MCQ
$\int_{}^{} {{e^{{x^2}}}x\;dx} $ is equal to
- A${e^{{x^2}}}$
- ✓$\frac{1}{2}{e^{{x^2}}}$
- C$2{e^{{x^2}}}$
- D$\frac{{{e^{{x^2}}} - {x^2}}}{2}$
✓
Answer
Correct option: B.
$\frac{1}{2}{e^{{x^2}}}$
b
(b)$\int_{}^{} {{e^{{x^2}}}.x\,dx} = \frac{1}{2}\int_{}^{} {(2x){e^{{x^2}}}dx} $
(b)$\int_{}^{} {{e^{{x^2}}}.x\,dx} = \frac{1}{2}\int_{}^{} {(2x){e^{{x^2}}}dx} $
(Put ${x^2} = t \Rightarrow dt = 2x\,dx)$.
$ = \frac{1}{2}\int_{}^{} {{e^t}dt} = \frac{1}{2}{e^t} = \frac{1}{2}{e^{{x^2}}}$,
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