1
$\begingroup$

Evaluate $ \int_{0}^{1} \sqrt{1+\sqrt[3] {x}} \,dx,$

So,i've tried doing the substitution $ u=\sqrt[3] {x}$( this was a suggestion by wolfram), but frankly i am reaching a point where it doesn't work . Could help me out ?

$\endgroup$
2
  • 1
    $\begingroup$ I'd have tried $u = 1+\sqrt[3]x$, which gives $dx=3\sqrt[3] {x^2}du= 3(u-1)^2du$ $\endgroup$
    – lulu
    Mar 22 '20 at 16:20
  • $\begingroup$ Try $u = 1+\sqrt[3] {x}$ $\endgroup$ Mar 22 '20 at 16:21
3
$\begingroup$

If $u = \sqrt{1+\sqrt[3]{x}}$, then the integral becomes $$\int_0^1 \sqrt{1+\sqrt[3]{x}} \ dx = 6 \int_1^\sqrt{2} u^2 \, (u^2-1)^2du.$$ Now, expand the integrand and integrate a polynomial.

$\endgroup$
1
$\begingroup$

Hint:

Let $u = 1 + \sqrt[3]x\implies\mathrm dx = 3x^{\frac23}\mathrm du$. Therefore,

$$\int_0^1\sqrt{1 + \sqrt[3]x}\,\mathrm dx\equiv3\int_1^2(u - 1)^2\sqrt u\,\mathrm du = 3\int_1^2 u^{\frac52}\,\mathrm du - 6\int_1^2 u^{\frac32}\,\mathrm du + 3 \int_1^2 u^{\frac 12}\,\mathrm du.$$

Solve the three integrals and plug-in limits.

$\endgroup$
-1
$\begingroup$

Hint:

Let $\sqrt{1+\sqrt[3]x}= y\implies x=((y^2-1)^3, dx=?$

If $x=0,y=1; x=1, y=?$

$\endgroup$

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.