Question......please answer thanks !!!?

Question

Find each of the two areas bounded by y=x^3-4x and y=x^2+2x ?
please help me with dis.....i dont know if may answer is correct

Answer 1

well dear you need to integrate of both functions

Part One;
F(x)=x^3-4x
∫F(x) dx = ∫x^3 - 4x dx ;
{ explanation ;
if f(x) = x^n so ∫F(x) dx = ∫x^n dx = (x^n+1) / n+1 }
∫x^3 - 4x = ∫x^3 - ∫ 4x =( x^3+1) / 3+1 - (4x^1+1)/1+1 + c
∫x^3 - 4x = x^4/ 4 - 4x^2/2 = x^4/4 - 2x^2 + c

Part Two;
F(x) =x^2+2x
∫F(x) dx = ∫x^2+2x
∫x^2 + ∫2x =[ (x^2+1) / 2+1] + [2x^1+1/1+1] =
x^3/3 + 2x^2 /2 = x^3/3 + x^2 + c


Good luck darling.

Answer 2

It is true that these two curves bound two different finite areas, but before you can calculate these areas, you must first find the "end points" of these curves (or simply the intersections of the curves).

y = x³ - 4x
y = x² + 2x

If you subtract the 2nd equation from the 1st equation,
0 = x³ - x² - 4x - 2x

Combining like terms,
x³ - x² - 6x = 0

Factoring out,
x(x² - x - 6) = 0

Factoring the trinomial,
x(x - 3)(x + 2) = 0

The solutions are:
x = -2, x = 0, and x = 3

Thus, these/three points bound the 2 "regions". The 1st region is bounded by [-2,0] and the 2nd region is bounded by [0,3].

The formula for the area between two curves [say f(x) and g(x)] in the interval [a,b] is:
∫ f(x)dx - ∫ g(x) dx .... [a.b]

or
= ∫ [f(x) - g(x)]dx .... [a,b]

If we let ∫ f(x)dx .... [a,b] = integral of f(x) between x = a to x = b

Thus, the first area is
∫ [x³ - 4x - (x² + 2x)]dx .... [-2,0]

Simplifying,
= ∫ (x³ - x² - 6x)dx .... [-2,0]

Using the Fundamental Theorem of Calculus,
= (1/4 x^4 - 1/3 x³ - 3x²)|(from -2 to 0)

Substituting,
= [1/4 (0)^4 - 1/3 (0)³ - 3(0)²] - [1/4 (-2)^4 - 1/3 (-2)³ - 3(-2)² ]

Simplifying,
= 0 - 0 - 0 - 4 - 8/3 + 12

Adding,
= 16/3

Thus, the first area is 16/3

The second area is given by
∫ [x³ - 4x - (x² + 2x)]dx .... [0,3]

Getting the integral,
= (1/4 x^4 - 1/3 x³ - 3x²)|(from 0 to 3)

Substituting,
= [1/4 (3)^4 - 1/3 (3)³ - 3(3)²] - [1/4 (0)^4 - 1/3 (0)³ - 3(0)² ]

Simpflying,
= 81/4 - 9 - 27 - 0 + 0 + 0

Adding,
= (81/4 - 36/4 - 108/4)
= (81 - 144)/4
= -63/4

Thus, the area (positive) is
= 63/4

Therefore,
the first area is 16/3
the second area is 63/4

^_^

Answer 3

x^3 - 4x = x^2 + 2x
x^3 - x^2 - 6x = 0
x( x^2 - x - 6 ) = 0
x( x - 3)( x + 2) = 0
x= 3 or -2

f(x) = 3 yields 15
f(x) = -2 yields 0

I'd go with 15

:)

Answer 4

area = x^5 + 2x^4 - 4x^3 - 8x^2
sadly, i don't actually get your question. hehe... =D
but, i think, if you multiply the 2 sides you'll know the area.

Answer 5

You can only solve this answer by plotting the graphs.
Only then you can find the areas.

Cheers.

Answer 6

7?

Answer 7

give me 2 points

Answer 8

I don't know

Answer 9

idk ask ur math teacher