solve indefinite integral

Question

I have this indefinite integral $\int 3 \sqrt{x}\,dx$ to solve.

My attempt:

$$\int 3 \sqrt{x}\,dx = 3 \cdot \frac {x^{\frac {1}{2} + \frac {2}{2}}}{\frac {1}{2} + \frac {2}{2}}$$

$$\int 3 \sqrt{x}\,dx = 3 \frac{x^{\frac {3}{2}}}{\frac {3}{2}} = \frac{2}{3} \cdot \frac{9}{3} x^{\frac {3}{2}}$$

$$\int 3 \sqrt{x}\,dx = \frac{18}{3} x^{\frac{3}{2}} = 6 x^{\frac{3}{2}}$$

But according to wolframalpha the answer should be $2 x^{\frac {3}{2}}$

Where did I make a error in my calculation?

Thanks!

First of all, try to write integration properly. – Oct 20 '14 at 17:30 — , Oct 20 '14 at 17:30

score 3 · Accepted Answer · answered Oct 20 '14 at 17:16

3

In your penultimate step, $\frac 23$.$\frac 93$ $=\frac {18}{9}$ $=2$, not 6.

answered Oct 20 '14 at 17:16

score 2 · Answer 2 · edited Oct 20 '14 at 17:51

2

$\frac{2}{3} \cdot \frac{9}{3}=2$. not $6$

edited Oct 20 '14 at 17:51

Michael Hardy

answered Oct 20 '14 at 17:15

user160823

score 2 · Answer 3 · edited Oct 20 '14 at 17:51

2

You failed in $\displaystyle\frac{2\cdot9}{3\cdot3} = \frac{18}{3}$, actually $\displaystyle\frac{2\cdot9}{3\cdot3} = \frac{2}{3}\cdot3 = 2$.

edited Oct 20 '14 at 17:51

Michael Hardy

answered Oct 20 '14 at 17:15

Rono

score 1 · Answer 4 · answered Oct 20 '14 at 17:52

1

$$ \frac{3}{\left(\frac 3 2\right)} = 3\cdot\frac 2 3 \ne 3\cdot \frac 3 2. $$

answered Oct 20 '14 at 17:52

Michael Hardy

4 Answers4