Find the area of the shaded region
y= 2x/(x^2+4)
from x= -2 and x= 2
Note: People have been telling me that the two lines do not intersect but they do there has to be an answer. please help me.
Thanks
ln(x^2 + 4) = 0
well of course the lines x=2 and x=-2 do not intersect; they are parallel vertical lines. Not sure what that has to do with anything; the just set bounds to the region.
Since y is an odd function, and the bounds are symmetric about the y-axis, the integral will be zero.
Let u = x^2+4 and you just have
∫ du/u
Note that the limits of integration are now 8 and 8, so the integral is zero -- the area is symmetric above and below the x-axis.
To find the area of the shaded region, we first need to determine the points of intersection between the function y = 2x/(x^2 + 4) and the horizontal lines.
Setting y equal to zero, we have:
0 = 2x/(x^2 + 4)
Now, since division by zero is undefined, we know that x^2 + 4 cannot be equal to zero. Thus, there are no x-values for which y is equal to zero. This implies that the function does not intersect the x-axis, and therefore, there is no shaded region to find the area of in this case.
Hence, the people were correct in stating that the two lines do not intersect.
To find the area of the shaded region, we first need to determine the points where the two lines intersect. In this case, you are correct that there are points of intersection between the line y = 2x/(x^2+4) and the x-axis. Let's proceed with finding those points.
Setting y = 0, we have:
0 = 2x/(x^2+4)
Cross-multiplying, we get:
2x = 0
This equation is satisfied when x = 0, meaning the line intersects the x-axis at x = 0.
Now, we can find the x-values where the line intersects the boundaries of the shaded region, which are x = -2 and x = 2.
Let's calculate the area of the shaded region step by step:
1. Find the area from x = -2 to x = 0:
Let's integrate the function from x = -2 to x = 0:
∫(2x/(x^2+4)) dx
To integrate this function, we can use the substitution u = x^2 + 4, du = 2x dx:
∫(1/u) du
= ln|u| + C
= ln|x^2 + 4| + C
Evaluating the integral from x = -2 to x = 0:
[ln|(0)^2 + 4| + C] - [ln|(-2)^2 + 4| + C]
= ln|4| - ln|8|
= ln(4/8)
= ln(1/2)
= -ln(2)
2. Find the area from x = 0 to x = 2:
Let's integrate the function from x = 0 to x = 2:
∫(2x/(x^2+4)) dx
Using the same substitution as before, u = x^2 + 4, du = 2x dx:
∫(1/u) du
= ln|u| + C
= ln|x^2 + 4| + C
Evaluating the integral from x = 0 to x = 2:
[ln|(2)^2 + 4| + C] - [ln|(0)^2 + 4| + C]
= ln|8| - ln|4|
= ln(8/4)
= ln(2)
3. Calculate the absolute value of the difference in these two area calculations:
|-ln(2) - (-ln(2))|
= |-ln(2) + ln(2)|
= |0|
= 0
Therefore, the area of the shaded region is 0. This means that the shaded region between the curve y = 2x/(x^2+4) and the x-axis from x = -2 to x = 2 has no area, as it collapses into a line.