Question

Evaluate the following integral

Answer 1

\[\int\limits_{-1}^{1}(2x^{1/3} + x ^{4/5})dx\]

Answer 2

coming up with imaginary numbers...is that right?

Answer 3

fun. lol.
This is a useful rule, called power rule for integration:
\[n \neq -1 \\ \int\limits_{}^{}x^{n} dx=\frac{x^{n+1}}{n+1}+C\]

Answer 4

so looking at the x^(1/3)

Answer 5

n is (1/3)

Answer 6

so replace the n's in the formula I gave with (1/3)

Answer 7

so, the entire antiderivative I came up with is (3/2)x^(4/3) + (5/9)x^(9/5)

Answer 8

(you used wolfram to check right; wolfram does consider complex numbers to roots; you will probably want to use the real-valued root )

Answer 9

anyways let me check your antiderivative

Answer 10

ok that part looks great

Answer 11

\[(\frac{3}{2}(1)^\frac{4}{3}+\frac{5}{9}(1)^\frac{9}{5})-(\frac{3}{2}(-1)^\frac{4}{3}+\frac{5}{9}(-1)^\frac{9}{5})\]

Answer 12

ok first 1 to any power is 1

Answer 13

yes, what about the neg 1

Answer 14

second \[(-1)^\frac{1}{n}=-1 \text{ when } n \text{ is odd } \]
both your 3 and 5 are odd

so after you consider the root part of the exponent
consider the power part of your root

(-1)^even=1
(-1)^odd=-1

Answer 15

for example
\[(-1)^\frac{24}{5}=[(-1)^\frac{1}{5}]^{24}=(-1)^{24}=1\]

Answer 16

wouldn't it be imaginary when (-1)^(1/n) and n is odd?

Answer 17

or for example:
\[(-1)^\frac{21}{5}=[(-1)^\frac{1}{5}]^{21}=(-1)^{21}=-1\]

Answer 18

it depends

Answer 19

are you suppose to consider the real valued root or the principal root

Answer 20

just to evaluate the expression, so I'm not sure

Answer 21

usually in calculus 1 or whatever I only have ever seen real value root

Answer 22

so I bet you are suppose to consider only real value root here

Answer 23

ok, that makes it alot easier then

Answer 24

got zero for the final answer then

Answer 25

\[(\frac{3}{2}(1)^\frac{4}{3}+\frac{5}{9}(1)^\frac{9}{5})-(\frac{3}{2}(-1)^\frac{4}{3}+\frac{5}{9}(-1)^\frac{9}{5}) \\ (\frac{3}{2}+\frac{5}{9})-(\frac{3}{2}-\frac{5}{9}) \\ \frac{3}{2} +\frac{5}{9}-\frac{3}{2}+\frac{5}{9}\]

Answer 26

the 3/2's cancel yep
but the 5/9's do not

Answer 27

oops, 10/9

Answer 28

yeah
now
you can see both answers if you want:
http://www.wolframalpha.com/input/?i=integrate%282x%5E%281%2F3%29%2Bx%5E%284%2F5%29%2Cx%3D-1..1%29&a=%5E_Real (one with real-value root considered)

http://www.wolframalpha.com/input/?i=integrate%282x%5E%281%2F3%29%2Bx%5E%284%2F5%29%2Cx%3D-1..1%29&a=%5E_Principal (one with principal root considered)

I have never in calculus used the complex solution to an integral
I honestly don't know if the complex solution is as important as the real solution

Answer 29

how is it possible to not consider the imaginary root, is it so miniscule it doesn't matter?

Answer 30

well you can
it just depends on how your teacher defined the rooted functions

Answer 31

If your teacher said f(x)=x^(1/n) where x^(1/n) is the principal value
then I suppose you would have to consider the complex solution

Answer 32

gotcha, one more down, and getting steadily easier...thanks again

Answer 33

alright