Questions about taking the area under curves

[hestabbedbasil]

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

 

[galactus]

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.

\(\displaystyle \int{cos(2x)}dx=\frac{sin(2x)}{2}\)

\(\displaystyle \frac{sin(2(\frac{{\pi}}{2}))}{2}-\frac{sin(2(\frac{{\pi}}{4}))}{2}\)

Your region is below the x axis.

 

[hestabbedbasil]

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.

\(\displaystyle \int{cos(2x)}dx=\frac{sin(2x)}{2}\)

\(\displaystyle \frac{sin(2(\frac{{\pi}}{2}))}{2}-\frac{sin(2(\frac{{\pi}}{4}))}{2}\)

Your region is below the x axis.

Thanks a million for your speedy response. Why are you dividing the sin(2x) by two? When you take the antiderivative, do you ignore the quantity in the parentheses?

Thanks again

 

[galactus]

\(\displaystyle \int{cos(2x)}dx\)

Let u=2x

du=2dx

du/2=dx

\(\displaystyle \int{cos(u)}du=sin(u)\)

Resub:

\(\displaystyle \frac{sin(2x)}{2}\)

 

[hestabbedbasil]

I'm sorry if I'm being dense, but I'm still not following you. Where did the du come from? What does the d there stand for?

Thanks again!

 

[galactus]

You've not done u-substitution?.