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

calculus integration definite-integrals area

6,337

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 $-3$.

Finally:

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

6,337

Road2PreSchool

BY DAY: Student BY NIGHT: I write code and create databases for school and freelancing projects. My friends are smarter than yours. FOR FUN: Dota2 player, YouTube videos, comics, movies based on comics, BATMAN, everything that is related to World of Warcraft, smoking. "To ask why we code is to ask why the leaves fall"- Anonymous coder(ME).

Updated on August 16, 2022

Comments

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 about 7 years
  
  Use definite integration.
- amWhy about 7 years
  
  The x-axis is given by $y=0$
- Road2PreSchool about 7 years
  
  yeah I understand that i must use definite integration but how to determine what should i integrate
Road2PreSchool about 7 years

could it be that the third one result is 8/3
Road2PreSchool about 7 years

yes thank you but could you explain how can i find the region bounded without a graph ?
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