definite Integrals (using fundamental Theorem)
Evaluate
from -1 to 2(x^2 - 4x)dx
[x^3/3 - 2 x^2] @ 2
MINUS
[x^3/3 - 2 x^2] @ -1
= [8/3 - 8] - [-1/3 -2]
= -16/3 + 7/3 = -3
Thank you so much. I had the integral part, but was off on the rest. :)
To evaluate the definite integral ∫[-1 to 2] (x^2 - 4x) dx using the Fundamental Theorem of Calculus, we will follow these steps:
Step 1: Find the antiderivative of the integrand.
Step 2: Substitute the upper and lower limits into the antiderivative.
Step 3: Evaluate the integral using the Fundamental Theorem of Calculus.
Let's break down each step in detail:
Step 1: Find the antiderivative of the integrand.
To find the antiderivative, we need to reverse the process of differentiation. The antiderivative of the function f(x) is denoted as F(x) and satisfies d/dx[F(x)] = f(x).
In this case, we have:
f(x) = x^2 - 4x.
To find F(x), we integrate each term separately:
∫(x^2 - 4x) dx = ∫ x^2 dx - ∫ 4x dx.
Integrating each term:
= (1/3) x^3 - 2x^2 + C,
where C is the constant of integration.
Step 2: Substitute the upper and lower limits into the antiderivative.
The definite integral involves finding the difference between the antiderivative evaluated at the upper limit and the antiderivative evaluated at the lower limit.
Substituting the upper limit (2) into the antiderivative:
F(2) = (1/3) (2)^3 - 2(2)^2 + C = (8/3) - 8 + C.
Substituting the lower limit (-1) into the antiderivative:
F(-1) = (1/3) (-1)^3 - 2(-1)^2 + C = (-1/3) - 2 + C.
Step 3: Evaluate the integral using the Fundamental Theorem of Calculus.
To evaluate the definite integral, we subtract the antiderivative at the lower limit from the antiderivative at the upper limit:
∫[-1 to 2] (x^2 - 4x) dx = F(2) - F(-1) = [(8/3) - 8 + C] - [(-1/3) - 2 + C].
Simplifying further:
= (8/3) - 8 - (-1/3) + 2 + C - C.
Finally, simplifying the expression:
= -6 + (7/3).
Therefore, the value of the definite integral from -1 to 2 of (x^2 - 4x) dx using the Fundamental Theorem of Calculus is -6 + (7/3).