How do you find the antiderivative of #8sqrtx#?

Answer 1

#16/3x^(3/2)+C#

the counteractive agent for

#8sqrtx#

is provided by

#int8sqrtxdx#

Utilize the power rule and switch to powers

#intx^ndx=x^(n+1)/(n+1)+C, " "n!=-1#
#int8sqrtxdx=8intx^(1/2)dx#
#=8xx(x^(1/2+1))/(1/2+1)+C#
#=8xxx^(3/2)/(3/2)+C#
#=8xx2/3x^(3/2)+C#
#=16/3x^(3/2)+C#
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Answer 2

To find the antiderivative of (8\sqrt{x}), use the power rule for integration in reverse. The power rule states that the antiderivative of (x^n) is (\frac{{x^{n+1}}}{{n+1}}), where (n) is any real number except for -1. Applying this rule to (8\sqrt{x}), which can be written as (8x^{1/2}), we get:

[ \int 8\sqrt{x} , dx = 8\int x^{1/2} , dx = 8 \cdot \frac{{x^{1/2+1}}}{{1/2+1}} + C ]

Simplify:

[ = 8 \cdot \frac{{x^{3/2}}}{{3/2}} + C ]

[ = \frac{{16x^{3/2}}}{{3}} + C ]

Where (C) is the constant of integration.

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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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