Find the area of the indicated region. We suggest you graph the curves to check whether one is above the other or whether they cross, and that you use technology to check your answer.
enclosed by y = ex, y = 5, and the y-axis
Oh, I see what you're trying to do here! You want me to find the area of the region enclosed by the curve y = ex, the line y = 5, and the y-axis. Well, let me put on my mathematical clown nose and calculate it for you.
To find the area, we need to set up an integral. First, let's find the x-values where the curve y = ex intersects with the line y = 5.
Setting ex = 5 and solving for x, we get x = ln(5). So, that's our upper limit of integration for our integral.
Now, we need to find the lower limit of integration. Since we're dealing with the y-axis, we know that x = 0 at the y-axis. So, our lower limit is 0.
Now, let's set up the integral to find the area:
∫[0, ln(5)] ex dx
We can integrate this using the power rule of integration. Integrating ex gives us ex, and our integral becomes:
[ex] from 0 to ln(5)
Plugging in our limits of integration, we get:
e^(ln(5)) - e^0
Simplifying further, we have:
5 - 1
And the final answer is:
4 square units of clown fun!
Remember to double-check my calculations using some mathematical technology, just to be safe. Happy calculating!
To find the area of the region enclosed by the curves y = ex, y = 5, and the y-axis, we can follow these steps:
Step 1: Graph the curves.
Start by graphing the curves y = ex and y = 5 on a coordinate plane. The y-axis is already given and we know it is vertical. The graph should look like this:
x-axis
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____|/___|____ y-axis
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The curve y = ex represents an exponential curve that starts at (0,1) and increases as x increases. The curve y = 5 is a horizontal line that intersects the y-axis at y = 5.
Step 2: Determine the points of intersection.
Next, find the points where the curves y = ex and y = 5 intersect. Since y = ex increases and y = 5 is constant, they will intersect at some x-value. Set the two functions equal to each other and solve for x:
ex = 5
Taking the natural logarithm (ln) of both sides:
ln(ex) = ln(5)
x = ln(5)
So the curves intersect at x = ln(5).
Step 3: Calculate the area.
To calculate the area, we need to integrate the difference between the two curves with respect to x from x = 0 to x = ln(5).
The equation of the region can be expressed as:
A = ∫[0, ln(5)] (5 - ex) dx
Integrating the expression:
A = [5x - e^x] from 0 to ln(5)
Evaluating the integral:
A = [5ln(5) - e^(ln(5))] - [5(0) - e^0]
A = 5ln(5) - 5 + 1
A = 5ln(5) - 4
So the area of the region enclosed by y = ex, y = 5, and the y-axis is 5ln(5) - 4.
To find the area of the region enclosed by the curves y = ex, y = 5, and the y-axis, we can start by graphing the curves and determining the points where they intersect.
1. Graph the curves: Go to a graphing calculator, software, or an online graphing tool. Plot the curves y = ex and y = 5. Make sure to set the appropriate range for the y-axis so that you can see the region clearly.
2. Determine the points of intersection: Look for the points where the curve y = ex intersects with the curve y = 5. These points will serve as the boundaries for the region.
3. Calculate the area: The area of the region can be calculated by finding the definite integral of the difference between the two curves over the interval of intersection. In this case, the integral will be taken with respect to y, as we want to find the area between the curves and the y-axis.
∫[lower bound, upper bound] (5 - ex) dy
4. Use technology to evaluate the integral: Input the definite integral into a symbolic math calculator, computer software, or an online integral calculator to find the value of the integral. The result will give you the area of the region enclosed by the curves.
Following these steps will help you find the area of the indicated region with the help of graphing technology and integral calculus.
So, having drawn the graphs, it should be clear that the area can be considered as a collection of vertical strips of width dx, making it
A = ∫[0,ln5] (5-e^x) dx = 5ln5 - 4
or, as a set of horizontal strips of width dy,
A = ∫[1,5] lny dy = 5ln5 - 4