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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}.

The given region is {(x, y): x² ≤ y ≤ x}
This region is the intersection of the following regions:

straight R subscript 1 space equals space open curly brackets open parentheses straight x comma space straight y right parenthesis close parentheses space colon space straight x squared space less or equal than space straight y close curly brackets straight R subscript 2 space equals space open curly brackets left parenthesis straight x comma space straight y right parenthesis space colon space straight y space less or equal than straight x close curly brackets

Consider the equations

straight y space equals straight x squared

...(1)
and y = x ...(2)
From (1) and (2), we get

straight x space equals space straight x squared space space or space space straight x squared minus straight x space equals space 0

rightwards double arrow space space space space straight x left parenthesis straight x minus 1 right parenthesis space equals space 0 space space space space rightwards double arrow space space space space straight x space equals space 0 comma space space 1

∴ from (2), y = 0, 1
∴ curves (1) and (2) intersect in the points O (0, 0) and A (1, 1)
Required area = area of the shaded region

equals space integral subscript 0 superscript 1 straight x space dx space minus space integral subscript 0 superscript 1 straight x squared space dx space equals space open square brackets straight x squared over 2 close square brackets subscript 0 superscript 1 space minus space open square brackets straight x cubed over 3 close square brackets subscript 0 superscript 1 equals space open parentheses 1 half minus 0 close parentheses space minus space open parentheses 1 third minus 0 close parentheses space equals space 1 half minus 1 third space equals space fraction numerator 3 minus 2 over denominator 6 end fraction space equals space 1 over 6 space sq. space units.

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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.

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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.