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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.

The equation of given line is
y = 3 x + 2    ...(1)
Consider the lines
x = -1    ...(2)
and    x = 1    ...(3)
Line (1) meets x-axis where y = 0

$therefore$ putting y = 0 in (1), we get,
$0 space equals space 3 straight x plus 2 space space space or space space straight x space space equals space minus 2 over 3$
$therefore$ line (1) meets x-axis in $straight A open parentheses negative 2 over 3 comma space 0 close parentheses$
Let line (1) meet lines (2) and (3) in B and D respectively. From B, draw BC ⊥ x-axis and from D, draw DE ⊥ x-axis.
Required area = Area of region ACBA + area of region ADEA
$equals space open vertical bar integral subscript negative 1 end subscript superscript fraction numerator negative 2 over denominator 3 end fraction end superscript left parenthesis 3 straight x plus 2 right parenthesis space dx close vertical bar space plus space integral subscript negative 2 over 3 end subscript superscript 1 left parenthesis 3 straight x plus 2 right parenthesis space dx equals space open vertical bar open square brackets fraction numerator 3 straight x squared over denominator 2 end fraction plus 2 straight x close square brackets subscript negative 1 end subscript superscript negative 2 over 3 end superscript close vertical bar space plus space open square brackets fraction numerator 3 straight x squared over denominator 2 end fraction plus 2 straight x close square brackets subscript negative 2 over 3 end subscript superscript 1 equals space open vertical bar open curly brackets 3 over 2 minus open parentheses negative 2 over 3 close parentheses squared plus 2 space open parentheses negative 2 over 3 close parentheses close curly brackets space minus space open curly brackets 3 over 2 left parenthesis negative 1 right parenthesis squared plus 2 left parenthesis negative 1 right parenthesis close curly brackets close vertical bar space space space space space space space space space space space space space space space space space$
   $plus open square brackets 3 over 2 left parenthesis 1 right parenthesis squared plus 2 left parenthesis 1 right parenthesis close square brackets space minus space open square brackets 3 over 2 open parentheses negative 2 over 3 close parentheses squared plus 2 open parentheses negative 2 over 3 close parentheses close square brackets$
$equals space open vertical bar open parentheses 2 over 3 minus 4 over 3 close parentheses minus open parentheses 3 over 2 minus 2 close parentheses close vertical bar plus open parentheses 3 over 2 plus 2 close parentheses minus open parentheses 2 over 3 minus 4 over 3 close parentheses space equals space open vertical bar negative 2 over 3 plus 1 half close vertical bar plus open parentheses 7 over 2 plus 2 over 3 close parentheses equals space open vertical bar negative 1 over 6 close vertical bar plus 25 over 6 space equals space 1 over 6 plus 25 over 6 equals space 26 over 6 equals space 13 over 3 space sq. space units.$

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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.