jonnburton
Junior Member
- Joined
- Dec 16, 2012
- Messages
- 155
I have had several attempts at integrating this problem but always come to the same conclusion. I believe the correct solution should be 1 but I'm not sure how.
Can anyone tell me what my mistake is?
\(\displaystyle e^{t^2} \cdot \frac{d}{dt}\int^t_0 e^{-s^2}ds\)
This is how I have dealt with the problem:
i)Evaluate Integral
ii)Differentiate result wrt 't'
iii)multiply by \(\displaystyle e^{t^2}\)
i) Evaulate Integral
\(\displaystyle \int^t_0 e^{-s^2}ds\) by substitution. \(\displaystyle u=-s^2\); \(\displaystyle \frac{du}{ds}= -2s\); \(\displaystyle ds =\frac{1}{-2s} du\)
\(\displaystyle \frac{1}{-2s} \int^t_0 e^u du\)
\(\displaystyle \frac{1}{-2s} \left[e^{u}\right]^t_0\)
\(\displaystyle \left[\frac{e^{-s^2}}{-2s}\right]^t_0\)
\(\displaystyle \left[\frac{e^{-t^2}}{-2t}-\frac{e^0}{0}\right]\) = \(\displaystyle \frac{e^{-t^2}}{-2t}\)
ii) Differentiate
\(\displaystyle \frac{d}{dt}\frac{e^{-t^2}}{-2t}]\), using quotient rule:
\(\displaystyle \frac{-2t(-2t e^{-2t^2}-e^{-t^2}(-2)}{4t^2}\)
\(\displaystyle \frac{4t^2e^{-t^2}+2e^{-t^2}}{4t^2}\) = \(\displaystyle \frac{2e^{-t^2}(2t^2+)}{4t^2}\)
iii) Multiply by \(\displaystyle e^{t^2}\)
\(\displaystyle e^{t^2}\cdot e^{-t^2}=1\)
so the result becomes:
\(\displaystyle \frac{2(2t^2+)}{4t^2}\)
Can anyone tell me what my mistake is?
\(\displaystyle e^{t^2} \cdot \frac{d}{dt}\int^t_0 e^{-s^2}ds\)
This is how I have dealt with the problem:
i)Evaluate Integral
ii)Differentiate result wrt 't'
iii)multiply by \(\displaystyle e^{t^2}\)
i) Evaulate Integral
\(\displaystyle \int^t_0 e^{-s^2}ds\) by substitution. \(\displaystyle u=-s^2\); \(\displaystyle \frac{du}{ds}= -2s\); \(\displaystyle ds =\frac{1}{-2s} du\)
\(\displaystyle \frac{1}{-2s} \int^t_0 e^u du\)
\(\displaystyle \frac{1}{-2s} \left[e^{u}\right]^t_0\)
\(\displaystyle \left[\frac{e^{-s^2}}{-2s}\right]^t_0\)
\(\displaystyle \left[\frac{e^{-t^2}}{-2t}-\frac{e^0}{0}\right]\) = \(\displaystyle \frac{e^{-t^2}}{-2t}\)
ii) Differentiate
\(\displaystyle \frac{d}{dt}\frac{e^{-t^2}}{-2t}]\), using quotient rule:
\(\displaystyle \frac{-2t(-2t e^{-2t^2}-e^{-t^2}(-2)}{4t^2}\)
\(\displaystyle \frac{4t^2e^{-t^2}+2e^{-t^2}}{4t^2}\) = \(\displaystyle \frac{2e^{-t^2}(2t^2+)}{4t^2}\)
iii) Multiply by \(\displaystyle e^{t^2}\)
\(\displaystyle e^{t^2}\cdot e^{-t^2}=1\)
so the result becomes:
\(\displaystyle \frac{2(2t^2+)}{4t^2}\)
