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Integrals

Question
CBSEENMA12032419

Evaluate
integral subscript 0 superscript 1 open parentheses xe to the power of 2 straight x end exponent plus sin πx over 2 close parentheses dx

Solution

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

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