$\begin{align}&f(x)=x^3-x^2-6x\end{align}$$'como find the area of the region between the curve of the previous function? and the X axis. The area is expressed in units of surface. Suggestion: Make the graph for a better understanding of the exercise.

# Do find the area of the region?

- Asked on 14 de November, 2017
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It is a parable cubic with positive coefficient for x^3, so that starts to the left -infinite and ends to the right at +infinity.
Factorizándola, has three real roots, that is to say: three crossings with the x axis:
(-2); 0; 3.
Has two regions comprised between the curve and the x-axis: to the left above the axis (positive), and right below it (area denial). To find the total area "enclosed" between the curve and the x-axis, we must add the "modules" of these areas (that is, to take both as positive).
Integrate first to get the indefinite:
(1/4)x^4 - (1/3)x^3 - 3x^2
I get the two areas separately:
Between (-2) and 0: 0 - \[4+(8/3) - 12\]: (16/3)
Between 3 and 2: \[(81/4) - 9 - 27\] - \[4-(8/3)-12\]; (-63/4) + (32/3); (-189+128)/12;
(-61/12). Let us remember that we must take it as a positive:
(16/3) + (61/12) = (64+61)/12;
\ #### 125/12 units^2

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