Find the area of the region bounded by y2 = 4 x, x = 1, x = 4
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    Application of Integrals
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    Find the area bounded by the curve y2 = 4 a (x-1) and the lines x = 1 and y = 4 a.
    The equation of parabola is y2 = 4 a (x-1)    ...(1)
    When y = 4 a , from (1) 16 a2 = 4 a (x-1)
    ⇒ x - 1 = 4 a ⇒ x = 4 a + 1
    therefore space space required space area space equals space integral subscript straight x space equals space 1 end subscript superscript straight x equals 4 straight a plus 1 end superscript straight y space dx space equals space integral subscript straight x equals 1 end subscript superscript straight x space equals space 4 straight a plus 1 end superscript 2 square root of straight a space square root of straight x minus 1 end root space dx open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
                           equals space 2 square root of straight a integral subscript 1 superscript 4 straight a plus 1 end superscript left parenthesis straight x minus 1 right parenthesis to the power of 1 half end exponent dx space equals space 2 square root of straight a open square brackets fraction numerator left parenthesis straight x minus 1 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 1 superscript 4 straight a plus 1 end superscript equals space fraction numerator 4 square root of straight a over denominator 3 end fraction open square brackets left parenthesis straight x minus 1 right parenthesis to the power of 3 over 2 end exponent close square brackets subscript 1 superscript 4 straight a plus 1 end superscript space equals space fraction numerator 4 square root of straight a over denominator 3 end fraction open square brackets left parenthesis 4 straight a right parenthesis to the power of 3 over 2 end exponent minus 0 close square brackets space equals space fraction numerator 4 square root of straight a over denominator 3 end fraction cross times 8 straight a to the power of 3 over 2 end exponent equals space 32 over 3 space straight a squared space sq. space units
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    Answer

    Find the area of the region bounded by x2 = 16 y, y = 1, y = 4 and the y-axis in the first quadrant.

    The equation of curve is
                      straight x squared space equals space 16 space straight y
    Required area = integral subscript 1 superscript 4 straight x space dy
                     equals space integral subscript 1 superscript 4 4 square root of straight y dy space equals space 4 integral subscript 1 superscript 4 straight y to the power of 1 half end exponent dy space equals space 4 open square brackets fraction numerator straight y 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 equals space 8 over 3 open square brackets straight y to the power of 3 over 2 end exponent close square brackets subscript 1 superscript 4 space equals space 8 over 3 open square brackets left parenthesis 4 right parenthesis to the power of 3 over 2 end exponent minus 1 close square brackets space equals space 8 over 3 left square bracket 8 minus 1 right square bracket space equals 8 over 3 cross times 7 space equals space 56 over 3 space sq. space units.


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

    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
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    Find the area of the region lying in the first quadrant and bounded by x2 = y - 3, y = 4, y = 6 and the y-axis in the first quadrant.
    The equation of curve is x2 = y - 3. which is an upward parabola whose vertex is (0,3).
    Lines are y = 4,  y = 6
    Required area = Area ABCD
                             equals space integral subscript 4 superscript 6 straight x space dy space equals space integral subscript 4 superscript 6 square root of straight y minus 3 end root space dy
                                equals space integral subscript 4 superscript 6 left parenthesis straight y minus 3 right parenthesis to the power of 1 half end exponent dy space equals space open square brackets fraction numerator left parenthesis straight y minus 3 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 equals space 2 over 3 open square brackets left parenthesis straight y minus 3 right parenthesis to the power of 3 over 2 end exponent close square brackets subscript 4 superscript 6 space equals space 2 over 3 open square brackets left parenthesis 3 right parenthesis to the power of 3 over 2 end exponent minus left parenthesis 1 right parenthesis to the power of 3 over 2 end exponent close square brackets space equals 2 over 3 left parenthesis 3 square root of 3 space minus space 1 right parenthesis space sq. space units.
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    Answer

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

    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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