# Thread: Finding area enclosed by integral and x-axis

1. ## Finding area enclosed by integral and x-axis

Find the area enclosed by the graph of definite integral [4(x^3-3x^2+2x)dx] (a=0, b=2) and the x-axis. The first part of this question asked to evaluate the definite integral for which i got 0 and i thought you would find the area the same way but the last part asks you to explain why the two answers aren't the same!

2. If you graph the equation, you will see that an equal portion lies in the negative y axis and the positive. When you run the integral from 0 to 1 and 1 to 2, you will get the area correct.

See graph

3. Note that area,whether above or below the axis, is always positive.

Integrating $\displaystyle \int_a^b f(x)dx$ will give the area between the graph of f(x) and the x-axis only if f(x) is non-negative between a and b.

By the way, your phrasing is rather peculiar. This is not the area between an integral and the x-axis, it is the area between a graph and the x-axis. And not the "graph of the definite integral". A definite integral is always a number and has no "graph". You are asking for the area between the graph of y= 4(x^3-3x^2+2x) and the x-axis and you find that by taking the definite integral (and being careful of the sign as I say).