Finding area under a curve

From [1,3]
2x^2 -4x +1

is your curve y = 2x^2 - 4x + 1 ?

Is your region between the curve and the x-axis within your stated domain ?
Did you notice that the parabola has an x-intercept within your domain?

its from 1 to 3 on the x axis including when it crosses underneath, you subtract that (its like negative area)

To find the area under a curve, you can use integration. In this case, let's integrate the function 2x^2 - 4x + 1 with respect to x over the interval [1, 3].

First, let's find the antiderivative (or integral) of the function. We can do this by applying the power rule of integration. The power rule states that ∫ x^n dx = (1/n+1) * x^(n+1) + C, where C is the constant of integration.

Applying the power rule to each term of the function individually, we get:
∫ 2x^2 dx = (2/3) * x^3 + C1 (integral of 2x^2)
∫ -4x dx = -2x^2 + C2 (integral of -4x)
∫ 1 dx = x + C3 (integral of 1)

Adding these integrals together, we get the overall antiderivative:
∫ (2x^2 - 4x + 1) dx = (2/3) * x^3 - 2x^2 + x + C

Next, we will evaluate this antiderivative at the upper and lower limits of the interval [1, 3]. Substituting x = 3, we have:
((2/3) * (3)^3 - 2(3)^2 + 3) - ((2/3) * (1)^3 - 2(1)^2 + 1)

Simplifying this expression, we get:
((2/3) * 27 - 18 + 3) - ((2/3) * 1 - 2 + 1)
= (54/3 - 18 + 3) - (2/3 - 2 + 1)
= (54/3 - 15) - (-4/3)
= 18 - (-4/3)
= 18 + 4/3
= (54/3 + 4/3)
= 58/3

Therefore, the area under the curve of the function 2x^2 - 4x + 1 over the interval [1, 3] is 58/3.