How do you find the definite integral for: #root3(x^2)# for the intervals [-1,1]?

Answer 1

#int_(-1)^1 root(3)(x^2)dx =6/5#

We know that:

#root(3)(x^2) = x^(2/3)#

and that in general:

#int x^adx = x^(a+1)/(a+1) + C#

So, applying the fundamental theorem of calculus:

#int_(-1)^1 root(3)(x^2)dx = int_(-1)^1 x^(2/3)dx#
#int_(-1)^1 root(3)(x^2)dx =[x^(2/3+1)/(2/3+1)]_(-1)^1#
#int_(-1)^1 root(3)(x^2)dx =[3/5 x^(5/3)]_(-1)^1#
#int_(-1)^1 root(3)(x^2)dx =3/5- (-3/5)#
#int_(-1)^1 root(3)(x^2)dx =6/5#
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Answer 2

To find the definite integral of ( \sqrt{3}x^2 ) over the interval ([-1, 1]), you integrate the function with respect to (x) and then evaluate the result at the upper and lower bounds of the interval, and finally, subtract the lower bound value from the upper bound value.

First, integrate ( \sqrt{3}x^2 ) with respect to (x):

[ \int_{-1}^{1} \sqrt{3}x^2 , dx ]

[ = \sqrt{3} \int_{-1}^{1} x^2 , dx ]

[ = \sqrt{3} \left[ \frac{x^3}{3} \right]_{-1}^{1} ]

Now, plug in the upper and lower bounds:

[ = \sqrt{3} \left( \frac{1^3}{3} - \frac{(-1)^3}{3} \right) ]

[ = \sqrt{3} \left( \frac{1}{3} - \frac{-1}{3} \right) ]

[ = \sqrt{3} \left( \frac{1 + 1}{3} \right) ]

[ = \sqrt{3} \left( \frac{2}{3} \right) ]

[ = \frac{2\sqrt{3}}{3} ]

So, the definite integral of ( \sqrt{3}x^2 ) over the interval ([-1, 1]) is ( \frac{2\sqrt{3}}{3} ).

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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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