Question

help understanding this answer:

Answer 1

sure, go ahead.

Answer 2

\[\int\limits_{0}^{4} \left| \sqrt(x)-1 \right|=2\]
Why?

Answer 3

um, to advanced for me lol sorry ask someone more advanced.

Answer 4

For \(0\le x < 1\) , \(\sqrt{x} -1\) <0
For \(1\le x \le 1\) , \(\sqrt{x} -1 \ge\)0

So,
\[\int\limits_{0}^{4} \left| \sqrt(x)-1 \right|dx\]\[=\int_0^1 (1-\sqrt{x}) dx + \int_1^4 (\sqrt{x}-1)dx\]Can you do it from here?

Answer 5

@callisto
\[\int\limits_{0}^{1} [x-\frac{ 2 }{ 3 }x^\frac{ 3 }{ 2 }] +\int\limits_{1}^{4} [\frac{ 2 }{ 3 }x^\frac{ 3 }{ 2 }-x]= [1/3+16/3-5/3]=4\]
But the answer supposedly is 2. Is this wrong?

Answer 6

It is wrong. You made some mistakes?!

Answer 7

Can you help me with it till the end? I can't seem to figure it out...

Answer 8

\[\int\limits_{0}^{1} [x-\frac{ 2 }{ 3 }x^\frac{ 3 }{ 2 }] +\int\limits_{1}^{4} [\frac{ 2 }{ 3 }x^\frac{ 3 }{ 2 }-x]\]\[= [1-\frac{2}{3}+\frac{16}{3}-4 - \frac{2}{3}+1]\]\[= [1-\frac{4}{3}+\frac{16}{3}-4+1]\]\[= [1+\frac{12}{3}-4+1]\]\[= [1+4-4+1]\]\[=2\]

Answer 9

The first line should be
\[ [x-\frac{ 2 }{ 3 }x^\frac{ 3 }{ 2 }]_0^1 +[\frac{ 2 }{ 3 }x^\frac{ 3 }{ 2 }-x]_1^4\]

Answer 10

thanks! You're amazing! Greatly appreciated it!

Answer 11

You're welcome :)