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Application of Integrals

Name: Zigya - For The Curious Learner
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Long Answer Type

31. Find the area of the region enclosed by the parabola y² = 4 a x and the line y = mx.

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32.

Find the area bounded by the curve y = x² and the line y = x.
OR
Find the area of the region {(x. y): x² ≤ y ≤ x}.

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33. Using the method of integration find the area bounded by the curve |x| + |y| = 1.
[Hint: The required region is bounded by lines x + y = 1, x– y = 1, – x + y = 1 and
– x – y = 1].

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34.

Find the area of the region bounded by the line y = 3 x + 2, the x-axis and the ordinates x = - 1 and x = 1.

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35. Find the area of the region included between the parabola

straight y space equals 3 over 4 straight x squared space and space the space line space 3 straight x space minus space 2 straight y space plus space 12 space equals space 0

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36. Find the area bounded by the parabola x² = 4 y and the straight line x = 4 y - 2.

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37. Find the area of the region included between the parabola y² = x and the line x + y = 2.

The equation of parabola is
y² = x    ...(1)
The equation of line is
x + y = 2    ...(2)
From (2), y = 2 - x    ...(3)
Putting this value of y in (1), we get,
(2 - x)²= x
or x² - 4 x + 4 = x or x² - 5 x + 4 = 0

∴ (x - 1) (x - 4) = 0
∴ x = 1, 4
∴ from (3), y = 1, - 2
∴ parabola (1) and line (2) intersect in the points A (1, 1), B (4, - 1)
Also line (2) meets x-axis in C (2,0)
Required area is shaded.
Area above x-axis = area AOL + area ALC
$equals space integral subscript 0 superscript 1 square root of straight x dx plus integral subscript 1 superscript 2 left parenthesis 2 minus straight x right parenthesis space dx space equals space 2 over 3 open square brackets straight x to the power of 3 over 2 end exponent close square brackets subscript 0 superscript 1 plus open square brackets 2 space straight x space minus space straight x squared over 2 close square brackets subscript 1 superscript 2 equals space 2 over 3 left parenthesis 1 minus 0 right parenthesis space plus space open square brackets left parenthesis 4 minus 2 right parenthesis space minus space left parenthesis 2 minus 1 half right parenthesis close square brackets space equals space 2 over 3 plus 2 minus 3 over 2 equals 2 over 3 plus 1 half equals 7 over 6 space sq. space units$
Area below x-axis = Area OBM - area CBM
$equals space integral subscript 0 superscript 4 square root of straight x minus integral subscript 2 superscript 4 left parenthesis 2 minus straight x right parenthesis space dx space equals space 2 over 3 open square brackets straight x to the power of 3 divided by 2 end exponent close square brackets subscript 0 superscript 4 space minus space open square brackets 2 straight x minus straight x squared over 2 close square brackets subscript 2 superscript 4 equals space 2 over 3 left square bracket 4 to the power of 3 divided by 2 end exponent minus 0 right square bracket space minus space open square brackets left parenthesis 8 minus 8 right parenthesis space minus space left parenthesis 4 minus 2 right parenthesis close square brackets equals space 16 over 3 minus left parenthesis negative 2 right parenthesis space equals space 16 over 3 minus 2 space space space space space space space space space space space space space space left square bracket because space space area space is space always space positive right square bracket equals space 10 over 3 space sq. space units$
$therefore space space total space area space equals 7 over 6 plus 10 over 3 equals fraction numerator 7 plus 20 over denominator 6 end fraction equals 27 over 6 equals 9 over 2 space sq. space units. space$

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38.

Find the area of the region enclosed by the parabola x² = y, the line y = x + 2 and the x-axis.
OR
Draw the rough sketch and find the area of the region:
{(x, y): x² < y < x + 2}

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Short Answer Type

39.

Draw a rough sketch of the curves y = sin x and y = cos x as x varies from 0 to $straight pi over 2$ and find the area of the region enclosed by them and the x-axis.