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Vector Algebra

Question
CBSEENMA12032728

Find the area bounded by the curve y = cos x between x = 0 and x = 2 straight pi.

Solution

The equation of curve is
y = cos x
Its rough sketch from x = 0 to x = 2 straight pi is shown in figure.

Required area = area of the region OABO + area of the region BCDB + area of the region DEFD.
equals space integral subscript 0 superscript straight pi over 2 end superscript cos space 2 straight x space dx space plus space open vertical bar integral subscript straight pi over 2 end subscript superscript fraction numerator 3 straight pi over denominator 2 end fraction end superscript cosx space dx close vertical bar space plus space integral subscript fraction numerator 3 straight pi over denominator 2 end fraction end subscript superscript 2 space straight pi end superscript space cos space straight x space dx
equals space open square brackets sin space straight x space close square brackets subscript 0 superscript straight pi over 2 end superscript space plus space open vertical bar open square brackets sin space straight x close square brackets subscript straight pi over 2 end subscript superscript fraction numerator 3 straight pi over denominator 2 end fraction end superscript close vertical bar space plus space open square brackets sin space straight x close square brackets subscript fraction numerator 3 straight pi over denominator 2 end fraction end subscript superscript 2 space straight pi end superscript
space equals space open parentheses sin space straight pi over 2 minus sin space 0 close parentheses space plus space open vertical bar open parentheses sin space fraction numerator 3 straight pi over denominator 2 end fraction minus sin space straight pi over 2 close parentheses close vertical bar space plus space open parentheses sin space 2 straight pi space minus space sin space fraction numerator 3 straight pi over denominator 2 end fraction close parentheses
equals space left parenthesis 1 minus 0 right parenthesis space plus space open vertical bar negative 1 minus 1 close vertical bar space plus space left parenthesis 0 plus 1 right parenthesis
equals space 1 plus open vertical bar negative 2 close vertical bar space plus space 1 space equals space 1 plus 2 plus 1 space equals space 4

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