- #1
1MileCrash
- 1,342
- 41
I received no credit, resulting in an 84 for a few integral problems. I had correct final answers for everything.
When I confronted my professor about this, he said it was because I didn't actually put "u" and "du" into the integral. Is that really always necessary? Why actually put the u in if I'm just going to replace it back after, if I'm perfectly comfortable working with the actual statement (as most people should be!)? I do define u and du in order to "jot down" the function that I'm applying the theorem to.
From the test:
[itex]\int\frac{3x}{\sqrt{x^2-15}}dx[/itex]
My work (first got rid of fraction)
[itex]\int 3x(x^2-15)^\frac{-1}{2}dx[/itex]
let u = x^2 - 15; du = 2x dx
[itex]\int \frac{3}{2}2x(x^2-15)^\frac{-1}{2}dx[/itex]
[itex]\frac{3}{2} [\frac{(x^2 - 15)^\frac{1}{2}}{\frac{1}{2}}][/itex]
[itex]3 \sqrt{(x^2 - 15)} + C[/itex]
Now, I know that as with most "test complaint" threads there are going to be the guys that rush in here, gung ho, adamant that everything I did is completely wrong and that my professor is the most wonderful, compassionate, and blessed man to ever walk the Earth and that I should burn in the fiery pits of hell for questioning him or thinking that I actually know a little bit about math.
I work the problems better, and more efficiently this way. I would have to stop my "rhythm" to write something as useless to me as:
[itex]\int u^\frac{-1}{2}\frac{3}{2}du[/itex]
[itex]\frac{3}{2}\int u^\frac{-1}{2}du[/itex]
[itex]\frac{3}{2} (2)u^\frac{1}{2}[/itex]
Just to put u back as x^2 - 15?
It just can't be that you guys feel the need to do that every time? Maybe for a long u that you just don't feel like writing out..
When I confronted my professor about this, he said it was because I didn't actually put "u" and "du" into the integral. Is that really always necessary? Why actually put the u in if I'm just going to replace it back after, if I'm perfectly comfortable working with the actual statement (as most people should be!)? I do define u and du in order to "jot down" the function that I'm applying the theorem to.
From the test:
[itex]\int\frac{3x}{\sqrt{x^2-15}}dx[/itex]
My work (first got rid of fraction)
[itex]\int 3x(x^2-15)^\frac{-1}{2}dx[/itex]
let u = x^2 - 15; du = 2x dx
[itex]\int \frac{3}{2}2x(x^2-15)^\frac{-1}{2}dx[/itex]
[itex]\frac{3}{2} [\frac{(x^2 - 15)^\frac{1}{2}}{\frac{1}{2}}][/itex]
[itex]3 \sqrt{(x^2 - 15)} + C[/itex]
Now, I know that as with most "test complaint" threads there are going to be the guys that rush in here, gung ho, adamant that everything I did is completely wrong and that my professor is the most wonderful, compassionate, and blessed man to ever walk the Earth and that I should burn in the fiery pits of hell for questioning him or thinking that I actually know a little bit about math.
I work the problems better, and more efficiently this way. I would have to stop my "rhythm" to write something as useless to me as:
[itex]\int u^\frac{-1}{2}\frac{3}{2}du[/itex]
[itex]\frac{3}{2}\int u^\frac{-1}{2}du[/itex]
[itex]\frac{3}{2} (2)u^\frac{1}{2}[/itex]
Just to put u back as x^2 - 15?
It just can't be that you guys feel the need to do that every time? Maybe for a long u that you just don't feel like writing out..