Question

find the area of the region between y=x^1/2 and y=x^1/3 for 0 less than equal to x less than equal to 1.

Answer 1

first find the intersections of y=x^(1/2) and y=x^(1/3)

Answer 2

by setting x^(1/2)=x^(1/3)
solve for x

Answer 3

so i get 1 and 0?

Answer 4

right so
now we need to figure out which curve is above the other curve between 0 and 1

Answer 5

for example
which is true:
\[(\frac{1}{2})^\frac{1}{2}> (\frac{1}{2})^\frac{1}{3}? \\ \text{ or is } \\ (\frac{1}{2})^\frac{1}{3} >(\frac{1}{2})^\frac{1}{2}?\]

Answer 6

i graphed it in my calculator and it appears x^1/2 is above the other

Answer 7

i actually think (1/2)^(1/3) is bigger than (1/2)^(1/2)

Answer 8

so on that interval from x=0 to x=1
x^(1/3)>x^(1/2)

Answer 9

ok i got that far and i have it set up as you showed but where do i go from there?

Answer 10

\[\int\limits_{0}^{1}(x^\frac{1}{3}-x^\frac{1}{2}) dx\]

Answer 11

not on that interval

Answer 12

My mistake you're right let me delete that!

Answer 13

oh yea your right because the interval is from 0 to 1

Answer 14

so were looking at that interval on the graph

Answer 15

how do i solve now?

Answer 16

but if you did choose the first way you could always take the | | of the answer

Answer 17

use the integration rule called power rule

Answer 18

so i get square root of x and cubed root of x?

Answer 19

\[\int\limits_{}^{}x^n dx=\frac{x^{n+1}}{n+1} +C , n \neq -1 \]

Answer 20

so e than have x^1.5/1.5?

Answer 21

and x^1.3/1.3

Answer 22

and i solve using the interval of 1 and 0 and plu in to those

Answer 23

plug*

Answer 24

is it -1/12

Answer 25

\[\int\limits\limits_{0}^{1}(x^\frac{1}{3}-x^\frac{1}{2}) dx \\ =[\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}-\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}]_{0}^{1} \\ =[\frac{x^{\frac{1}{3}+\frac{3}{3}}}{\frac{1}{3}+\frac{3}{3}}-\frac{x^{\frac{1}{2}+\frac{2}{2}}}{\frac{1}{2}+\frac{2}{2}} ]_0^1 \\=[\frac{x^\frac{4}{3}}{\frac{4}{3}} -\frac{x^\frac{3}{2}}{\frac{3}{2}}]^1_0 \\ =[\frac{3}{4}x^\frac{4}{3}-\frac{2}{3}x^\frac{3}{2}]_0^1\]
and plug in the limits

Answer 26

hmm you should got a positive answer

Answer 27

did you do x^(1/2)-x^(1/3)?

Answer 28

we are suppose to do greater-smaller

Answer 29

yea i had them switched whoops!

Answer 30

ok well you can just take the | | of the result

Answer 31

so then its just 1/12?

Answer 32

yes 3/4-2/3 is 1/2

Answer 33

1/12*

Answer 34

thanks!

Answer 35

what other integration rules are there like the power rule you showed me?

Answer 36

http://www.mathsisfun.com/calculus/integration-rules.html

Answer 37

thank you very much for your help! Have a great day :)

Answer 38

could you help me with one more?

Answer 39

find the area of the region between y=x^2 and y=x^3 for 0 less than equal to x less than equal to 1.