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

The equations of the given curves are

straight y equals space 3 over 4 straight x squared

...(1)
and

3 straight x minus 2 straight y plus 12 space equals space 0

...(2)
From (2),

2 straight y space equals space 3 straight x plus 12

therefore space space space straight y space equals space fraction numerator 3 straight x plus 12 over denominator 2 end fraction

Putting this value of y in (1), we get,

fraction numerator 3 straight x plus 12 over denominator 2 end fraction space equals 3 over 4 straight x squared

therefore space space 6 straight x space plus 24 space equals space 3 straight x squared space space space space space rightwards double arrow space space straight x squared minus 2 straight x minus 8 space equals space 0 space space space space rightwards double arrow space space space space left parenthesis straight x plus 2 right parenthesis thin space left parenthesis straight x minus 4 right parenthesis space equals space 0

rightwards double arrow space space space space straight x space equals space minus 2.4 therefore space space space space straight y space equals space 3 comma space 12

therefore

curves (1) and (2) intersect in points A (4, 12) and B(-2, 3).
From A. draw AM ⊥ x -axis and from B, draw BN ⊥ x-axis.
Required area = area of trapezium BNMA - (area BNO + area OMA)

equals space 1 half left parenthesis 3 plus 12 right parenthesis space cross times 6 space minus space integral subscript negative 2 end subscript superscript 4 3 over 4 straight x squared dx

open square brackets because space of space left parenthesis 1 right parenthesis close square brackets

equals space 45 space minus space 3 over 4 integral subscript negative 2 end subscript superscript 4 straight x squared dx space equals space 45 minus 3 over 4 open square brackets straight x cubed over 3 close square brackets subscript negative 2 end subscript superscript 4 equals space 45 minus 1 fourth open square brackets straight x cubed close square brackets subscript negative 2 end subscript superscript 4 space equals space 45 minus 1 fourth left square bracket 64 plus 8 right square bracket space equals space 27 space sq. space units. space

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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.

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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.

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40.

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

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