Vector Algebra

  • Question 1
    CBSEENMA12032684

    Find the area of the region bounded by the curve y= x and the lines x = 1, x = 4 and the x-axis in the first quadrant.

    Solution

    The equation of curve is y2 = x
    Required area = integral subscript 1 superscript 4 space straight y space dx space equals space integral subscript 1 superscript 4 square root of straight x space end root dx space equals integral subscript 1 superscript 4 straight x to the power of 1 half end exponent dx
                            equals space open square brackets fraction numerator straight x to the power of begin display style 3 over 2 end style end exponent over denominator begin display style 3 over 2 end style end fraction close square brackets subscript 1 superscript 4 space equals 2 over 3 open square brackets straight x to the power of 3 over 2 end exponent close square brackets subscript 1 superscript 4 space equals space 2 over 3 open square brackets open parentheses 4 close parentheses to the power of 3 over 2 end exponent minus 1 close square brackets
equals space 2 over 3 left parenthesis 8 minus 1 right parenthesis space equals space 2 over 3 cross times 7 space equals space 14 over 3 sq. space units.


     
    Question 2
    CBSEENMA12032685

    Find the area of the region bounded by y2 = 4 x, x = 1, x = 4 and the x-axis in the first quadrant.   

    Solution
    The equation of curve is y2 = 4 x
    Required area  = integral subscript 1 superscript 4 straight y space dx
                equals space integral subscript 1 superscript 4 2 square root of straight x space dx space equals space 2 integral subscript 1 superscript 4 straight x to the power of 1 half end exponent dx
equals space 2 open square brackets fraction numerator straight x to the power of begin display style 3 over 2 end style end exponent over denominator begin display style 3 over 2 end style end fraction close square brackets subscript 1 superscript 4 space equals space 4 over 3 open square brackets straight x to the power of 3 over 2 end exponent close square brackets subscript 1 superscript 4 space equals space 4 over 3 open square brackets open parentheses 4 close parentheses to the power of 3 over 2 end exponent minus 1 close square brackets
equals space 4 over 3 left parenthesis 8 minus 1 right parenthesis space equals space 4 over 3 cross times 7 space equals space 28 over 3 space sq. space units

    Question 3
    CBSEENMA12032686

    Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant.

    Solution
    The equation of curve is y2 = 9x, which is right handed parabola.
    Two lines are x = 2,  x = 4.
        Required area = Area ABCD
                                equals space integral subscript 2 superscript 4 straight y space dx space equals space integral subscript 2 superscript 4 3 square root of straight x space dx space equals space 3 space integral subscript 2 superscript 4 straight x to the power of 1 half end exponent dx
space equals space 3 open square brackets fraction numerator straight x to the power of begin display style 3 over 2 end style end exponent over denominator begin display style 3 over 2 end style end fraction close square brackets subscript 2 superscript 4 space equals space 3 cross times 2 over 3 open square brackets straight x to the power of 3 over 2 end exponent close square brackets subscript 2 superscript 4
equals space 2. space open square brackets left parenthesis 4 right parenthesis to the power of 3 over 2 end exponent minus left parenthesis 2 right parenthesis to the power of 3 over 2 end exponent close square brackets space equals space 2 space left square bracket space 8 space minus square root of 8 right square bracket space equals space 2 left parenthesis 8 minus 2 square root of 2 right parenthesis
equals space left parenthesis 16 minus 4 square root of 2 right parenthesis space sq. space units.

    Question 4
    CBSEENMA12032687

    Find the area of the region bounded by y2 = x - 2, x = 4, x = 6 and the x-axis in the first quadrant.

    Solution

    The equation of curve is y2 = x - 2. which is right handed parabola with vertex at (2, 0).
    Two lines are x = 4 and x = 6
    Required area  = Area ABCD
                             equals space integral subscript 4 superscript 6 straight y space dx space equals space integral subscript 4 superscript 6 square root of straight x minus 2 end root dx
equals space integral subscript 4 superscript 6 left parenthesis straight x minus 2 right parenthesis to the power of 1 half end exponent dx
equals space open square brackets fraction numerator left parenthesis straight x minus 2 right parenthesis to the power of begin display style 3 over 2 end style end exponent over denominator begin display style 3 over 2 end style end fraction close square brackets subscript 4 superscript 6 space equals space 2 over 3 open square brackets left parenthesis straight x minus 2 right parenthesis to the power of 3 over 2 end exponent close square brackets subscript 4 superscript 6
equals space 2 over 3 open square brackets 4 to the power of 3 over 2 end exponent minus 2 to the power of 3 over 2 end exponent close square brackets space equals space space 2 over 3 left parenthesis 8 minus 2 square root of 2 right parenthesis space sq. space units


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