Finding the Area of a Triangular Region in the First Quadrant

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In summary: Also note that the definite integral can be evaluated using the trapezoidal rule.In summary, the area of the triangular region in the first quadrant that is bounded above by the curve y=e^(2x), below by y=e^x, and on the right by the line x=ln(3) is 8.6.
  • #1
shadow5449
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This one has me stumped, I don't even know where to start.
-Find the area of the triangular region in the first quadrant that is bounded above by the curve y=e^(2x), below by y=e^x, and on the right by the line x=ln(3).
Thanks for any help
 
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  • #2
Try plotting the curves, and check definite integrals.
 
  • #3
How would I go about checking for definate integrals? The book we use is not very descriptive on this topic. Thanks.
 
  • #4
Plot the area, then find your integration limits. That's what he means by definite integral. Your final result will be a number, not an expression, representing the area in the shape.
 
  • #5
If you are doing a problem like this, then you certainly should be familiar with area given by integrals- that's the whole point of a problem like this. If you graph this, you will see that the two curves intersect at x= 0 and are bounded on the right by the vertical line x= 3. If you divide the area into thin rectangles, the height of each rectangle will be e3x- ex and the width will be "dx". Now, what integral is that?
 
  • #6
have you tried estimating it? i.e. can you show the area is less than 1,000?

or 10?
 
  • #7
Since this has been around for a couple of days now:

The graph of y= e3x is always above y= ex for x> 0. The two graphs cross at x= 0 and we are told that the are is bounded on the right by x= ln 3.

The area is [tex]\int_0^{ln 3}(e^{3x}- e^x)dx[/tex].

Can you do that integral?
 
  • #8
first draw the graph of all three fuctions. you'll get the picture. find the point of intersection of e^x, e^2x and x = ln3. this can be done easily...

one point is (0,1)...trivial solution
put x=ln3 in e^x and get y=3...(ln3,3)
put x=ln3 in e^2x and get y=9...(ln3,9)

now, for the next part you have to draw the graph.

integrate e^x limits 0 to 3 to get e^3 - 1
integrate e^2x limits 0 to 9 to get (e^9 - 1)/2

The area required is [(e^9 - 1)/2 - (e^3 - 1)]
note: area under graph of a function between a and b is its definite integral with a and b as lower and upper limits.
 

What is the formula for finding the area of a triangular region in the first quadrant?

The formula for finding the area of a triangular region in the first quadrant is A = 1/2 * base * height, where A is the area, base is the length of the triangle's base, and height is the height of the triangle from the base to the opposite vertex.

What are the steps for finding the area of a triangular region in the first quadrant?

The steps for finding the area of a triangular region in the first quadrant are:
1. Identify the length of the triangle's base and the height from the base to the opposite vertex.
2. Plug these values into the formula A = 1/2 * base * height.
3. Simplify the equation.
4. The resulting value is the area of the triangular region.

Can the area of a triangular region in the first quadrant be negative?

No, the area of a triangular region in the first quadrant cannot be negative. The area of a region is always a positive value, representing the amount of space within the boundaries of the region.

What units should be used for the base and height when finding the area of a triangular region in the first quadrant?

The base and height should be in the same units when finding the area of a triangular region in the first quadrant. For example, if the base is in inches, the height should also be in inches to ensure accurate results.

Can the formula for finding the area of a triangular region in the first quadrant be used for triangles in other quadrants?

No, the formula for finding the area of a triangular region in the first quadrant is specifically designed for triangles in the first quadrant. Triangles in other quadrants may require different formulas or adjustments to the existing formula to accurately calculate the area.

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