# I'm having trouble finding the area under curves

• January 24th 2010, 10:55 AM
cdlegendary
I'm having trouble finding the area under curves
Find the area of the region bounded by the parabola http://hw.math.ucsb.edu/webwork/math...777517img2.gif the tangent line to this parabola at (1, 1) that you found in (a), and the
x-axis.

The tangent line is y= 2x-1.

Here's a graph of what it looks like..
http://www4c.wolframalpha.com/Calcul...image/gif&s=45

I can't seem to find the right answer. I found the integral of x^2-2x+1 from 0 to 1, and get the answer 1/3.
• January 24th 2010, 11:01 AM
1005
Quote:

Originally Posted by cdlegendary
Find the area of the region bounded by the parabola http://hw.math.ucsb.edu/webwork/math...777517img2.gif the tangent line to this parabola at (1, 1) that you found in (a), and the
x-axis.

The tangent line is y= 2x-1.

Here's a graph of what it looks like..
http://www4c.wolframalpha.com/Calcul...image/gif&s=45

I can't seem to find the right answer. I found the integral of x^2-2x+1 from 0 to 1, and get the answer 1/3.

I get 1/3 for that integral too. Everything looks right to me if you want the area between the two curves from 0 to 1.
• January 24th 2010, 11:03 AM
VonNemo19
Quote:

Originally Posted by cdlegendary
Find the area of the region bounded by the parabola http://hw.math.ucsb.edu/webwork/math...777517img2.gif the tangent line to this parabola at (1, 1) that you found in (a), and the
x-axis.

The tangent line is y= 2x-1.

Here's a graph of what it looks like..
http://www4c.wolframalpha.com/Calcul...image/gif&s=45

I can't seem to find the right answer. I found the integral of x^2-2x+1 from 0 to 1, and get the answer 1/3.

If we are to do this with respect to x, then we wil need two integrals.

Note that the tangent line crosses the x axis at $x=\frac{1}{2}$

And the two graphs intersext at $x=1$

This implies that the Area in question may be found by

$A=\int_0^\frac{1}{2}x^2dx+\int_\frac{1}{2}^1[x^2-(2x-1)]dx$
• January 24th 2010, 11:04 AM
cdlegendary
Yeah, I really don't know whats wrong. The website I'm supposed to submit this to just isn't accepting that answer (Headbang)(Headbang)
• January 24th 2010, 11:06 AM
cdlegendary
Quote:

Originally Posted by VonNemo19
If we are to do this with respect to x, then we wil need two integrals.

Note that the tangent line crosses the x axis at $x=\frac{1}{2}$

And the two graphs intersext at $x=1$

This implies that the Area in question may be found by

$A=\int_0^\frac{1}{2}x^2dx+\int_\frac{1}{2}^1[x^2-(2x-1)]dx$

Ahh thanks! I see why it's like that now
• January 24th 2010, 11:09 AM
cdlegendary
• January 24th 2010, 11:33 AM
Alternatively, calculate the area under the curve from x=0 to x=1.
Then subtract the area of the half-rectangle from x=0.5 to x=1

$\int_0^1{x^2}dx-0.5(0.5)=\frac{1^3}{3}-0.25=\frac{1}{3}-\frac{0.75}{3}=\frac{0.25}{3}=\frac{1}{12}$

The integral will include the area under the line and above the x-axis from x=0.5 to x=1, hence we must subtract the area under the line in order to find the area between the line and curve, bounded by the positive x and y axes, or take the route shown by Von Nemo.
• January 24th 2010, 11:34 AM
drumist
Can you show us some intermediate work? Like what do you get for the two antiderivatives?

The solution VonNemo19 gave doesn't add up to 1/3, so there must be a mistake somewhere if you are still getting 1/3.
• January 24th 2010, 11:38 AM
drumist
Quote:

Then subtract the area of the rectangle from x=0.5 to x=1

Surely you mean to subtract the area of the triangle, which is (1/2)*0.5*1 = 0.25.

I'm not sure if you arrived at the correct area by accident (the shape isn't a rectangle!) or just halved the height of the triangle in your head first.
• January 24th 2010, 11:59 AM
of course i meant a triangle
• January 24th 2010, 12:01 PM
VonNemo19
Quote:

Originally Posted by cdlegendary

Let's do it with respect to y, It'll be easier.

Recall that

$A=\int_c^d(\text{right curve - left curve})dy$

when we want to find area between curves with respect to y.

Step 1. Find limits of integration.

We need to know over what interval y will move along. This is easy. From the context of the question (i.e find the are between curves from y=0 to y=1).

So, we see that $c=0$ and $d=1$.

Step 2 Determine which is the $\mbox{right curve}$ and which is the $\mbox{left curve}$.

If we were to draw a line which is parallel to the x-axis we would see that the curve $y=x^2$ would intersct this line to the left of where $y=2x-1$ would intersect said line. Therefore $\mbox{rightcurve}:y=2x-1$ and $\mbox{leftcurve}:y=x^2$.

Step 3. Get every thing in terms of y

$y=x^2\mbox{ implies that }x=\sqrt{y}$

And

$y=2x-1\mbox{ implies that }x=\frac{y+1}{2}$

Step 4. Substitute the relevant information into the formula

$A=\int_c^d(\mbox{right curve - left curve})dy=\int_0^1\left(\frac{y+1}{2}-\sqrt{y}\right)dy$

Step 5. Evaluate

If you need help with step 5, let me know.
• January 24th 2010, 01:44 PM
cdlegendary
Quote:

Originally Posted by VonNemo19
Let's do it with respect to y, It'll be easier.

Recall that

$A=\int_c^d(\text{right curve - left curve})dy$

when we want to find area between curves with respect to y.

Step 1. Find limits of integration.

We need to know over what interval y will move along. This is easy. From the context of the question (i.e find the are between curves from y=0 to y=1).

So, we see that $c=0$ and $d=1$.

Step 2 Determine which is the $\mbox{right curve}$ and which is the $\mbox{left curve}$.

If we were to draw a line which is parallel to the x-axis we would see that the curve $y=x^2$ would intersct this line to the left of where $y=2x-1$ would intersect said line. Therefore $\mbox{rightcurve}:y=2x-1$ and $\mbox{leftcurve}:y=x^2$.

Step 3. Get every thing in terms of y

$y=x^2\mbox{ implies that }x=y^2$

And

$y=2x-1\mbox{ implies that }x=\frac{y+1}{2}$

Step 4. Substitute the relevant information into the formula

$A=\int_c^d(\mbox{right curve - left curve})dy=\int_0^1\left(\frac{y+1}{2}-y^2\right)dy$

Step 5. Evaluate

If you need help with step 5, let me know.

thanks, that worked out. Got the answer 1/12.