find the area between y=x^3–14x^2+45x

and y=–x^3+14x^2–45x

i am having trouble how the graphs look like and how to do this...

As a practical matter, you will have to find the graphs. You have a problem of "negative" areas if your ytop becomes ybottom on your integration, so the total a area integrated becomes less (postive and negative areas combined.). So I would graph them (graphing calcs have a few nice advantages).

Just looking at the equations, where y1-y1 ( a probable crossing point)
x^3-14x+45=-x^3+14x-45 or
2x(x^2-14x+45)=0 which after factoring, give you three roots
2x(x-9)(x-5)=0. So knowing nothing else, you could look at from -inf to zero, zero to 5, 5 to 9, and 9 to inf.
But at - inf,the curves go to y +- unbounded,so that does bound an areas, and the same is true on the upper end. So your area of interest is
zero to 5, 5 to 9
Without graphing, you might be wondering which one is y top?
WEll, it wont matter if you integrate both as absolutes, then add them

area= abs INT ytop-ybottom dx from x=0 to 5 PLUS abs INT ytop-ybottom dx from x=5 to 9.

Which is ytop and y bottom? Machts nichts.

But it is nicer, and more clear, if you have a graph first.

factor the first into

y = x(x-5)(x-9)
so you have a cubic which crosses the x-axis at 0, 5, and 9

your second equation is simply the reflection of that in the x-axis.

So the closed regions are between 0 and 5 and between 5 and 9, with the bottom half of each equal to its matching top half.

This is an integration problem in calculus

I would then take
2[integral of](x^3–14x^2+45x)dx from 0 to 5 + 2[integral of](-x^3+14x^2-45x)dx from 5 to 9

I hope you can take it from there.

To find the area between two curves, we first need to understand how the graphs look like. Let's start by plotting the two equations on a graph.

Equation 1: y = x^3 – 14x^2 + 45x
Equation 2: y = –x^3 + 14x^2 – 45x

To get a rough idea of the shape of the graphs, we can plug in some values for x and see what values of y we get. Let's start with x = 0, and increment by 1.

For Equation 1:
- When x = 0, y = (0^3) – 14(0^2) + 45(0) = 0.
- When x = 1, y = (1^3) – 14(1^2) + 45(1) = 32.
- When x = 2, y = (2^3) – 14(2^2) + 45(2) = 42.
- And so on...

For Equation 2:
- When x = 0, y = (–0^3) + 14(0^2) – 45(0) = 0.
- When x = 1, y = (–1^3) + 14(1^2) – 45(1) = -31.
- When x = 2, y = (–2^3) + 14(2^2) – 45(2) = -41.
- And so on...

From these values, we can see that both equations produce similar parabolic shapes, but with opposite orientations. Equation 1 opens upward, while Equation 2 opens downward.

To find the area between the two curves, we need to find the points where they intersect. By setting the two equations equal to each other, we can solve for x:

x^3 – 14x^2 + 45x = –x^3 + 14x^2 – 45x

By simplifying, we get:

2x^3 – 28x^2 + 90x = 0

Factoring out 2x, we have:

2x(x^2 – 14x + 45) = 0

To find the roots of this equation, we can factorize further:

2x(x – 5)(x – 9) = 0

From this, we see that x = 0, x = 5, and x = 9 are the x-intercepts. These are the points where the two curves intersect.

Next, we need to find the y-values at these x-intercepts by substituting these values back into either equation. Let's use Equation 1:

When x = 0, y = (0^3) – 14(0^2) + 45(0) = 0.
When x = 5, y = (5^3) – 14(5^2) + 45(5) = 100.
When x = 9, y = (9^3) – 14(9^2) + 45(9) = 612.

Now that we have these points, we can sketch the graph of the two curves. Plot the points (0, 0), (5, 100), and (9, 612), and draw curves that resemble parabolas passing through these points. One opens upward and the other opens downward.

To find the area between the curves, we need to find the definite integral of the expression that represents the difference between the two curves within the given x-interval.

∫[(Equation 1) - (Equation 2)] dx evaluated from x = 0 to x = 5

This will give you the area between the curves within the interval from x = 0 to x = 5.

Similarly,

∫[(Equation 2) - (Equation 1)] dx evaluated from x = 5 to x = 9

This will give you the area between the curves within the interval from x = 5 to x = 9.

Finally, you can add the two calculated areas to obtain the total area between the curves from x = 0 to x = 9.