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Question
Solution
\[Let\ I = \int_0^1 x\ e^{x^2} d\ x . \]
\[Let\ x^2 = t . Then, 2x\ dx = dt\]
\[When\ x = 0, t = 0\ and\ x\ = 1\, t = 1\]
\[ \therefore I = \frac{1}{2} \int_0^1 e^t\ dt\]
\[ \Rightarrow I = \frac{1}{2} \left( e^t \right)_0^1 \]
\[ \Rightarrow I = \frac{1}{2}\left( e - 1 \right)\]
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