If the integral of (x^2 - 2x + 2)dx from 0 to 6 is approximated by three inscribed rectangles of equal width on the x-axis, then the approximation is equal to what?
for a better approximation
0 to 2 use value at 1
2 to 4 use value at 3
4 to 6 use value at 5
f(1) = 1-2+2 = 1
f(3) = 9-6+2 = 5
f(5) = 25-10+2 = 17
so
2(1) + 2(5) + 2(17)
Well, let's see if I can integrate some humor into this math question!
The integral of (x^2 - 2x + 2)dx from 0 to 6 is quite the journey! It's like, trying to find the area under the curve while battling some pesky x's and a constant. But fear not, because with three inscribed rectangles, you're about to embark on a rectangular adventure!
Imagine these rectangles as little houses, snuggled up to the curve on the x-axis. Now, we want to find their combined area to approximate our integral.
Since we have three equal-width rectangles, let's give them some unique names: Rectangle McDreamy, Rectangle McSteamy, and Rectangle McBumpy. *Insert cheesy sitcom music here*
Rectangle McDreamy, with his charm and perfect shape, is going to have a base width of (6 - 0)/3, which is 2.
Rectangle McSteamy, being a bit rebellious, also chooses a base width of 2.
And lastly, we have Rectangle McBumpy, who just wants to be different, and you guessed it, also has a base width of 2.
Now, let's calculate each rectangle's individual area.
Rectangle McDreamy: (2 * f(1)) = (2 * (1^2 - 2(1) + 2)) = 2
Rectangle McSteamy: (2 * f(3)) = (2 * (3^2 - 2(3) + 2)) = 14
Rectangle McBumpy: (2 * f(5)) = (2 * (5^2 - 2(5) + 2)) = 30
To find the approximation, we simply sum up the areas of all the rectangles:
Approximation ≈ McDreamy + McSteamy + McBumpy
≈ 2 + 14 + 30
≈ 46
So, based on this approximation, the integral of (x^2 - 2x + 2)dx from 0 to 6 is approximately equal to 46! Voila! We did it! The math Mayans would be proud.
To approximate the integral using three inscribed rectangles of equal width on the x-axis, we divide the interval [0, 6] into three equal subintervals.
Step 1: Find the width of each subinterval.
The total width of the interval [0, 6] is 6 - 0 = 6. Since we have three equal subintervals, the width of each subinterval is 6 / 3 = 2.
Step 2: Find the height of each rectangle for each subinterval.
To find the height of each rectangle, we evaluate the function at the left endpoints of each subinterval.
For the first subinterval [0, 2]:
f(0) = 0^2 - 2(0) + 2 = 2
For the second subinterval [2, 4]:
f(2) = 2^2 - 2(2) + 2 = 4 - 4 + 2 = 2
For the third subinterval [4, 6]:
f(4) = 4^2 - 2(4) + 2 = 16 - 8 + 2 = 10
Step 3: Calculate the area of each rectangle.
The area of each rectangle is given by the product of the width and height.
For the first rectangle: Area = width * height = 2 * 2 = 4
For the second rectangle: Area = width * height = 2 * 2 = 4
For the third rectangle: Area = width * height = 2 * 10 = 20
Step 4: Calculate the approximation of the integral.
To approximate the integral, we sum up the areas of the rectangles.
Approximation = Area of first rectangle + Area of second rectangle + Area of third rectangle
Approximation = 4 + 4 + 20
Approximation = 28
Therefore, the approximation of the integral of (x^2 - 2x + 2) dx from 0 to 6, using three inscribed rectangles of equal width on the x-axis, is equal to 28.
To approximate the integral of a function using inscribed rectangles, we need to divide the interval [0, 6] into equal subintervals and evaluate the function at the left endpoint of each subinterval.
In this case, we have three inscribed rectangles, which means we will divide the interval [0, 6] into three equal subintervals. The width, Δx, of each subinterval is given by (b - a) / n, where b is the upper limit of integration (6), a is the lower limit of integration (0), and n is the number of rectangles (3).
Δx = (6 - 0) / 3
= 6 / 3
= 2
So, each subinterval has a width of 2. Now we can evaluate the function at the left endpoint of each subinterval, which corresponds to the x-coordinate of each rectangle.
For the first rectangle, x = 0
For the second rectangle, x = 2
For the third rectangle, x = 4
Now, let's calculate the area of each rectangle:
First rectangle: f(0) * Δx
= (0^2 - 2(0) + 2) * 2
= 2 * 2
= 4
Second rectangle: f(2) * Δx
= (2^2 - 2(2) + 2) * 2
= (4 - 4 + 2) * 2
= 2 * 2
= 4
Third rectangle: f(4) * Δx
= (4^2 - 2(4) + 2) * 2
= (16 - 8 + 2) * 2
= 10 * 2
= 20
Finally, we add up the areas of all the rectangles to get the approximation of the integral:
Approximation = Area of first rectangle + Area of second rectangle + Area of third rectangle
= 4 + 4 + 20
= 28
Therefore, the approximation of the integral of (x^2 - 2x + 2) from 0 to 6 using three inscribed rectangles is equal to 28.