Verify that F(x) is the antiderivative of the integrand f(x) and use Part 2 of the Fundamental Theorem to evaluate the definite integral.
Integral 3x^2 dx from 1 to 4 , F(x) = x^3 + 2
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To verify that F(x) is the antiderivative of the integrand f(x), we can differentiate F(x) and check if it matches the original integrand.
Taking the derivative of F(x), we get:
F'(x) = d/dx (x^3 + 2)
= 3x^2 + 0
= 3x^2
The derivative of F(x) is indeed equal to the integrand f(x) = 3x^2.
Now, using Part 2 of the Fundamental Theorem of Calculus, we can evaluate the definite integral using the antiderivative F(x).
∫[1 to 4] (3x^2) dx = F(x) | [1 to 4]
= F(4) - F(1)
= (4^3 + 2) - (1^3 + 2)
= (64 + 2) - (1 + 2)
= 66 - 3
= 63
Therefore, the value of the definite integral ∫[1 to 4] (3x^2) dx is 63.
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