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How do you evaluate the definite integral #int x^2sqrt(1-x^3)# from #[0,1]#?

Answer 1

Ezra Adams

#= 2/9#

spot the pattern

#d/dx (f^(x))^n = n (f(x))^(n-1) f'(x)#

#int_0^1 x^2sqrt(1-x^3) dx#

#= int_0^1 d/dx(-2/9 root3(1-x^3) )dx#

#= -2/9[ root3(1-x^3) ]_0^1#

#= -2/9 (0 - 1) = 2/9#

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Answer 2

Scarlett Allison

To evaluate the definite integral (\int_0^1 x^2\sqrt{1-x^3} , dx), you can use the substitution method. Let (u = 1 - x^3), then (du = -3x^2 , dx). Solving for (dx), we get (dx = -\frac{1}{3x^2} , du). Substituting these into the integral, we get:

[\begin{split}\int_0^1 x^2\sqrt{1-x^3} , dx &= \int_{u(0)}^{u(1)} -\frac{1}{3x^2} \cdot u^{1/2} , du \ &= -\frac{1}{3} \int_{u(0)}^{u(1)} \frac{u^{1/2}}{x^2} , du \ &= -\frac{1}{3} \int_{u(0)}^{u(1)} \frac{u^{1/2}}{\sqrt{1 - u}} , du\end{split}]

Now we can integrate this using a basic integral formula for (\int \frac{u^{1/2}}{\sqrt{1 - u}} , du), and evaluate it from (u = 0) to (u = 1). After finding the antiderivative, you can evaluate it at (u = 1) and subtract the value at (u = 0). This will give you the result of the definite integral.

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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.