I started by integrating it this way:
$$(\frac{3}{7} x^{7/3} + \frac{16}{4}x^{4/3})$$
What is wrong with it?
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What's wrong with it is that it is a definite integral and you just calculated the indefinite integral. You now have to plug in $1$ and $-1$ and subtract to get the final answer. – Gregory Grant May 07 '15 at 19:01
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1@GregoryGrant The OP stated "I started by ..." There is nothing wrong with the start. – Mark Viola May 07 '15 at 19:03
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Hint: $3 \times 4=12$! so your $16/4$ is wrong :) – Emilio Novati May 07 '15 at 19:06
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I took it to mean what should they do next. But if they simply want to know if that's the anti-derivative, then obviously the answer is "nothing". – Gregory Grant May 07 '15 at 19:06
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use that $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ if $n\ne -1$ – Dr. Sonnhard Graubner May 07 '15 at 19:08
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If I replace x by -1 I'd have $\sqrt[3]{-1^7}$ – juliano.net May 07 '15 at 19:08
4 Answers
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The answer is
$\frac{6}{7}$
Integrating,
$ \int^1_{-1} x^{4/3} 4 x^{1/3} \,dx = (\frac{3}{7}x^{7/3} + 3 x^{4/3})^{1}_{-1} = \frac{3}{7} + 3 + \frac{3}{7} - 3 = \frac{6}{7}. $
In the OP as it reads now, it appears that you increased the power on the second term and multiplied by the new power instead of dividing.
Sean Henderson
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Simple way for me is to split the integral up and look at in that respect. $\int_{-1}^{1} x^{4/3} + 4x^{1/3}dx=\int x^{4/3}dx + \int 4x^{1/3}dx = \frac{6}{7} + 0 = \frac{6}{7}$
spun
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Hint: for the indefinite integral we get $$3/7\,{x}^{4/3} \left( x+7 \right)$$
Dr. Sonnhard Graubner
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You are asking "What is wrong with it?"
$$(\frac{3}{7} x^{7/3} + \frac{16}{4}x^{4/3})$$
Check the derivative
$$(\frac{3}{7}\frac{7}{3} x^{4/3} + \color{red}{\frac{16}{4}\frac{4}{3}}x^{1/3}).$$
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Ironically, the error in the coefficient of the second term is harmless as the term vanishes anyway. – May 07 '15 at 19:30