# How do you find the indefinite integral of #int -(5root4(x))/2dx#?

We will use the rules:

We see that:

Simplifying completely gives:

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To find the indefinite integral of ( -\frac{5\sqrt{4x}}{2}dx ), you can use the substitution method. Let ( u = 2x ). Then, ( du = 2dx ). Rewrite the integral in terms of ( u ):

[ \int -\frac{5\sqrt{u}}{2} \frac{du}{2} ]

Now, rewrite the integral with the constant outside:

[ -\frac{5}{4} \int \sqrt{u} du ]

Integrate ( \sqrt{u} ) with respect to ( u ):

[ -\frac{5}{4} \left( \frac{2}{3}u^{3/2} \right) + C ]

Substitute back ( u = 2x ):

[ -\frac{5}{4} \left( \frac{2}{3}(2x)^{3/2} \right) + C ]

Simplify:

[ -\frac{5}{6}(2x)^{3/2} + C ]

[ = -\frac{5}{6}(2\sqrt{2}x)^{3} + C ]

Thus, the indefinite integral of ( -\frac{5\sqrt{4x}}{2}dx ) is ( -\frac{5}{6}(2\sqrt{2}x)^{3} + C ), where ( C ) is the constant of integration.

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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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