Somewhat tricky integral: (sqrt[x] + 3cbrt[x]) / 6th-rt[x]

[sigma]

Had this on a test yesterday. Not sure if I worked it out right.

\(\displaystyle \
\L\
\int_1^{64} {\frac{{\sqrt x + 3\sqrt[3]{x}}}{{\sqrt[6]{x}}}} dx
\\)

I was suppose to simplify as much as possible, without a calculator. Also we haven't learned substitution yet. What I did was moved the \(\displaystyle \
\L\
\sqrt[6]{x}
\\) up to the top and changed every roots to their fractional exponent equivalent and went from there. Just want to see if I was right in the end. I had one number of \(\displaystyle \
\L\
\sqrt[6]{{64^8 }}
\\) which is impossible to do without a calculator so I left it as that. There was a bunch of fractions as well.

 

[stapel]

I would agree with your first step:

. . . . .[x^1/2 + 3x^1/3] / x^1/6

. . . . .[x^3/6 + 3x^2/6] / x^1/6

. . . . .x^{3/6 - 1/6} + 3x^{2/6 - 1/6}

. . . . .x^2/6 + 3x^1/6

. . . . .x^1/3 + 3x^1/6

I'm not sure how you arrived at an answer involving 8/6-th powers...? It would be helpful to see your work and reasoning, in order to be able to check your work.

Thank you.

Eliz.

Note: 8/6 = 4/3, and 4³ = 64, so 64^1/3 = 4.

 

[skeeter]

fyi ... \(\displaystyle \L \sqrt[6]{64^8} = 64^{\frac{8}{6}} = 64^{\frac{4}{3}} = 4^4 = 256\)

\(\displaystyle \L \frac{x^{\frac{1}{2}} + 3x^{\frac{1}{3}}}{x^{\frac{1}{6}}} = x^{\frac{1}{3}} + 3x^{\frac{1}{6}}\)

\(\displaystyle \L \int_1^{64} x^{\frac{1}{3}} + 3x^{\frac{1}{6}} dx =\)

\(\displaystyle \L \left[\frac{3}{4}x^{\frac{4}{3}} + \frac{18}{7}x^{\frac{7}{6}} \right]_1^{64} =\)

\(\displaystyle \L \left(192 + \frac{2304}{7}\right) - \left(\frac{3}{4} + \frac{18}{7}\right)=\frac{14499}{28}\)

 

[sigma]

Yea it was something like that only I didn't realize to reduce the 8/6 to 4/3 that would have helped me allot to see that. I had that same answer just not reduced to that level. Just didn't have time and its hard to deal with large numbers without a calculator especially when your running out of time.

 

[stapel]

...its hard to deal...when your running out of time.

I know the feeling!

See if it helps to do some extra practice before the next test. Sometimes having reworked the homework exercises (or additional odd-numbered exercises, so you can check your answers in the back of the book) the day or two before the test can help you get "in the groove" for working those sorts of exercises, and can make the test a little less painful.

Eliz.

 

[sigma]

Yea usually thats what I do. This was somewhat of a curve ball question as every test has one and its impossible to prepare for. I tend to study too much and no matter how much I know, they always, always put a question in that's suppose to confuse you. I'm sure they research all the study materials they think most people will go over and put in the test of what was not covered in the study materials.