    Each of these terms can be integrated using the Power Rule for integration, which is:

    #int x^ndx = x^(n+1)/(n+1) + C#

    Plugging our 3 terms into this formula, we have:

    #int x^3dx = x^(3+1)/(3+1) = x^4/4#

    #int 4x^2dx = (4x^(2+1))/(2+1) = (4x^3)/3#

    #int 5dx = int 5x^0dx = (5x^(0+1))/(0+1) = (5x^1)/1 = 5x#

    Now we arrive at our final answer by adding these together, remembering to add our constant (#C#) on the end:

    #int x^3 + 4x^2 + 5dx = x^4/4 + (4x^3)/3 + 5x + C#

    Collin C. · · Aug 24 2014
  • How do you evaluate the integral #int_0^4x^3+2x^2-8x-1#?

    First you integrate the function:

    #intx^3+2x^2-8x-1=1/4x^4+2/3x^3-4x^2-x#

    Then you substitute in your values for the upper and lower bounds. Start with 4:

    #1/4(4)^4+2/3(4)^3-4(4)^2-4= 1/4(256)+2/3(64)-4(16)-4#

    Solving that out yields:

    #64+128/3-64-4= 116/3 (or 38.66666)#

    Next you would substitute in 0, but looking at the equation, you can see that subbing 0 in will just yield zero. So last you do #116/3 - 0#, which of course is just #116/3#, and that's your answer.

    Johnny L. · · Oct 14 2014
  • What is the antiderivative of a polynomial?

    Let

    #f(x)=a_nx^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0#.

    An antiderivative #F(x)#of #f(x)# can be found by

    #F(x)=int f(x)dx#

    #=int(a_nx^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0)dx#

    #=a_n/{n+1}x^{n+1}+a_{n-1}/nx^n+cdots+a_1/2x^2+a_0x+C#.

    I hope that this was helpful.

    Wataru · · Oct 17 2014

Questions

Introduction to Integration