If [ x ] stands for integral part of x, then show that $integral subscript 0 superscript 1 left square bracket 5 space straight x right square bracket space dx space equals space 2.$

Solution

Since $open square brackets 5 space straight x close square brackets$ has integral values at x = 0, $1 fifth comma space 2 over 5 comma space 3 over 5 comma space 4 over 5 comma space 1$
$therefore space space space space left square bracket 5 space straight x right square bracket$ is discontinuous at x = 0, $1 fifth comma space 2 over 5 comma space 3 over 5 comma space 4 over 5 comma space 1$
$therefore space space straight I space equals space integral subscript 0 superscript 1 left square bracket space 5 space straight x right square bracket space dx space equals space integral subscript 0 superscript 1 fifth end superscript left square bracket space 5 space straight x right square bracket space dx space plus space integral subscript 1 fifth end subscript superscript 2 over 5 end superscript left square bracket space 5 space straight x right square bracket space dx space plus space integral subscript 2 over 5 end subscript superscript 3 over 5 end superscript left square bracket 5 space straight x right square bracket space dx plus integral subscript 3 over 5 end subscript superscript 4 over 5 end superscript left square bracket 5 space straight x right square bracket space dx space plus space integral subscript 4 over 5 end subscript superscript 1 left square bracket 5 space straight x right square bracket space dx$
$equals space integral subscript 0 superscript 1 fifth end superscript left parenthesis 0 right parenthesis space dx space plus space integral subscript 1 fifth end subscript superscript 2 over 5 end superscript left parenthesis 1 right parenthesis space dx space plus space integral subscript 2 over 5 end subscript superscript 3 over 5 end superscript left parenthesis 2 right parenthesis space dx space plus space integral subscript 3 over 5 end subscript superscript 4 over 5 end superscript left parenthesis 3 right parenthesis space dx space plus space integral subscript 4 over 5 end subscript superscript 1 left parenthesis 4 right parenthesis space dx$
$equals 0 plus open square brackets straight x close square brackets subscript 1 fifth end subscript superscript 2 over 5 end superscript plus 2 open square brackets straight x close square brackets subscript 2 over 5 end subscript superscript 3 over 5 end superscript plus 3 open square brackets straight x close square brackets subscript 3 over 5 end subscript superscript 4 over 5 end superscript plus space 4 open square brackets straight x close square brackets subscript 4 over 5 end subscript superscript 1 equals space 0 plus open parentheses 2 over 5 minus 1 fifth close parentheses plus 2 open parentheses 3 over 5 minus 2 over 5 close parentheses plus 3 open parentheses 4 over 5 minus 3 over 5 close parentheses plus 4 open parentheses 1 minus 4 over 5 close parentheses equals space 0 plus 1 fifth plus 2 over 5 plus 3 over 5 plus 4 over 5 equals space space 2$

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Integrals

If [ x ] stands for integral part of x, then show that $integral subscript 0 superscript 1 left square bracket 5 space straight x right square bracket space dx space equals space 2.$

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