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Can you help me plz
- Thread starter Rufat2003
- Start date
D
Deleted member 4993
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Did you sketch the area of integration?
Please post a picture of your work.
The integral is:
\(\displaystyle \int_{x=0}^{x=2} \int_{y=0}^{y=4-x^2} \dfrac{xe^{2y}}{4-y}\hspace1ex dy\,dx\)
The region is within \(\displaystyle y=0 \) and \(\displaystyle y=4-x^2\) (both of these are easy to draw)
and \(\displaystyle x=0\) and \(\displaystyle x=2\) (both of these are easy to draw).
\(\displaystyle \int_{x=0}^{x=2} \int_{y=0}^{y=4-x^2} \dfrac{xe^{2y}}{4-y}\hspace1ex dy\,dx\)
The region is within \(\displaystyle y=0 \) and \(\displaystyle y=4-x^2\) (both of these are easy to draw)
and \(\displaystyle x=0\) and \(\displaystyle x=2\) (both of these are easy to draw).
HallsofIvy
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- Jan 27, 2012
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x goes from 0 to 2. For each x, y goes from 0 to [imath]4- x^2[/imath]. [imath]y= 4- x^2[/imath] is a parabola, opening downward, with vertex at (0, 4) and crossing the x-axis at x= 2 and x= -2. Since x is between 0 and 2, for this problem we only have the part of the parabola in the first quadrant.
To change the order of integration, note that y goes from 0 to 4. For each y, x goes from 0 to the graph, [imath]y= 4- x^2[/imath] so [imath]x^2= 4- y[/imath], and, since we are in the first quadrant, [imath]x= \sqrt{4- y}[/imath].
The integral is
[math]\int_0^4\int_0^{\sqrt{4- y}}\frac{xe^{2y}}{4- y} dxdy[/math][math]=\left[\frac{1}{2}\int_0^4 \frac{x^2e^{2y}}{4- y}dy\right]_0^{\sqrt{4- y}}[/math][math]= \frac{1}{2}\int_0^4 \frac{(4- y)e^{2y}}{4- y}dy[/math][math]= \frac{1}{2}\int_0^4 e^{2y}dy[/math]
That should be easy now.
To change the order of integration, note that y goes from 0 to 4. For each y, x goes from 0 to the graph, [imath]y= 4- x^2[/imath] so [imath]x^2= 4- y[/imath], and, since we are in the first quadrant, [imath]x= \sqrt{4- y}[/imath].
The integral is
[math]\int_0^4\int_0^{\sqrt{4- y}}\frac{xe^{2y}}{4- y} dxdy[/math][math]=\left[\frac{1}{2}\int_0^4 \frac{x^2e^{2y}}{4- y}dy\right]_0^{\sqrt{4- y}}[/math][math]= \frac{1}{2}\int_0^4 \frac{(4- y)e^{2y}}{4- y}dy[/math][math]= \frac{1}{2}\int_0^4 e^{2y}dy[/math]
That should be easy now.