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