H hank Junior Member Joined Sep 13, 2006 Messages 209 Feb 14, 2007 #1 I have the following S [ 0 to 1] (x^3 - 2x^2) - (2x^2 - 3x) dx + S [1 to 3] (2x^2 - 3x) - (x^3 - 2x^2) dx The answer I get is 37/12. Can someone just verify if my answer is correct?
I have the following S [ 0 to 1] (x^3 - 2x^2) - (2x^2 - 3x) dx + S [1 to 3] (2x^2 - 3x) - (x^3 - 2x^2) dx The answer I get is 37/12. Can someone just verify if my answer is correct?
M mark07 Junior Member Joined Feb 3, 2007 Messages 84 Feb 14, 2007 #2 It is correct. Here's a useful link... "Integrator" powered by Mathematica: http://integrals.wolfram.com/ This free web version doesn't do definite integrals but it still comes in handy when you just want to check your answers.
It is correct. Here's a useful link... "Integrator" powered by Mathematica: http://integrals.wolfram.com/ This free web version doesn't do definite integrals but it still comes in handy when you just want to check your answers.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Feb 14, 2007 #3 Hello, Hank! \(\displaystyle \L\int^1_0\left[(x^3\,-\,2x^2)\,-\,(2x^2\,-\,3x)\right] dx\:+\:\int^3_1\left[(2x^2\,-\,3x)\,-\,(x^3\,-\,2x^2)\right]\,dx\) The answer I get is: \(\displaystyle \L\frac{37}{12}\;\;\) . . . Me too! Click to expand... You're finding the area between \(\displaystyle \,y\:=\:x^3\,-\,2x^2\,\) and \(\displaystyle \,y\:=\:2x^2\,-\,3x\;\) . . . right?
Hello, Hank! \(\displaystyle \L\int^1_0\left[(x^3\,-\,2x^2)\,-\,(2x^2\,-\,3x)\right] dx\:+\:\int^3_1\left[(2x^2\,-\,3x)\,-\,(x^3\,-\,2x^2)\right]\,dx\) The answer I get is: \(\displaystyle \L\frac{37}{12}\;\;\) . . . Me too! Click to expand... You're finding the area between \(\displaystyle \,y\:=\:x^3\,-\,2x^2\,\) and \(\displaystyle \,y\:=\:2x^2\,-\,3x\;\) . . . right?
