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Forum index » Science and Technology » Math » Undergraduate
Help with a few problems about area between curves
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Hestabbedbasil
science forum beginner


Joined: 22 May 2006
Posts: 1

PostPosted: Mon May 22, 2006 11:51 pm    Post subject: Help with a few problems about area between curves Reply with quote
So, I've tried and retried these problems without getting an answer close to the one in the back of the book. Here goes:

Find the area of the region enclosed by the curves:

y=cos 2x, y=0, x= (pi)/4, x= (pi)/2

My work went like this:

The antiderivative I need is sin(x^2), I think.

I worked it out as -sin (pi/2)^2 + sin (pi/4)^2

This didn't give me the right answer, though. The right answer is 1/2.

Find the area of the region enclosed by the curves:

y=e^x , y=e^2x , x=0, x= ln2

I graphed both y=, then took their antiderivatives, coming up with:

(e^x)/x for y=e^x and (e^2x)/2x for y=e^2x.

This didn't land me with the right answer either, which is 1/2.

Find the area of the region enclosed by the curves:

y=2/(1+x^2) and y= |x|

I figured out that the intercept points were 1,1 and -1,1. I really wasn't sure on the antiderivatives here. For y=2/(1+x^2), I thought it would be

2x(1+x)^-2

for |x|, I figured it would be (x^2)/2

After putting these in, then doing the upper limit-lower limit as I did in the other problems, I didn't get the right answer, which was supposed to be (pi)-1. I ended up with 1/2. Don't know what I did wrong.

If anyone could help me with these, I would be really thankful. Thanks for your time, in any case. I really appreciate the help, here.
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Lynn Kurtz
science forum Guru


Joined: 02 May 2005
Posts: 603

PostPosted: Tue May 23, 2006 12:36 am    Post subject: Re: Help with a few problems about area between curves Reply with quote
On Mon, 22 May 2006 19:51:47 EDT, Hestabbedbasil
<Hestabbedbasil@hotmail.com> wrote:

Quote:
So, I've tried and retried these problems without getting an answer close to the one in the back of the book. Here goes:

Find the area of the region enclosed by the curves:

y=cos 2x, y=0, x= (pi)/4, x= (pi)/2

My work went like this:

The antiderivative I need is sin(x^2), I think.

No, it isn't.

If you got that because the integral of 2x is x^2 you may need more
help than I can give you here. Review in your text about
substitutions, sometimes called u-substitutions. Try letting u = 2x.

Quote:
I worked it out as -sin (pi/2)^2 + sin (pi/4)^2

This didn't give me the right answer, though. The right answer is 1/2.


Find the area of the region enclosed by the curves:

y=e^x , y=e^2x , x=0, x= ln2

I graphed both y=, then took their antiderivatives, coming up with:

(e^x)/x for y=e^x and (e^2x)/2x for y=e^2x.


Both incorrect.

If you haven't mastered the derivatives and antiderivatives of e^x
yet, there is no point in trying to work applications such as areas.
Have you studied your text?

Quote:
This didn't land me with the right answer either, which is 1/2.

Find the area of the region enclosed by the curves:

y=2/(1+x^2) and y= |x|

I figured out that the intercept points were 1,1 and -1,1. I really wasn't sure on the antiderivatives here. For y=2/(1+x^2), I thought it would be

2x(1+x)^-2


Wrong again. Hint: What is the derivative of arctan(x)? You need to
know differentiation before you can master antidifferentiation.


Quote:
for |x|, I figured it would be (x^2)/2

Nope. What is it if x > 0? What if x < 0?


Quote:
After putting these in, then doing the upper limit-lower limit as I did in the other problems, I didn't get the right answer, which was supposed to be (pi)-1. I ended up with 1/2. Don't know what I did wrong.

If anyone could help me with these, I would be really thankful. Thanks for your time, in any case. I really appreciate the help, here.

Study, study, study!

--Lynn
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The World Wide Wade
science forum Guru


Joined: 24 Mar 2005
Posts: 790

PostPosted: Tue May 23, 2006 12:50 am    Post subject: Re: Help with a few problems about area between curves Reply with quote
In article
<15726749.1148341937517.JavaMail.jakarta@nitrogen.mathforum.org>,
Hestabbedbasil <Hestabbedbasil@hotmail.com> wrote:

Quote:
So, I've tried and retried these problems without getting an answer close to
the one in the back of the book. Here goes:

Find the area of the region enclosed by the curves:

y=cos 2x, y=0, x= (pi)/4, x= (pi)/2

My work went like this:

The antiderivative I need is sin(x^2), I think.

I worked it out as -sin (pi/2)^2 + sin (pi/4)^2

This didn't give me the right answer, though. The right answer is 1/2.

Find the area of the region enclosed by the curves:

y=e^x , y=e^2x , x=0, x= ln2

I graphed both y=, then took their antiderivatives, coming up with:

(e^x)/x for y=e^x and (e^2x)/2x for y=e^2x.

This didn't land me with the right answer either, which is 1/2.

Find the area of the region enclosed by the curves:

y=2/(1+x^2) and y= |x|

I figured out that the intercept points were 1,1 and -1,1. I really wasn't
sure on the antiderivatives here. For y=2/(1+x^2), I thought it would be

2x(1+x)^-2

for |x|, I figured it would be (x^2)/2

After putting these in, then doing the upper limit-lower limit as I did in
the other problems, I didn't get the right answer, which was supposed to be
(pi)-1. I ended up with 1/2. Don't know what I did wrong.

If anyone could help me with these, I would be really thankful. Thanks for
your time, in any case. I really appreciate the help, here.

Hint: All your antiderivatives are wrong.
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Virgil
science forum Guru


Joined: 24 Mar 2005
Posts: 5536

PostPosted: Tue May 23, 2006 1:56 am    Post subject: Re: Help with a few problems about area between curves Reply with quote
In article
<15726749.1148341937517.JavaMail.jakarta@nitrogen.mathforum.org>,
Hestabbedbasil <Hestabbedbasil@hotmail.com> wrote:

Quote:
So, I've tried and retried these problems without getting an answer close to
the one in the back of the book. Here goes:

Find the area of the region enclosed by the curves:

y=cos 2x, y=0, x= (pi)/4, x= (pi)/2

My work went like this:

The antiderivative I need is sin(x^2), I think.

One can always check an anti-derivative by differentiation.

d(sin(x^2))/dx = cos(x^2)d(x^2)/dx = cos(x^2) (2x) dx/dx = 2 x cos(x^2)
which is not equal to cos(2x)
Quote:

I worked it out as -sin (pi/2)^2 + sin (pi/4)^2

This didn't give me the right answer, though. The right answer is 1/2.
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