Question

Evaluate the definite integral:

Answer 1

Amistre, how do you write a fractional exponent using equation editor?

Answer 2

depends on how you want it to look:

^{\frac{top}{bottom}}

^{top/bottom}

Answer 3

Okay, thanks

Answer 4

any complicated exponents are just encased in the {...}

Answer 5

we can do this by parts if need be, but the x^29 is simply the derivative of the innards

Answer 6

all your missing out on is a -1
\[-\int_{0}^{a^{^p/_q}}-x^{29}\sqrt{a^2-x^{30}}dx\]
is apossibility, then gotta determine the missing parts; may be able to get it looking like a trig function thro thru substition

Answer 7

amistre64, do you have an answer by any chance?

Answer 8

If I'm not mistaken a is a constant, so:
\[\int\limits_{0}^{\frac{a}{15}} x^{29} \sqrt{a^2-x^{30}}dx, u=a^2-x^{30}, du=-30x^{29}dx, \frac{du}{-30x^{29}}=dx\]
\[\int\limits \frac{x^{29} \sqrt{u}}{-30x^{29}}du \rightarrow -\frac{1}{30} \int\limits \sqrt{u}du \rightarrow -\frac{1}{30}\frac{2}{3}u^\frac{3}{2}\rightarrow \left[ -\frac{2}{90}(a^2-x^{30}) \right]_{0}^{\frac{a}{15}}\]

Answer 9

an answer by chance, no :) nadeems got a good one tho ;)

Answer 10

just account for the upper limit being a^{9/15} if your original was not typoed maybe?

Answer 11

INTEGRAL HERE!!

Answer 12

is that upper limit a to the power of one over 15?!

Answer 13

its like online tourrets syndrome lol

Answer 14

on the attachment it looks like a/15.. or a*1/15...

Answer 15

a^(1/15) is what i see, but then again I am blind

Answer 16

tourrets... hahaha

Answer 17

wwwwwhhh wwhhhaaa wwwwwhhhhaatt mmmm akkess yyyyou sssaaay ttthhhhat?

Answer 18

hey amistre whats up!!!!

Answer 19

Howdy sath

Answer 20

\[\sqrt[15]{a}\] anti derivative is ...oh nadeem has it all. plug in to get answer i got back to dungeon now

Answer 21

.... my question is, who keeps leaving the dungeon door open :)

Answer 22

Then this would be it:
\[\left[ -\frac{2}{90}(a^2−x^{30}) \right]_{0}^{a^{\frac{1}{15}}}\]

Answer 23

Where's the dungeon door...

Answer 24

i think its in the shop getting rekeyed ...

Answer 25

Stupid shop.... they need to hurry up

Answer 26

Hahaha

Answer 27

I'm so confused, I cant view all this:

\[\int\limits_{0}^{\frac{a}{15}} x^{29} \sqrt{a^2-x^{30}}dx, u=a^2-x^{30}, du=-30x^{29}dx, \frac{du}{-30x^{29}}=dx\]
\[\int\limits \frac{x^{29} \sqrt{u}}{-30x^{29}}du \rightarrow -\frac{1}{30} \int\limits \sqrt{u}du \rightarrow -\frac{1}{30}\frac{2}{3}u^\frac{3}{2}\rightarrow \left[ -\frac{2}{90}(a^2-x^{30}) \right]_{0}^{\frac{a}{15}}\]


its all raw input when i view it, i tried viewing in IE , firefox, and chrome but it looks like gobbledegook either way ...anyone have a solution?

Answer 28

use Chrome

Answer 29

The answer is :
(-2/90)(a^3)= (-1/45)(a^3)

Answer 30

There is a typo on the last part:
\[\left| −\frac{1}{45}(a^2−x^{30})^\frac{3}{2} \right|_{0}^{\sqrt[15]{a}}\]

Answer 31

the answer -1/45(a^3) is showing up as incorrect :/

Answer 32

Whats the answer?

Answer 33

What is your upper limit of integration?
is it a/15 or a^(1/15)

Answer 34

a^(1/15)