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How do you evaluate the definite integral #int (x+6)^2dx# from [0,4]?

Answer 1

Aurora Alvarado

#784/3 = 261.333#

#int (x+6)^2 = (x+6)^3 /3 | 0, 4 = (4+6)^3 /3 - 6^3 / 3 = 261.333 #

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Answer 2

Adam Adkins

To evaluate the definite integral ∫(x + 6)^2 dx from 0 to 4, you first expand the integrand, integrate the resulting polynomial function, and then apply the limits of integration. The steps are as follows:

Expand the integrand: (x + 6)^2 = x^2 + 12x + 36.
Integrate the expanded polynomial term by term: ∫(x^2 + 12x + 36) dx = (1/3)x^3 + 6x^2 + 36x + C.
Evaluate the antiderivative at the upper and lower limits of integration: F(4) - F(0), where F(x) = (1/3)x^3 + 6x^2 + 36x.
Substitute the upper limit: F(4) = (1/3)(4)^3 + 6(4)^2 + 36(4).
Substitute the lower limit: F(0) = (1/3)(0)^3 + 6(0)^2 + 36(0).
Subtract the value at the lower limit from the value at the upper limit: F(4) - F(0).

After performing these calculations, you will obtain the value of the definite integral ∫(x + 6)^2 dx from 0 to 4.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.