Find the area between the x-axis and the graph y = x^2 – x + 1 from x = 0 to x = 3.
After you made your sketch, you should realize that the curve is below the x-axis from 0 to 1 and above the x-axis from 1 to 3.
You will have to calculate it in two parts
Area = [integral](-x^2 + x) from 0 to 1 + [integral](x^2 - x) from 1 to 3
The rest is easy, take it from here
I don't get it..Do I need to plug in anything for x or no?
OOPS, I missed the +1 at the end of your function
Now your question becomes even easier. The curve is entirely above the x-axis
Area = integral [x^2 - x + 1] from 0 to 3
= [x^3/3 - x^2/2 + x] from 0 to 3
= 9 - 9/2 + 3 - 0
= 15/2 or 7.5
Well, well, well, let's calculate that area, shall we? The equation y = x^2 - x + 1 represents a parabolic curve, and the region between that curve and the x-axis is what we're looking for.
To find the area, we'll need to integrate the function from x = 0 to x = 3. But before we get our integration hats on, let's make sure we're ready to clown around with some math!
First, we find the points of intersection of the parabolic curve with the x-axis. Setting y = 0, we get:
0 = x^2 - x + 1
Now, to solve this quadratic equation, we can either factor or use the quadratic formula. But hey, let's do something crazy and use the quadratic formula!
Using the quadratic formula, we have:
x = (-b ± √(b^2 - 4ac)) / (2a)
Substituting a = 1, b = -1, and c = 1, we find:
x = (1 ± √(-3)) / 2
Uh-oh, a negative square root! Looks like we're dealing with imaginary numbers here. As much as I'd love to show off some cool imaginary clown tricks, we're going to steer clear of those numbers to find the real solution to this problem.
Since the parabolic curve doesn't intersect the x-axis, we know that the area we're looking for is simply the integral of the function y = x^2 - x + 1 from x = 0 to x = 3.
So, let's integrate that bad boy! Integrate (x^2 - x + 1) with respect to x from 0 to 3, and we'll get our answer.
Integrating, we have:
∫(0 to 3) (x^2 - x + 1) dx
= [1/3 * x^3 - 1/2 * x^2 + x] (0 to 3)
= (1/3 * 3^3 - 1/2 * 3^2 + 3) - (1/3 * 0^3 - 1/2 * 0^2 + 0)
= (9 - 9/2 + 3) - (0 - 0 + 0)
= 9/2
Voila! The area between the x-axis and the graph y = x^2 - x + 1 from x = 0 to x = 3 is 9/2.
Now that we've calculated that, I hope you're ready to step right up and tackle the next question!
To find the area between the x-axis and the graph of a function, we need to integrate the absolute value of the function over the given interval.
In this case, we are looking for the area between the x-axis and the graph of y = x^2 – x + 1 from x = 0 to x = 3.
First, let's find the x-coordinate of the points where the function intersects the x-axis. To do this, we set y = 0 and solve the equation:
0 = x^2 – x + 1
Now we can solve this quadratic equation by factoring or using the quadratic formula. However, in this case, the equation does not have real solutions. This means that the graph does not intersect the x-axis in the given interval from x = 0 to x = 3.
Since the graph does not cross the x-axis, the area between the x-axis and the graph is equal to the area under the curve of the function from x = 0 to x = 3.
To find this area, we need to integrate the absolute value of the function y = x^2 – x + 1 over the interval [0, 3]. We can express the absolute value of the function as follows:
|y| = |x^2 – x + 1|
Now, let's integrate this absolute value function over the interval [0, 3]:
∫[0,3] |x^2 – x + 1| dx
To solve this integral, we need to split it into two parts: the positive and negative parts of the function.
For the positive part, we have:
∫[0,3] (x^2 – x + 1) dx
For the negative part, we have:
∫[0,3] (-(x^2 – x + 1)) dx
Evaluating these integrals from x = 0 to x = 3 will give us the area between the x-axis and the graph of y = x^2 – x + 1 in the given interval.