# Evaluate the Following Integral

1. Oct 16, 2012

### Northbysouth

1. The problem statement, all variables and given/known data
Evaluate the following integral.

I have attached an image of the question.

2. Relevant equations

3. The attempt at a solution
I know that I need to switch the limits, but I'm not sure how to do this.

Do I just solve y = x/2 for x which would give me:

x = 2y and substitute this in place of the x/2?

Help would be appreciated.

#### Attached Files:

• ###### math 2224 14.2 1.png
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2. Oct 16, 2012

### BloodyFrozen

Where does it say $y = \frac{x}{2}$? I believe in your case, y is independent of x.

3. Oct 16, 2012

### Northbysouth

I assumed that that was what the lower bound of the integral with respect to y is.

4. Oct 16, 2012

### BloodyFrozen

That is part of the limit of integration. It doesn't tell us that y is a function of x. y in this case is just like x; it's a dummy variable. Just integrate first with respect to y, then evaluate the inner integral. At this point, the integrand should be in terms of x. Then just integrate again and evaluate.

5. Oct 16, 2012

### Northbysouth

The problem is, I can't figure out how to integrate sin(2y^2) because u substitution doesn't work. When I spoke with my professor he said to switch the limits but I am not sure how to do this.

6. Oct 16, 2012

### Northbysouth

I've got it now. Using the boundaries for y and x I draw a graph of the boundaries which gave me a triangle bounded by the sides X=0, Y=1 and y=x/2

Then I rearranged the integral so that I integrated with respect to x before I integrated integrated with respect to y.

In the end I got the answer to be

-1/2 + cos(2)/2

7. Oct 16, 2012

### Ray Vickson

I could not read your attachment on my i-Phone 4, so here it is again in readable form:
$$\int_0^2 \int_{x/2}^1 sin(2y^2)\, dy \, dx.$$

RGV

8. Oct 16, 2012

### BloodyFrozen

Sorry, my misinterpretation/ignorance.