Evaluate this integral. I think I'm missing something.

Question

$$\int_1^{18} \sqrt{\frac{3}{z}}dz $$

So I have to find the antiderivative of $\sqrt{\frac{3}{z}}dz$ = $\frac{2}{3} \cdot \frac{3}{z}^{\frac{3}{2}}$

So the integral is:

$$\left[ \frac{2}{3} \cdot \frac{3}{z}^{\frac{3}{2}} \right ]_1^{18}$$

But I am missing something because when I take the derivative of the antiderivative, it seems like I still have to use chain rule. I missed something I think right?

Answer 1

Write your integrand as $$\sqrt{\frac 3z}=\sqrt3(z^{-1/2})$$ and integrate using the rule that $$\int z^a\,dz=\frac{z^{a+1}}{a+1}+C$$ where $a\neq-1$ and $C$ is a constant.

Answer 2

Substitute $$t=\sqrt{\frac{3}{z}}$$ then we get $$dz=\frac{-6}{t^3}dt$$ and $$z_1=1$$ we get $$t_1=\sqrt{3}$$ and $$z_2=18$$ then we get $$t_2=\sqrt{\frac{3}{18}}$$

Answer 3

There is a problem with your antiderivative:

Writing $\sqrt{\frac{3}{z}}=\sqrt{3} z^\frac{-1}{2}$ you obtain for an antiderivative: $$\sqrt{3}\left( -\frac{1}{2}+1 \right)^{-1} z^{-\frac{1}{2}+1}$$ So the integral is: $$\left[2\sqrt{3}\sqrt{z} \right]_1^{18}$$

Answer 4

Without substitution:

$$\int_1^{18} \sqrt{\frac{3}{z}}dz = \sqrt{3}\int_1^{18} z^{-\frac{1}{2}}dz.$$

You can use the following formula:

$$\int z^a dz = \frac{z^{a+1}}{a+1} + c.$$

Then, in your case:

$$\sqrt{3}\int_1^{18} z^{-\frac{1}{2}}dz = \sqrt{3}\left[\frac{z^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]_{z=1}^{z=18} = \\ = \sqrt{3}\left[2z^{\frac{1}{2}}\right]_{z=1}^{z=18} = \sqrt{3}\left[2(18)^{\frac{1}{2}} - 2(1)^{\frac{1}{2}}\right] = \\ = 2\sqrt{3}\left[3\sqrt{2}-1\right].$$