How do you evaluate the definite integral #int sqrt(2/x)dx# from [1,8]?

Answer 1

#8-2sqrt2#

firstly change the function of #x# from roots to powers.
#int_1^8sqrt(2/x)dx=int_1^8sqrt2(x)^(-1/2)dx#

Apply the power rule now, and assess.

#=sqrt2[x^(-1/2+1)/(-1/2+1)]_1^8#
#=sqrt2[2x^(1/2)]_1^8#
#=2sqrt2[sqrt8-sqrt1]#
#2sqrt2[2sqrt2-1]#
#=8-2sqrt2#
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Answer 2

To evaluate the definite integral [ \int_{1}^{8} \sqrt{\frac{2}{x}} , dx ]

We can rewrite the integral as: [ \int_{1}^{8} 2x^{-\frac{1}{2}} , dx ]

Now, integrate with respect to ( x ): [ \int 2x^{-\frac{1}{2}} , dx = 2 \int x^{-\frac{1}{2}} , dx ]

Integrate ( x^{-\frac{1}{2}} ) with respect to ( x ): [ \int x^{-\frac{1}{2}} , dx = \int x^{-\frac{1}{2}} \cdot x^1 , dx = \int x^{\frac{1}{2}-1} , dx ]

This gives: [ \int x^{-\frac{1}{2}} , dx = 2x^{\frac{1}{2}} + C ]

Now, evaluate the definite integral from 1 to 8: [ \int_{1}^{8} 2x^{-\frac{1}{2}} , dx = \left[ 2x^{\frac{1}{2}} \right]_{1}^{8} ]

[ = 2(8^{\frac{1}{2}}) - 2(1^{\frac{1}{2}}) ] [ = 2(2\sqrt{2}) - 2(1) ] [ = 4\sqrt{2} - 2 ]

So, [ \int_{1}^{8} \sqrt{\frac{2}{x}} , dx = 4\sqrt{2} - 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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