let R be the region bounded by the x-axis, the graph of y=sqrt(x+1), and the line x=3.
Find the area of the region R
Well, it seems like you're asking me to calculate the area of region R. But hey, who needs numbers and calculations when you can just ask a clown for a silly answer? So, here's my clownish response:
The area of region R can be best measured by the number of laughs it brings. So, I'd say it's approximately equal to the number of people tickled by a feather multiplied by the sound of a clown horn. It's an unconventional unit, you see.
But if you're looking for a more serious answer, you can calculate it by integrating the function y = √(x+1) from x = 0 to x = 3. That's the standard way to find the area under a curve. Good luck with the math!
To find the area of the region R, we need to find the area between the x-axis, y = sqrt(x+1), and the vertical line x = 3.
First, we need to find the x-coordinate of the point where the graph y = sqrt(x+1) intersects with the line x = 3.
To do this, we set x + 1 = 9 (since x = 3), and solve for x:
x + 1 = 9
x = 8
So the point of intersection is (8, sqrt(8+1)).
Next, we need to find the area of the portion of the region R to the left of x = 3.
This area can be found by integrating y = sqrt(x+1) from x = 0 to x = 3.
∫[0 to 3] sqrt(x+1) dx
Using the power rule of integration, we can rewrite the integral as:
∫[0 to 3] (x+1)^(1/2) dx
Integrating using the power rule, we get:
∫[0 to 3] (x+1)^(1/2) dx = [2/3 * (x+1)^(3/2)] evaluated from 0 to 3
Evaluating the integral, we get:
[2/3 * (3+1)^(3/2)] - [2/3 * (0+1)^(3/2)]
= [2/3 * 4^(3/2)] - [2/3 * 1^(3/2)]
= [2/3 * 8] - [2/3]
= 16/3 - 2/3
= 14/3
So, the area of the portion of the region R to the left of x = 3 is 14/3 square units.
Finally, we need to find the area of the triangle formed by the line x = 3, the x-axis, and the x-coordinate of the point of intersection.
The base of the triangle is the difference in the x-coordinates: 8 - 3 = 5 units.
The height of the triangle is the y-coordinate of the point of intersection: sqrt(8+1) = sqrt(9) = 3 units.
The area of the triangle is given by the formula: 1/2 * base * height.
So, the area of the triangle is 1/2 * 5 * 3 = 15/2 square units.
The total area of the region R is the sum of the areas of the triangle and the portion to the left of x = 3:
Total area = 14/3 + 15/2
= (28/6) + (45/6)
= 73/6
Therefore, the area of the region R is 73/6 square units.
To find the area of region R, we need to calculate the integral of the function that represents the top curve (y=sqrt(x+1)) minus the bottom curve (the x-axis) within the bounds of the region.
1. Let's start by identifying the points where the top curve and the bottom curve intersect. In this case, the graph of y=sqrt(x+1) intersects the x-axis when y=0. So, we need to find the value of x when y=0.
0 = sqrt(x+1)
Square both sides:
0^2 = (sqrt(x+1))^2
0 = x+1
x = -1
Therefore, the region R is bounded by x=-1, x=3.
2. Next, we need to set up the integral to find the area. Since the top curve is y=sqrt(x+1) and the bottom curve is the x-axis, the integral is:
Area = ∫[from -1 to 3] (sqrt(x+1) - 0) dx
= ∫[from -1 to 3] sqrt(x+1) dx
3. We can simplify the integral and solve it. Let's use a substitution method:
Let u = x+1
Then, du = dx
As the new variable is u, we need to change the bounds accordingly:
When x = -1, u = (-1)+1 = 0
When x = 3, u = 3+1 = 4
So, our integral becomes:
Area = ∫[from 0 to 4] sqrt(u) du
4. Integrate the function sqrt(u) with respect to u:
Area = ∫[from 0 to 4] u^(1/2) du
= [2/3 * u^(3/2)] [from 0 to 4]
= 2/3 * (4^(3/2) - 0^(3/2))
= 2/3 * (8 - 0)
= 2/3 * 8
= 16/3
= 5.33 (rounded to two decimal places)
Therefore, the area of the region R is 5.33 square units.
the curve starts at (-1,0)
area = [integral] (x+1)^(1/2) dx from -1 to 3
= (2/3)(x+1)^(3/2) from -1 to 3
= (2/3)(4)^(3/2)
= 16/3
check my arithmetic