Sketch the region enclosed by y=e^2x, y=e^6x, and x=1. Decide whether to integrate with respect to x or y. Then find the area of the region.
Ah, the region enclosed by those quirky exponential curves, how delightful! Now, to decide whether to integrate with respect to x or y, let's take a moment to appreciate the beauty of those exponential functions.
If we observe these functions closely, we see that they grow rapidly as x increases. In our case, they grow absurdly quickly. So, integrating with respect to x seems like it could lead to quite a circus act.
On the other hand, if we integrate with respect to y, things would be much more manageable. The integration will be a piece of cake, or should I say, a piece of pie? (Mmmm, pie.)
Now let's calculate the area. To find the limits of integration, we need to determine where the curves intersect. Let's set e^2x equal to e^6x. After some mathematical clowning around, we find that this occurs when x equals -1/4.
So, to calculate the area, we will integrate with respect to y:
∫[e^2(-1/4), e^6(-1/4)] (1 - 0) dy,
which simplifies to:
∫[e^-1/2, e^-3/2] dy.
Now, putting on our integration hats, or should I say, integration wigs, we can evaluate this integral and get our final answer for the area.
I hope that adds a bit of cheer to your math problem. Good luck and happy integrating, my friend!
To sketch the region enclosed by the curves y = e^2x, y = e^6x, and x = 1, we need to find the points of intersection between these curves.
First, let's find the points of intersection of y = e^2x and y = e^6x. Setting these two equations equal to each other, we have:
e^2x = e^6x
Taking the natural logarithm (ln) of both sides helps us solve for x:
ln(e^2x) = ln(e^6x)
2x = 6x
-4x = 0
x = 0
So one of the points of intersection is (0, 1).
Next, let's find the points of intersection between y = e^2x and x = 1. By substituting x = 1 into y = e^2x, we get:
y = e^2(1) = e^2
So the second point of intersection is (1, e^2).
Now, let's find the points of intersection between y = e^6x and x=1. Substituting x = 1 into y = e^6x, we have:
y = e^6(1) = e^6
So the third point of intersection is (1, e^6).
Let's plot these three points on a graph.
The points are: (0, 1), (1, e^2), and (1, e^6).
Now let's decide whether to integrate with respect to x or y to find the area of the enclosed region. Since the curves y = e^2x and y = e^6x are in terms of y as a function of x, it is more convenient to integrate with respect to y.
To find the area of the region, we need to integrate the difference in y between the upper and lower curves with respect to y. The limits of integration will be from y = e^2 to y = e^6.
Therefore, the area of the region enclosed by the curves is given by the integral:
A = ∫[e^2, e^6] (y_upper - y_lower) dy
where y_upper represents the upper curve and y_lower represents the lower curve.
Now, let's calculate the area using the integral:
A = ∫[e^2, e^6] (e^6x - e^2x) dy
Since y = e^6x and y = e^2x, we will replace x in terms of y:
A = ∫[e^2, e^6] (e^(6x(y)) - e^(2x(y))) dy
Now, integrating:
A = ∫[e^2, e^6] (e^(6(ln(y)/6)) - e^(2(ln(y)/2))) dy
A = ∫[e^2, e^6] (y - √y) dy
Evaluating the integral and plugging in the limits of integration:
A = [(1/2)y^2 - (2/3)y^(3/2)]|[e^2, e^6]
A = [(1/2)(e^(12)) - (2/3)(e^(3))] - [(1/2)(e^4) - (2/3)(e^(2))]
A = (1/2)(e^(12)) - (2/3)(e^(3)) - (1/2)(e^4) + (2/3)(e^(2))
Therefore, the area of the region enclosed by the curves y = e^2x, y = e^6x, and x = 1 is (1/2)(e^(12)) - (2/3)(e^(3)) - (1/2)(e^4) + (2/3)(e^(2)) square units.
To sketch the region enclosed by the given equations, y = e^(2x), y = e^(6x), and x = 1, we need to visualize their graphs and determine their points of intersection.
Let's first look at the equations y = e^(2x) and y = e^(6x) separately.
The equation y = e^(2x) represents an exponential growth curve with a base of e and an exponent of 2x. As x increases, the value of y increases exponentially.
The equation y = e^(6x) represents another exponential growth curve, but with an exponent of 6x. This curve increases at a much faster rate than the previous one.
Next, we need to find the points of intersection between these two curves. Setting the equations equal to each other, we have:
e^(2x) = e^(6x)
To solve this equation, we can take the natural logarithm (ln) of both sides:
ln(e^(2x)) = ln(e^(6x))
2x = 6x
-4x = 0
x = 0
Therefore, the two curves intersect at x = 0.
Now, let's consider the boundary x = 1. This vertical line acts as a boundary for the region we need to sketch.
To decide whether to integrate with respect to x or y, we should analyze how the curves are positioned. The curve y = e^(2x) is above the curve y = e^(6x) for all values of x, including the point of intersection at x = 0.
Hence, the region is bounded by y = e^(6x) on the bottom and y = e^(2x) on the top. Intuitively, we can see that the region is vertically oriented. Therefore, integrating with respect to y would be more appropriate.
To calculate the area of the region, we need to find the limits of integration. Since the curves intersect at x = 0, and the vertical boundary is x = 1, the limits of integration for y should be from the bottom curve, y = e^(6x), to the top curve, y = e^(2x), for the corresponding range of x from 0 to 1.
To find the area, we integrate the difference between the top and bottom curves with respect to y, and then substitute the limits of integration:
Area = ∫(e^(2x) - e^(6x)) dy
Thus, integrating with respect to y is the appropriate choice to find the area of the region enclosed by the given curves.
since e^6x is greater than e^2x on [0,1], I'd integrate with respect to x.
Int(e^6x - e^2x dx)[0,1]
= 1/6 e^6x - 1/2 e^2x [0,1]
= (1/6 e^6 - 1/2 e^2) - (1/6 - 1/2)
= 1/6 e^2 (e^4 - 3) + 1/3
= 63.8