How to find the area bounded by the curve and x-axis


I'll tell you how to do the second one.

$$y = -x^3 - 3x^2$$

First of all let's find the extrema of the integration by solving those two equations:

1) Set $x = 0$ we find

$$ y = 0 $$

2) Set $y = 0$ we find

$$-x^3 - 3x^2 = 0 ~~~~~ \to ~~~~~ -x^2(x + 3) = 0 ~~~~~ \to ~~~~~ x = -3$$

Thence your extrema are $0$ and $-3$.


$$A = \int_{-3}^0 -x^3 - 3x^2\ \text{d}x = -\frac{1}{4}x^4 - x^3 \bigg|_{-3}^0 = 0 - \left(-\frac{(-3)^4}{4} - (-3)^3\right) = 0 + \frac{81}{4} - 27 = -6.75$$


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Updated on August 16, 2022


  • Road2PreSchool
    Road2PreSchool about 1 year

    The text says: Determine the area bounded by the curve and $x$-axis. Could somebody solve or at least explain how I should solve these problems.

    • $y=x^3+1.5x^2$
    • $y=-x^3-3x^2$
    • $y=x^2+x+2$ and $y=-x^2+x+4$
    • gambler101
      gambler101 about 7 years
      Use definite integration.
    • amWhy
      amWhy about 7 years
      The x-axis is given by $y=0$
    • Road2PreSchool
      Road2PreSchool about 7 years
      yeah I understand that i must use definite integration but how to determine what should i integrate
  • Road2PreSchool
    Road2PreSchool about 7 years
    could it be that the third one result is 8/3
  • Road2PreSchool
    Road2PreSchool about 7 years
    yes thank you but could you explain how can i find the region bounded without a graph ?
  • Ahmad Bazzi
    Ahmad Bazzi about 7 years
    You will have to know which part is bounded between the curve and x-axis..Plotting or guessing should be done