Ryan Rigdon
Junior Member
- Joined
- Jun 10, 2010
- Messages
- 246
I was having a lot of trouble with this one. is my work sufficient?
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Integration By Parts
- Thread starter Ryan Rigdon
- Start date
Hello, Ryan!
\(\displaystyle \text{You integrated }\int t^{\frac{2}{7}}\ln t\,dt\;\text{ and got }\,\tfrac{7}{81}t^{\frac{9}{7}}(9\ln t - 7) + C\)
\(\displaystyle \text{I checked it . . . it's absolutely correct!}\)
. . . . . . \(\displaystyle \text{Good work!}\)
\(\displaystyle \text{You integrated }\int t^{\frac{2}{7}}\ln t\,dt\;\text{ and got }\,\tfrac{7}{81}t^{\frac{9}{7}}(9\ln t - 7) + C\)
\(\displaystyle \text{I checked it . . . it's absolutely correct!}\)
. . . . . . \(\displaystyle \text{Good work!}\)
BigGlenntheHeavy
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle Evaluate \ by \ I \ by \ P \ \int_{1}^{e}t^{2/7}ln|t|dt.\)
\(\displaystyle Let \ u \ = \ ln|t|, \ then \ du \ = \ \frac{dt}{t}.\)
\(\displaystyle Let \ dv \ = \ t^{2/7}dt, \ then \ v \ = \ \frac{7t^{9/7}}{9}.\)
\(\displaystyle Ergo \ \int_{1}^{e}t^{2/7}ln|t|dt \ = \ \frac{7}{9}t^{9/7}ln|t|\bigg]_{1}^{e}-\frac{7}{9}\int_{1}^{e} t^{2/7}dt\)
\(\displaystyle = \ \frac{7e^{9/7}}{9}-\frac{49t^{9/7}}{81}\bigg]_{1}^{e} \ = \ \frac{7e^{9/7}}{9}-\frac{49e^{9/7}}{81}+\frac{49}{81}\)
\(\displaystyle = \ \frac{14e^{9/7}+49}{81} \ \dot= \ 1.23014211103.\)
\(\displaystyle Let \ u \ = \ ln|t|, \ then \ du \ = \ \frac{dt}{t}.\)
\(\displaystyle Let \ dv \ = \ t^{2/7}dt, \ then \ v \ = \ \frac{7t^{9/7}}{9}.\)
\(\displaystyle Ergo \ \int_{1}^{e}t^{2/7}ln|t|dt \ = \ \frac{7}{9}t^{9/7}ln|t|\bigg]_{1}^{e}-\frac{7}{9}\int_{1}^{e} t^{2/7}dt\)
\(\displaystyle = \ \frac{7e^{9/7}}{9}-\frac{49t^{9/7}}{81}\bigg]_{1}^{e} \ = \ \frac{7e^{9/7}}{9}-\frac{49e^{9/7}}{81}+\frac{49}{81}\)
\(\displaystyle = \ \frac{14e^{9/7}+49}{81} \ \dot= \ 1.23014211103.\)
Ryan Rigdon
Junior Member
- Joined
- Jun 10, 2010
- Messages
- 246
soroban said:Hello, Ryan!
\(\displaystyle \text{You integrated }\int t^{\frac{2}{7}}\ln t\,dt\;\text{ and got }\,\tfrac{7}{81}t^{\frac{9}{7}}(9\ln t - 7) + C\)
\(\displaystyle \text{I checked it . . . it's absolutely correct!}\)
. . . . . . \(\displaystyle \text{Good work!}\)
thank you soroban. doesnt get any easier LOL. happy happy joy joy :shock:
Ryan Rigdon
Junior Member
- Joined
- Jun 10, 2010
- Messages
- 246
BigGlenntheHeavy said:\(\displaystyle Evaluate \ by \ I \ by \ P \ \int_{1}^{e}t^{2/7}ln|t|dt.\)
\(\displaystyle Let \ u \ = \ ln|t|, \ then \ du \ = \ \frac{dt}{t}.\)
\(\displaystyle Let \ dv \ = \ t^{2/7}dt, \ then \ v \ = \ \frac{7t^{9/7}}{9}.\)
\(\displaystyle Ergo \ \int_{1}^{e}t^{2/7}ln|t|dt \ = \ \frac{7}{9}t^{9/7}ln|t|\bigg]_{1}^{e}-\frac{7}{9}\int_{1}^{e} t^{2/7}dt\)
\(\displaystyle = \ \frac{7e^{9/7}}{9}-\frac{49t^{9/7}}{81}\bigg]_{1}^{e} \ = \ \frac{7e^{9/7}}{9}-\frac{49e^{9/7}}{81}+\frac{49}{81}\)
\(\displaystyle = \ \frac{14e^{9/7}+49}{81} \ \dot= \ 1.23014211103.\)
its helpful when you dont have a tutor to be right next to you, thats why i like coming here a lot and showing my work. thanx BigGlenntheHeavy. onto the next one. :shock: