Decide whether to integrate with respect to x or y. Find the area of the region.
y=e^2x y=e^6x x=1
I have no idea what to do
If x=0 then e^(2*0)=e^(6*0)
If 0<x<1 then e^2x<e^6x.
The area=integral from 0 to 1
(e^6x-e^2x)dx =
(1/6)e^6x-(1/2)e^2x from 0 to 1 =
(1/6)e^6-(1/2)e^2-(1/6)+(1/2)
The computer homework program says that is incorrect. Any other suggestions?
Your think your computer program is incorrect
I think in "anonymous' " last line he/she should have had
(1/6)e^6-(1/2)e^2-(1/6) - (1/2))
or
e^6/6 - e^2/2 + 1/3
I wrote ...-(1/6)+(1/2)
...-(1/6)+(1/2)=...+1/3.
You mean -((1/6)-(1/2))
Yay!! Thank you both very much! It worked.
To find the area of a region bounded by curves, we need to set up and evaluate an integral. In this case, we have two curves defined by their equations: y = e^2x and y = e^6x. Additionally, there is a vertical boundary line at x = 1.
To determine which variable to integrate with respect to, we should compare the equations of the curves and analyze their relationship. In this case, notice that both curves are in terms of y as a function of x. Therefore, it is more convenient to integrate with respect to y.
To find the bounds of the integral, we need to determine the y-values at which the curves intersect and the y-value at the upper boundary. Set the equations of the curves equal to each other to find their intersection points:
e^2x = e^6x
Now, we can cancel out the e^2x term by dividing both sides of the equation by e^2x:
1 = e^4x
To solve for x, we take the natural logarithm (ln) of both sides:
ln(1) = ln(e^4x)
0 = 4x
Solving for x, we find x = 0. Hence, the curves intersect at x = 0.
Next, we can find the y-value of the upper boundary by substituting x = 1 into either of the equations:
y = e^2(1)
y = e^2
So, the upper boundary is y = e^2.
Now that we have determined the bounds of the integral, we can set up the integral to find the area. Recall that the area bounded by the curves can be found by integrating the difference between the upper curve and the lower curve.
The integral to find the area is:
∫[lower bound, upper bound] (upper curve - lower curve) dy
In this case, the lower curve is y = e^2x, and the upper curve is y = e^6x. Therefore, the integral becomes:
∫[e^2, e^2] (e^6x - e^2x) dy
Simplifying the integral:
∫[e^2, e^2] (e^6x - e^2x) dy
∫[e^2, e^2] (d/dy)(e^6x - e^2x) dy
∫[e^2, e^2] d/dy [e^6x - e^2x] dy
Since we are integrating with respect to y, we need to change the variable of integration from x to y. To do this, we can solve either of the two curves for x:
From y = e^2x, we have: x = ln(y)/2
From y = e^6x, we have: x = ln(y)/6
Let's substitute the expression for x in terms of y into the integral:
∫[e^2, e^2] (d/dy) [e^6(ln(y)/6) - e^2(ln(y)/2)] dy
Now, we can calculate this integral to find the area of the region.