∫ 0 2   x e x d x =

02xexdx=
  1. e2+1
  2. e21
  3. e11
  4. e1+1

Solution

02xexdx=xexdx-dxdxexdxdx02

=xex-ex02=2e2-e2-0×e0-e0=e2+1

Asked in: AP EAMCET 2021 (19 Aug Shift 2)

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