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Integrals

Question
CBSEENMA12032417

Evaluate
integral subscript 0 superscript 1 open parentheses straight x space straight e to the power of straight x plus cos πx over 4 close parentheses dx

Solution

Let I = integral subscript 0 superscript 1 open parentheses straight x space straight e to the power of straight x plus cos πx over 4 close parentheses dx space equals integral subscript 0 superscript 1 xe to the power of straight x dx plus integral subscript 0 superscript 1 cos πx over 4 dx
       equals left square bracket space straight x space straight e to the power of straight x right square bracket subscript 0 superscript 1 space minus space integral subscript 0 superscript 1 1. space straight e to the power of straight x dx plus open square brackets fraction numerator sin space begin display style fraction numerator straight pi space straight x over denominator 4 end fraction end style over denominator begin display style straight pi over 4 end style end fraction close square brackets subscript 0 superscript 1 space equals space left square bracket straight x space straight e to the power of straight x right square bracket subscript 0 superscript 1 minus open square brackets straight e to the power of straight x close square brackets subscript 0 superscript 1 plus 4 over straight pi open square brackets sin space πx over 4 close square brackets subscript 0 superscript 1
       equals space left parenthesis straight e minus 0 right parenthesis space minus space left parenthesis straight e minus straight e to the power of 0 right parenthesis space plus space 4 over straight pi open parentheses sin space straight pi over 4 minus sin space 0 close parentheses
      equals straight e minus straight e plus 1 plus 4 over straight pi open parentheses fraction numerator 1 over denominator square root of 2 end fraction minus 0 close parentheses space equals space 1 plus fraction numerator 2 square root of 2 over denominator straight pi end fraction

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