The curve y=3x-x^2 cuts the x axis at the points O and A and meets the line y=-3x at the point B, as in the diagram. a) calculate the coordinate of A and B? B)FIND THE AREA OF SHADED REGION. i got the coordinates but the area is the issue

Answer 1

#color(blue)[A=int_0^6((3x-x^2)-(-3x))*dx=108-72=36]#

a)
to calculate the point A ,O and B

#3x-x^2=-3x#

#x^2-6x=0 rArr (x^2-6x+9)-9=0#

#(x-3)^2=9 rArr (x-3)=3#

"when" # x=6 rArr y=-18 #

#B=(6,-18)#

#O=(0,0) and A=(3,0)#

b)
Now lets calculate the area of the shaded region.

The Area due to x-axis given by:

#color(red)[A=int_a^b(y_2-y_1)*dx#

#A=int_0^6((3x-x^2)-(-3x))*dx#

#A=int_0^6(6x-x^2)*dx=[3x^2-1/3*x^3]_0^6#

#=108-72=36#

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Answer 2

To find the area of the shaded region, you need to integrate the difference between the two curves over the interval where they intersect. Set up the integral as follows:

[ \text{Area} = \int_{x_1}^{x_2} (3x - x^2 - (-3x)) , dx ]

where ( x_1 ) and ( x_2 ) are the x-coordinates of the intersection points A and B, respectively. Solve the integral to find the area of the shaded region.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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