0
$\begingroup$

How can I solve this problem correctly. I already know the answer but I am confused on how they got it.

$$ \int_{1}^{6} 4 \sqrt[5]{y^4} dy $$

Answer

$$ \frac{120 {\sqrt[5]{1296}} - 20} {9} $$

I apologize for the crazy looking code but I am new here so this gonna take some time.

$\endgroup$
4
  • 1
    $\begingroup$ Hint: $ \sqrt[5]{y^4} = y^{\frac{4}{5}}$ $\endgroup$
    – This Play Name
    Dec 4, 2015 at 4:58
  • $\begingroup$ O.O okay you have totally lost me. I don't see that anywhere $\endgroup$
    – Chynna Doll
    Dec 4, 2015 at 5:00
  • 1
    $\begingroup$ That's just algebra think power rule. $\endgroup$
    – This Play Name
    Dec 4, 2015 at 5:02
  • $\begingroup$ Sorry, guess I was wrong about the power rule >.< My teacher didn't go over this in class so when he gave us the homework I was so confused $\endgroup$
    – Chynna Doll
    Dec 4, 2015 at 5:46

4 Answers 4

Reset to default
0
$\begingroup$

This can be solved using simple integration methods. By factoring and using $n_{th}$ root rules, we get: $$I=4\int_1^6 y^{\frac{4}{5}}\,dy.$$ Add $\frac{5}{5}$ to the exponent and divide by the reciprocal $\big(\frac{5}{9}\big)$ to get: $$I=4\bigg[\frac{5y^\frac{9}{5}}{9}\bigg]_1^6=\frac{20}{9}\big[y^\frac{9}{5}\big]_1^6=\frac{20}{9}\bigg[6^\frac{9}{5}-1\bigg]=\frac{120\sqrt[5]{6^4}-20}{9}=\frac{120\sqrt[5]{1296}-20}{9}.$$ $\blacksquare$
EDIT: It seems your having a little difficulty with turning the $\sqrt[5]{n^4}$ into $y^\frac{4}{5}.$ A rule of algebra is: $$\sqrt[m]{n^p}=n^\frac{p}{m}$$ For example, $\sqrt{2}$ can be written as $2^\frac{1}{2}.$

$\endgroup$
4
  • $\begingroup$ I think I got it. So if I were to come across another problem similar to this but different numbers, the way to handle the problem would be this way? $\endgroup$
    – Chynna Doll
    Dec 4, 2015 at 5:44
  • $\begingroup$ @ChynnaDoll Yep! There's two main things you need to remember: the rule about rewriting nth-roots, and how to integrate x^(a/b) (by adding one to the exponent and dividing by the resulting reciprocal). Remember to mark the question resolved if it's been answered :) $\endgroup$
    – Eli Berkowitz
    Dec 4, 2015 at 6:29
  • $\begingroup$ Oh I have one more question. Why does the 6 need to be inside the square root? $\endgroup$
    – Chynna Doll
    Dec 4, 2015 at 6:40
  • $\begingroup$ @ChynnaDoll It's the same thing from the beginning of the problem, except in reverse. n^(p/m) can likewise be written as p_root(n^p). Here, 6^(9/5)=5_root(6^9). $\endgroup$
    – Eli Berkowitz
    Dec 4, 2015 at 7:03
0
$\begingroup$

HINT: using $\int x^n\ dx=\frac{x^{n+1}}{n+1}$, one should get $$\int_{1}^{6}4\sqrt[5]{y^4}\ dy=4\int_{1}^{6}y^{4/5}\ dy=4\left[\frac{y^{\frac{4}{5}+1}}{\frac{4}{5}+1}\right]_1^6=4\frac{5}{9}\left[y^{9/5}\right]_1^6$$

$\endgroup$
0
$\begingroup$

Think of exponents as fractions. So, $\sqrt{x} = x^{\frac{1}{2}}$. Then apply the power rule for integration.

$\endgroup$
1
  • 1
    $\begingroup$ Alright power rule. Got it thank you $\endgroup$
    – Chynna Doll
    Dec 4, 2015 at 5:45
0
$\begingroup$

$$\sqrt[5]{y^4} = \left(y^4\right)^{1/5} = y^{4/5}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .