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network profile
| bio | website | |
|---|---|---|
| location | ||
| age | 21 | |
| visits | member for | 10 months |
| seen | Jan 28 at 18:48 | |
| stats | profile views | 26 |
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Apr 19 |
awarded | Notable Question |
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Feb 11 |
awarded | Popular Question |
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Jan 28 |
revised |
using the washer method to find the volume of an object added 6 characters in body |
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Jan 28 |
accepted | using the washer method to find the volume of an object |
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Jan 28 |
comment |
using the washer method to find the volume of an object I don't understand which two are identical. |
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Jan 28 |
revised |
using the washer method to find the volume of an object added 6 characters in body |
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Jan 28 |
comment |
using the washer method to find the volume of an object Ok, that last edit should do it. |
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Jan 28 |
revised |
using the washer method to find the volume of an object edited body |
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Jan 28 |
comment |
using the washer method to find the volume of an object Ah man, so many mistakes I overlooked. I am editing this right now. |
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Jan 28 |
revised |
using the washer method to find the volume of an object edited body |
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Jan 28 |
comment |
using the washer method to find the volume of an object Just edited the main post. |
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Jan 28 |
revised |
using the washer method to find the volume of an object added 47 characters in body |
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Jan 28 |
asked | using the washer method to find the volume of an object |
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Jan 27 |
comment |
Finding the Anti-derivative can you please explain to me how i would tackle this using substitution from the stage $du = 2dx$ back into $\int u^2$ ? |
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Jan 27 |
comment |
Finding the Anti-derivative well, if i were to tackle this using substitution, i am to let $u$ = $2x+5^2$. i know that $\frac{du}{dx} 2$ then i solve for $du$ to get $du$ = $2dx$. so i have $\int u^2\$ , now how do i go about plugging $du$ = $2dx$ back into $\int u^2$ ? |
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Jan 27 |
comment |
Finding the Anti-derivative where does the $\frac{1}{2}$ come from? |
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Jan 27 |
comment |
Finding the Anti-derivative Where does the $10$ come from, and why did you place $\int$ before all the $x$'s in $4$, and $10$, but after $25$? |
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Jan 27 |
revised |
Finding the Anti-derivative deleted 100 characters in body; edited tags |
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Jan 27 |
comment |
Finding the Anti-derivative well, using substitution I can get it down to $du$ = $2dx$ by letting $u$ = $2x+5$ how would i plug that back into $\int u^2$ |
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Jan 27 |
comment |
Finding the Anti-derivative ok, and this is where I hit a wall. how would I go about putting $du$ = $2dx$ back into $\int u^2$ , if i let $u$ = $2x+5$ |
