How do you find the integral of #x^2-6x+5# from the interval [0,3]?

1 Answer
May 13, 2018

Answer:

-3

Explanation:

We can use the reverse power rule:

#f(x) = x^2 -6x + 5x^0#

Therefore,

#int_0^3x^2 - 6x + 5x^0dx = [x^3/3-6x^2/2+5x^1/1]_0^3#

This is equivalent to:

#(27/3-3(9)+5(3)-(0/3-6*0/2+5(0))#

Therefore, the answer is -3. Note that the definite integral gives the net area under the graph. The negative sign implies that there is more area below the x-axis (than above) for the interval from 0 to 3 (inclusive).

If you are unfamiliar with the reverse power rule, this might help: https://www.khanacademy.org/math/ap-calculus-ab/ab-antiderivatives-ftc/ab-reverse-power-rule/v/indefinite-integrals-of-x-raised-to-a-power