2
$\begingroup$

Here is the equation:

$$\int 3x \sqrt{1-2x^2}dt$$

Here is my answer:

$$ \dfrac14 \int (1-2x^2)^{1/2} . 3x = -\dfrac14 \dfrac{(1-2x^2)^{3/2}}{3/2} = -\dfrac14 \cdot \dfrac23 (1-2x^2)^{3/2} + c$$

correct answer:

it should be -1/2 instead of -1/4 . 2/3 at the end. what am I doing wrong?

$\endgroup$
3
  • $\begingroup$ You missed a factor of 3. Try to derivative your result and see where's missing. $\endgroup$ – Shuchang Dec 7 '13 at 16:47
  • 2
    $\begingroup$ It should be $dx$, not $dt$. $\endgroup$ – egreg Dec 7 '13 at 16:52
  • $\begingroup$ It is not an equation. $\endgroup$ – JP McCarthy Dec 9 '13 at 14:49
1
$\begingroup$

let us denote $s=1-2*x^2$,then $ds=-4*xdx$ ,because you have $3$ in your equation,outside of integrat will come $-3/4$,so it would be

$$\int (-3/4)*\sqrt{s}ds$$

can you continue from this?

$\endgroup$
6
  • $\begingroup$ so it would be 12x.... $\endgroup$ – Cash Vai Dec 7 '13 at 16:50
  • $\begingroup$ no look please ,i have updated $\endgroup$ – dato datuashvili Dec 7 '13 at 16:51
  • $\begingroup$ you are welcome, good lucks $\endgroup$ – dato datuashvili Dec 7 '13 at 16:56
  • $\begingroup$ @CashVai any question ?if yes you are welcome $\endgroup$ – dato datuashvili Dec 7 '13 at 17:02
  • $\begingroup$ thank you very much! I will, stay tuned :DD $\endgroup$ – Cash Vai Dec 7 '13 at 17:11
5
$\begingroup$

let $$\sqrt{1-2x^2}=t$$ then $$-4xdx=dt$$ the integral becomes $$\int \frac{-3}{4}\sqrt{t}dt$$ $$= \frac{-3}{4}.\frac{2}{3}t^\frac{3}{2}$$

$\endgroup$
0
$\begingroup$

Another way:

Denote $t^2 = 1-(\sqrt{2}x)^2$. Then you get $dx=-\frac{tdt}{\sqrt{2}x}$. So you get cancellations and the integral becomes $$ -\frac{3}{\sqrt{2}} \int |t|tdt= -\frac{t^3 \ sign (t)}{\sqrt{2}}+C $$ or just $-\frac{t^3}{\sqrt{2}}+C$ assuming $t>0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.