Please show me how to work this problem:
Give area of region bounded by F(x)=X^2 - 3, x=-2x and vertical line x= 2.
I worked the problem and I got 11/9. This is how I did the problem
Integrate (-2x - x^2-3) (limit -3 to 1) + integrate (-2x-x^2-3) (limit 1 to 2).
Please explain me my mistake and please go step by step.
thanks.
2007-12-02 11:25:31 · 3 answers · asked by Avi 1 in Science & Mathematics ➔ Mathematics
From your working out you have correctly found the points of intersection.
In your integrals you need to be careful with minus signs. They should be:
∫(-3 to 1) -2x-(x²-3)dx + ∫(1 to 2) x²-3 - (-2x) dx
=∫(-3 to 1) -2x-x²+3 dx + ∫(1 to 2) x²-3+2x dx
=-x²-x³/3+3x [-3 to 1] + x³/3+3x-x² [1 to 2]
=32/3 + 7/3
=39/3
=13
2007-12-02 11:35:09 · answer #1 · answered by Anonymous · 0⤊ 0⤋
area = â«x^2-3+2x dx, x from 1 to 2 = 7/3
2007-12-02 19:32:17 · answer #2 · answered by sahsjing 7 · 0⤊ 0⤋
Taking the area positive
A = Int (1,2) (x^2+2x -3)dx = x^3/3 +x^2-3x (1,2)=
8/3+4-6 -1/3-1+3 =7/3
2007-12-02 19:47:46 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋