Find the area of the region enclosed by these graphs and the vertical lines x = 3 and x = 5
f(x)=x^2+5 g(x)=1
ʃ (x^2+5)-1)dx
ʃ (x^2+4)dx
x^3/3 + 4x from 3 to 5
5^3/3 +4(5) -(3^3/3 + 4(3))
125/3 +20 -27/3 -12
=122/3
To find the area of the region enclosed by the graphs and the vertical lines x = 3 and x = 5, we need to calculate the definite integral of the difference between the two functions over the given interval.
First, let's find the points where the two functions intersect:
Setting f(x) = g(x), we have: x^2 + 5 = 1
Simplifying, we get: x^2 = -4
Taking the square root of both sides, we get: x = ±√(-4)
Since the square root of a negative number is undefined in the real number system, there are no real intersecting points between the two functions. Therefore, the region enclosed by the graphs and the vertical lines x = 3 and x = 5 is empty and has an area of 0.
To find the area of the region enclosed by the graphs of f(x) = x^2 + 5 and g(x) = 1, and the vertical lines x = 3 and x = 5, we need to calculate the definite integral of the difference between the two functions between these limits.
First, let's plot the graphs of the functions f(x) = x^2 + 5 and g(x) = 1 on a coordinate system:
Graph of f(x) = x^2 + 5:
I│
│
6│ .........
│ : :
5│.....:.........:...........
│ : :
1│ ... : :
───────────────────────
3 5
Graph of g(x) = 1:
I│
│
2│ .........
│ : :
1│.....:.........:...........
│ : :
0│ ... : :
───────────────────────
3 5
The region enclosed by the graphs and the vertical lines x = 3 and x = 5 is the area between the graphs above the x-axis. It is a rectangle with the base along the x-axis and the height equal to the difference between the two functions.
The first step is to subtract the two functions to find the height of the rectangle at any given x:
h(x) = f(x) - g(x)
= (x^2 + 5) - 1
= x^2 + 4
Now, we can calculate the area by integrating the height function between x = 3 and x = 5:
Area = ∫[3, 5] h(x) dx
= ∫[3, 5] (x^2 + 4) dx
Evaluating this definite integral will give us the area of the region enclosed by the graphs and the vertical lines x = 3 and x = 5.