# Finding the integral of $x^2\sqrt[3]{1-x}$

$$\int x^2\sqrt[3]{1-x}dx$$ I'm just starting out integration and found the above question on a textbook. It is meant to be worked out by substitution but even after a day of struggling, I can't get the answer given in the book. I tried substituting u = $\sqrt[3]{1-x}$ and then tried 1-x = u. On both the cases I am getting an answer but both are different from the answer given in the book. The answer given in the book is : $$-\frac{3}{140}\left(35-40x+14x^2 \right)\left( 1-x\right)^{4/3}$$ I'm getting a common factor -3 and $(1-x)^{4/3}$ but not 140 and the terms inside the other bracket are different.

• It would be great if you could actually show your steps. We can perhaps find out where you went wrong. Your suggested u-subs are certainly in the right direction... Commented Aug 26, 2016 at 15:05
• @imranfat I'll edit the question now. Commented Aug 26, 2016 at 15:08
• The answer you have given is incorrect. @heropup below has the correct answer. Commented Aug 26, 2016 at 15:11
• @IanMiller Answer is given in the book problems in calculus by I A Maron ia600206.us.archive.org/17/items/… It is question number 4.2.14 (a) and I too believe that the answer is wrong Commented Aug 26, 2016 at 15:20
• @HGSur If you look at the second last line of heropop's answer you can see where the typo in the book came from. Commented Aug 27, 2016 at 7:05

With the substitution $u = 1-x$, $du = -dx$, we obtain $$\int x^2 (1-x)^{1/3} \, dx = -\int (1-u)^2 u^{1/3} \, du = -\int u^{7/3} - 2u^{4/3} + u^{1/3} \, du.$$ Integrating term by term, we get $$-\frac{3u^{10/3}}{10} + \frac{6u^{7/3}}{7} - \frac{3u^{4/3}}{4} + C.$$ Factoring out a common term of $-\frac{3}{140}u^{4/3}$ we get $$-\frac{3u^{4/3}}{140} (14u^2 - 40u + 35) + C.$$ Expressing this in terms of $x$ gives $$-\frac{3(1-x)^{4/3}}{140} (14x^2 + 12x + 9) + C.$$

You should note that in the last step, expressing the antiderivative in terms of $x$ gives a different quadratic factor: perhaps this is the one you obtained? The answer that you claim your book has given is not correct.

• @HGSur If you want to be sure an anti derivative is right, make the integral a definite integral (you can use the interval) and figure out the exact area under the curve. Then use an online tool or TI84 to match the decimals...There are plenty of situations where book answers are incorrect. Commented Aug 26, 2016 at 15:23
• One question : How do you know that the common term to be taken out is 3/140 ? Commented Aug 27, 2016 at 7:37
• The greatest common factor of the numerator terms $3$, $6$, $3$ is obviously $3$. The least common multiple of the denominator terms $10$, $7$, $4$ is $140$. Therefore, factoring out $3/140$ guarantees that the quadratic factor will have integer coefficients, and that these coefficients will be "minimal" in the sense that they are relatively prime (though not necessarily pairwise relatively prime). Commented Aug 27, 2016 at 7:51

It would be nicer if $x$ carried the exponent $\frac13$ and $(1-x)$ carried the exponent $2$, because you could then just expand the latter and multiply through by the former to produce a simple sum of terms.

So, let's write $u=1-x$ (the thing under the radical). Then we can also get the part outside of the radical in terms of $u$ as well, because then $x=1-u$ and $dx = -du$. So the integral can be written as $$\int \underbrace{(1-u)^2}_{x^2}\underbrace{u^{1/3}}_{\sqrt[3]{1-x}}\underbrace{(-1)\; du}_{dx}$$ $$=\int -(1-2u + u^2)u^{1/3}\; du$$ $$=-\int (u^{1/3}-2u^{4/3}+u^{7/3})\; du$$ $$=-(\tfrac34u^{4/3} - \tfrac67u^{7/3} + \tfrac3{10}u^{10/3})+C$$ $$=- \left( \tfrac{105}{140}-\tfrac{120}{140}u+\tfrac{42}{140}u^2\right)u^{4/3}+C$$ $$=\tfrac3{140}(35 +40 u + 14 u^2)u^{4/3}+C$$ $$=\tfrac3{140}[35 +40(1-x) + 14 (1-x)^2](1-x)^{4/3}+C$$ $$=\tfrac3{140}[35 -(40-40x) + (14-28x+14x^2)](1-x)^{4/3}+C$$ $$=\boxed{\tfrac3{140}(9 -12 x + 14x^2)(1-x)^{4/3}+C}$$

setting $$t=(1-x)^\frac13$$ then we get $$x=1-t^3$$ and $$dx=-3t^2dt$$ and our integral is $$\int (1-t^3)^2\cdot t\cdot (-3t^2)dt$$ which is simplier.

• I did that too, but as it seems the answer in the book turns out to be wrong. Commented Aug 26, 2016 at 15:22