Question

Find the area of the region bounded by the parabola y = 5x^2, the tangent line to this parabola at (4, 80), and the x-axis.

Answer 1

You'd do well to first find the equation of the aforementioned tangent line ^_^

Answer 2

is the equation y=40x-80

Answer 3

Yes, it seems right to me.

Answer 4

then what do you do to get the area? I dont know how to set it up

Answer 5

Try to sketch it, maybe it'll give you an idea...

Answer 6

Crud... the draw option doesn't work...
>.>

Answer 7

Here's a rough sketch:

Answer 8

We need to find the area of that yellow bit... catch me so far?

Answer 9

yea

Answer 10

Okay, so this is an integration problem, one that will have to be divided into two parts because of the nature of the area that you need to, well, find the area of.
First find the area of this shaded part, because it's the more straightforward integral:

Answer 11

I believe, it can be done in one integral, if we take slices parallel to the x-axis and integrate with respect to y. There is nothing wrong with doing it the way @terenzreignz suggested. May be both take the same amount of time.

Answer 12

That is... true :3
But I'm allergic to radicals XD

Answer 13

You only have to deal with \( \sqrt y \)

Answer 14

how would i set the integral up?

Answer 15

Well, going on from the method I put up, the area of the black shaded part is simply the integral of the parabola (quadratic)

The only tricky part is figuring out the limits... have you figured it out?

Answer 16

is it 0 to 80

Answer 17

0 to somewhere, surely, but where does that tangent line touch the x-axis?

Hint: set y = 0, and solve for x.

Answer 18

5x^2=0

Answer 19

No, not the parabola, silly ^_^
the TANGENT LINE :D

Answer 20

x=20

Answer 21

x=2 my bad

Answer 22

Okay, so x runs from 0 to 2, now just integrate the parabola with those limits.

Answer 23

what do you mean?

Answer 24

Integrate 5x^2, with limits from 0 to 2

Answer 25

is it 40/3

Answer 26

Okay, so that's the area of the black shaded part, what about the rest of the yellow area?

Answer 27

do we have to get other limits?

Answer 28

Not only other limits, but another function, too.
You notice that this is bounded by the parabola and the LINE, and not the x-axis, so...

Answer 29

true so what other function do we need to use?

Answer 30

Well we can certainly take the area under the parabola, but we must take AWAY the area under the line, any ideas?

Answer 31

ummmm not really?

Answer 32

Take the area under the parabola, which is its integral, and then from that, subtract the area under the line.

So what you get is the parabola minus the line:
5x^2 - 40x + 80

Now integrate this... with limits...
....
what are the limits?
It should start from x = 2 to...?

Answer 33

2 to 80?

Answer 34

No.... you want to get where the line intersects the parabola, and it intersect the parabola at the point where x = ?

Answer 35

5?? sorry im not sure

Answer 36

is it 4?

Answer 37

It's 4, yes. So integrate that new function from 2 to 4

Answer 38

is it 40/3

Answer 39

Yup. So just add that to the first integral, and that's your answer.

Answer 40

so its 80/3

Answer 41

\[
\int_0^{80}
\left(\left(\frac{y}{40}+2\right)-\frac{\sqrt{y}}{\sqrt{5}}\right) \, dy=\frac{1}{40}
\left[-\frac{1}{3} 16
\sqrt {5}
y^{3/2}+\frac{y^2}{2}+80
y\right]_0^{80}=\frac{80}3

\]

Answer 42

Here is the way you did it with @terenzreignz
\[
\int_0^2
5 x^2 \, dx+
\int_2^4 \left(5 x^2-40
x+80\right) \, dx=\frac{80}{3}
\]