How do you find the definite integral of #x^2 * sqrt (x^3 + 1) * dx# in the interval #[1,2]#?

Answer 1
We wish to evaluate #int_1^2 x^2sqrt(x^3+1)dx#.
Let #u = x^3 + 1#. Then #du = 3x^2dx# and #dx = (du)/(3x^2)#. Change the bounds of integration.
#I = int_2^9 x^2sqrt(u) * (du)/(3x^2)#
#I = 1/3int_2^9 sqrt(u)#
# I = 1/3[2/3u^(3/2)]_2^9#
#I = 2/9(9)^(3/2) - 2/9(2)^(3/2) = 6 - 4/9sqrt(2) ~~ 5.37#

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Answer 2

To find the definite integral of (x^2 \cdot \sqrt{x^3 + 1}) with respect to (x) over the interval ([1,2]), you can use integration by substitution. Let (u = x^3 + 1), then (du = 3x^2 , dx). Rearranging, we have (dx = \frac{du}{3x^2}).

Substituting these into the integral, we get:

[\int_{1}^{2} x^2 \sqrt{x^3 + 1} , dx = \frac{1}{3} \int_{2}^{9} \sqrt{u} , du.]

Integrating (\sqrt{u}) with respect to (u), we get:

[\frac{1}{3} \left[\frac{2}{3}u^{\frac{3}{2}}\right]_{2}^{9} = \frac{2}{9}(9^{\frac{3}{2}} - 2^{\frac{3}{2}}).]

Finally, evaluating this expression, we find:

[\frac{2}{9}(27 - 2\sqrt{2}).]

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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.

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.

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.

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.

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