Area Under a Curve from Two Lines

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Jason76

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What is the area of the region in the first quadrant that is bounded by the line \(\displaystyle y = 6x\) and the parabola \(\displaystyle y = 3x^{2}\)

Hint ??
emoticon-0128-hi.gif
Answer = 4 How did it get there?

First of all, you set the two lines equal to each other.

\(\displaystyle 6x = 3x^{2}\)

Next, get the \(\displaystyle x\) latex value. Now, you plug \(\displaystyle x\) into each equation to find \(\displaystyle y\) values. But will the \(\displaystyle y\) values be the same or different? After that step, what is the next step? :confused:
 
Jason76 said:
What is the area of the region in the first quadrant that is bounded by the line \(\displaystyle y = 6x\) and the parabola \(\displaystyle y = 3x^{2}\)

Hint ??
emoticon-0128-hi.gif
Answer = 4 How did it get there?

First of all, you set the two lines equal to each other.

\(\displaystyle 6x = 3x^{2}\)

Next, get the \(\displaystyle x\) value. Now, you plug \(\displaystyle x\) into each equation to find \(\displaystyle y\) values. But will the \(\displaystyle y\) values be the same or different? After that step, what is the next step? :confused:
Click to expand...
The two x-values you found will be the limits of a definite integral in the x-direction. The function to be integrated is the difference of the two y-curves.
 
DrPhil said:
The two x-values you found will be the limits of a definite integral in the x-direction. The function to be integrated is the difference of the two y-curves.
Click to expand...

Ok makes sense. But how do we find the function to integrate?. We know the bounds of it, but what is it? As of now, we only have two functions.
 
Jason76 said:
What is the area of the region in the first quadrant that is bounded by the line \(\displaystyle y = 6x\) and the parabola \(\displaystyle y = 3x^{2}\)
Click to expand...
Jason76 said:
Ok makes sense. But how do we find the function to integrate?. We know the bounds of it, but what is it? As of now, we only have two functions.
Click to expand...
Draw the two functions. Shade in the "region bounded by the line and the parabola." consider a tiny slice, of width \(\displaystyle dx\), at some \(\displaystyle x\) between the two end points. The height of the slice is the difference of the two curves, and the incremental area is height × width.
\(\displaystyle dA = (y_1 - y_2)\ dx\)
Add up (that is, integrate) all the incremental slices.
 
Jason76 said:
What is the area of the region in the first quadrant that is bounded by the line \(\displaystyle y = 6x\) and the parabola \(\displaystyle y = 3x^{2}\)

Hint ??
emoticon-0128-hi.gif
Answer = 4 How did it get there?

First of all, you set the two lines equal to each other.

\(\displaystyle 6x = 3x^{2}\)

Next, get the \(\displaystyle x\) latex value. Now, you plug \(\displaystyle x\) into each equation to find \(\displaystyle y\) values. But will the \(\displaystyle y\) values be the same or different? After that step, what is the next step? :confused:
Click to expand...
That is a quadratic equation so has two x values. What are they? You have y= 6x and \(\displaystyle y= 3x^2\) and because x satisfies \(\displaystyle y= 6x= 3x^2\), each x, in either equation, gives the same y value. But the two different "x"s will give two different "y"s. Each (x, y) is one of the points where y= 6x and \(\displaystyle y= 3x^2\) intersect. The "function to integrate" is the height of an "infinitesimally" thin rectangle that sums to give the area- and that is the difference between the two y values, \(\displaystyle 6x- 3x^2\). \(\displaystyle 6x- 3x^2= 3x(2- x)= 0\) when x= 0 and x= 2 so the area is \(\displaystyle \int_0^2(6x- 3x^2)dx\).
 
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HallsofIvy said:
That is a quadratic equation so has two x values. What are they? You have y= 6x and \(\displaystyle y= 3x^2\) and because x satisfies \(\displaystyle y= 6x= 3x^2\), each x, in either equation, gives the same y value. But the two different "x"s will give two different "y"s. Each (x, y) is one of the points where y= 6x and \(\displaystyle y= 3x^2\) intersect. The "function to integrate" is the height of an "infinitesimally" thin rectangle that sums to give the area- and that is the difference between the two y values, \(\displaystyle 6x- 3x^2\). \(\displaystyle 6x- 3x^2= 3x(2- x)= 0\) when x= 0 and x= 2 so the area is \(\displaystyle \int_0^2(6x- 3x^2)dx\).
Click to expand...

Sorry, fuzzy on algebra.

How is \(\displaystyle 6x = 3x^{2}\) solved for \(\displaystyle x\) to get \(\displaystyle 0\) and \(\displaystyle 2\)?

\(\displaystyle 6 = \dfrac{3x^{2}}{x}\)

\(\displaystyle 6 = 3x\)

\(\displaystyle x = 2\) :confused: - But how about the \(\displaystyle 0\)?
 
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