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How do you evaluate #int (x^2(x^3 + 1)^3)# for [0, 1]?

Answer 1

Aurora Alvarado

#5/4#

#x^3+1=t => 3x^2dx=dt => x^2dx= dt/3#

#t_1 = 0^3+1 = 1# #t_2 = 1^3+1 = 2#

#I = int_0^1 x^2(x^3+1)^3dx = int_1^2 t^3 dt/3 = 1/3 t^4/4= 1/12 t^4#

#I = 1/12 (2^4-1^4) = 1/12 (16-1) = 15/12 = 5/4#

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Answer 2

Isabella Ackerman

To evaluate (\int_{0}^{1} x^2(x^3 + 1)^3) for the interval ([0, 1]), we first integrate the function with respect to (x) and then evaluate the result at the upper and lower limits of integration. This involves using integration techniques such as substitution and power rule.

After integrating, we substitute the upper limit of integration (1) into the antiderivative and subtract the result of substituting the lower limit of integration (0) into the antiderivative.

So, the steps are as follows:

Find the antiderivative of (x^2(x^3 + 1)^3) with respect to (x).
Substitute (x = 1) into the antiderivative.
Substitute (x = 0) into the antiderivative.
Subtract the result from step 3 from the result from step 2.

Following these steps will give us the value of the definite integral over the interval ([0, 1]).

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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.