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Integrate WITHOUT using Integration by Parts
- Thread starter tkvictim
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BigGlenntheHeavy
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\(\displaystyle Hey, \ bottomless \ pit, \ do \ us \ and \ yourself \ a \ favor \ and \ hit \ the \ bricks.\)
D
Deleted member 4993
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tkvictim said:Integrate this indefinite integral without using integration by parts.
Show all work. [and probably, by not using the calculator either]
Is that even possible?
10te[sup:h7gxruom]-t/3[/sup:h7gxruom]
Yes it is possible.
Do you know how to expand 10te[sup:h7gxruom]-t/3[/sup:h7gxruom] as a polynomial function of 't'?
.
BigGlenntheHeavy
Senior Member
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- Mar 8, 2009
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\(\displaystyle 10te^{-t/3} \ = \ 10 \sum_{n=0}^{\infty}\frac{(-1)^n t^{n+1}}{3^{n}n!} \ converges \ for \ all \ t \ (Maclauren \ Series).\)
\(\displaystyle \int [10te^{-t/3}]dt \ = \ \int\bigg[10 \sum_{n=0}^{\infty}\frac{(-1)^n t^{n+1}}{3^{n}n!} \bigg]dt\)
\(\displaystyle Hence, \ \int [10te^{-t/3}]dt \ = \ 10 \sum_{n=0}^{\infty}\frac{(-1)^n t^{n+2}}{3^{n}n!(n+2)}\)
\(\displaystyle Check: \ \int_{0}^{15}[10te^{-t/3}]dt \ = \ 10 \sum_{n=0}^{\infty}\frac{(-1)^n 15^{n+2}}{3^{n}n!(n+2)} \ \dot= \ 86.3615086205\)
\(\displaystyle However, \ since \ "the \ bottomless \ pit" \ doesn't \ understand \ I \ by \ P, \ we \ are \ obviously \ dealing \ with\)
\(\displaystyle \ a \ fisherman.\)
\(\displaystyle \int [10te^{-t/3}]dt \ = \ \int\bigg[10 \sum_{n=0}^{\infty}\frac{(-1)^n t^{n+1}}{3^{n}n!} \bigg]dt\)
\(\displaystyle Hence, \ \int [10te^{-t/3}]dt \ = \ 10 \sum_{n=0}^{\infty}\frac{(-1)^n t^{n+2}}{3^{n}n!(n+2)}\)
\(\displaystyle Check: \ \int_{0}^{15}[10te^{-t/3}]dt \ = \ 10 \sum_{n=0}^{\infty}\frac{(-1)^n 15^{n+2}}{3^{n}n!(n+2)} \ \dot= \ 86.3615086205\)
\(\displaystyle However, \ since \ "the \ bottomless \ pit" \ doesn't \ understand \ I \ by \ P, \ we \ are \ obviously \ dealing \ with\)
\(\displaystyle \ a \ fisherman.\)