Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Application of Integrals

Name: Zigya - For The Curious Learner
Rating: 4.4 (740 reviews)

Long Answer Type

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

176 Views

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}.

202 Views

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].

146 Views

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.

112 Views

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

127 Views

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

The equation of curve is x² = 4 y    ...(1)
which is upward parabola with vertex O.
The equation of line is
x = 4 y - 2    ...(2)
Let us solve (1) and (2)
Putting x = 4y - 2 in (1), we get
   $left parenthesis 4 space straight y space minus 2 right parenthesis squared space equals space 4 space straight y$
$therefore space space space 16 space straight y squared minus space 16 space straight y plus space 4 space equals space 4 straight y therefore space space 16 space straight y squared minus 20 straight y space plus space 4 space equals space 0 or space 4 straight y squared minus 5 straight y plus 1 space equals space 0 therefore space space space space straight y space equals space fraction numerator 5 plus-or-minus square root of 25 minus 16 end root over denominator 8 end fraction space equals space fraction numerator 5 plus-or-minus 3 over denominator 8 end fraction space equals space 8 over 8 comma space 2 over 8 therefore space space space space straight y space equals space 1 comma space space 1 fourth therefore space space from space left parenthesis 2 right parenthesis comma space space straight x space equals space 4 minus 2 comma space 1 minus 2 space equals space 2 comma space minus 1 therefore space space curve space left parenthesis 1 right parenthesis space and space line space left parenthesis 2 right parenthesis space intersect space in space two space points space straight A left parenthesis 2 comma space 1 right parenthesis space and space straight B space open parentheses negative 1 comma space 1 fourth close parentheses$

From A, draw AM ⊥ x-axis and from B. draw BN ⊥ x-axis.
Required area = area AOB
= Area of trapezium BNMA - (area BNO + area OMA)
= $1 half open parentheses 1 plus 1 fourth close parentheses cross times 3 space minus space integral subscript negative 1 end subscript superscript 2 straight y space dx space equals space 1 half cross times 5 over 4 cross times 3 space minus space integral subscript negative 1 end subscript superscript 2 straight x squared over 4. dx$    $open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets$
$equals space 15 over 8 minus 1 fourth integral subscript negative 1 end subscript superscript 2 straight x squared space dx space equals space 15 over 8 minus 1 fourth open square brackets straight x cubed over 3 close square brackets subscript negative 1 end subscript superscript 2 space equals space 15 over 8 minus fraction numerator 1 over denominator 4 cross times 3 end fraction open square brackets straight x close square brackets subscript negative 1 end subscript superscript 2 equals space 15 over 8 minus 1 over 12 open square brackets left parenthesis 2 right parenthesis cubed minus left parenthesis negative 1 right parenthesis cubed close square brackets space equals space 15 over 8 minus 1 over 12 open square brackets 8 minus left parenthesis negative 1 right parenthesis close square brackets equals space 15 over 8 minus 1 over 12 left parenthesis 8 plus 1 right parenthesis space equals space 15 over 8 minus 9 over 12 space equals space 9 over 8 space sq. space units. space$

201 Views

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

1184 Views

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}

518 Views

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.