Problem with Integration

**Paymemoney** · May 30th 2010, 03:44 PM

Hi

I am having trouble trying to integrate this equation:

$\int \sqrt(1+2x+2x^3)$

how would i approach an expression like this?

Do i integrate the inside and then the outside?

P.S

**TheCoffeeMachine** · May 30th 2010, 04:04 PM

Are you sure it's $\int{\sqrt{1+2x+3x^3}}\;{dx}$ ?

**Paymemoney** · May 30th 2010, 05:50 PM

Well, the original question asked to find the length of arc of $f(x) = \frac{2}{3}(1+x^2)^(\frac{3}{2})<br />$
$y=2-\frac{1}{2}x$ between x=0, y=3

I used the length of arc rule:

$\int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx$

This is where i was stuck:

$\int_0^3 \sqrt{1+2x+2x^3}dx$

**Sudharaka** · May 30th 2010, 06:20 PM

Originally Posted by Paymemoney

Well, the original question asked to find the length of arc of $f(x) = \frac{2}{3}(1+x^2)^(\frac{3}{2})<br />$
$y=2-\frac{1}{2}x$ between x=0, y=3

I used the length of arc rule:

$\int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx$

This is where i was stuck:

$\int_0^3 \sqrt{1+2x+2x^3}dx$

Dear Paymemoney,

I have some problems regarding your question. You want to find the arc length of $f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}$ is'nt?

What is the second equation $y=2-\frac{1}{2}x$ ??? How did you get this? Do you want to find the arc length of this curve too??

**Paymemoney** · May 30th 2010, 06:21 PM

Originally Posted by Sudharaka

Dear Paymemoney,

I have some problems regarding your question. You want to find the arc length of $f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}$ is'nt?

What is the second equation $y=2-\frac{1}{2}x$ ??? How did you get this? Do you want to find the arc length of this curve too??

opps srry my mistake the equation is only $f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}$

**Sudharaka** · May 30th 2010, 06:32 PM

Originally Posted by Paymemoney

opps srry my mistake the equation is only $f(x) = \frac{2}{3}(1+x^2)^{\frac{3}{2}}$

Dear Paymemoney,

You have simplified incorrectly,

$\int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx$

$=\int_0^3\sqrt{1+4x^2(1+x^2)}dx$

$=\int_0^3\sqrt{1+4x^2+4x^4}dx$

$=\int_0^3\sqrt{(1+2x^2)^2}dx$

Can you continue from here??

**Paymemoney** · May 30th 2010, 07:53 PM

Originally Posted by Sudharaka

Dear Paymemoney,

You have simplified incorrectly,

$\int_0^3 \sqrt{1+(2x(1+x^2)^{0.5})^2}dx$

$=\int_0^3\sqrt{1+4x^2(1+x^2)}dx$

$=\int_0^3\sqrt{1+4x^2+4x^4}dx$

$=\int_0^3\sqrt{(1+2x^2)^2}dx$

Can you continue from here??

yes,
thank you, i worked it out

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