Application of Integrals Class 12 OP Malhotra Exe-25A Area Under a Curve ISC Maths Solutions Ch-25. In this article you would learn How to find the area under a curve example and Practice questions / Problems with Answer / Solutions. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Application of Integrals Class 12 OP Malhotra Exe-25A Area Under a Curve ISC Maths Solutions Ch-25

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-25	Application of Integrals
Writer	OP Malhotra
Exe-25(a)	Calculation of Area Under a Curve

Calculation of Area Under a Curve

Application of Integrals Class 12 OP Malhotra Exe-25A ISC Maths Solutions Ch-25

Que-1: Using integration, find the area of the region bounded by
(i) the line 3 y = 2 x + 4, the x-axis and the lines x = 1 and x = 3.
(ii) 2 y = -x + 6, the x-axis and the lines x = 2 and x = 4. 

Sol: (i) The eqn. of given line be
3 y = 2 x + 4 ……………..(1)
eqn. (1) meets x-axis at A(-2, 0) and y-axis at B (0,4/3)

= 1/12 [(10)² – 6²]
= 1/12(100 – 36)
= 64/12 = 16/3 sq. units

(ii) The eqn. of given line be
2y = -x + 6
it meets coordinate axes at A(6, 0) and B(0, 3).

= -1/4 [4 – 16]
= 3 sq. units

Que-2: (i) Using integration, find the area of the region bounded between the line x = 2 and the parabola y² = 8 x.
(ii) Find the area of the region bounded by the parabola y² = 12 x and its latus rectum.
(iii) Find the area enclosed between the co-ordinate axes and the curve y² = 4 a(x + λ) in the second

Sol: (i) Clearly the figure is symmetrical about x-axis.
∴ area enclosed by the parabola
= 2 × area enclosed by parabola
and x = 2 in Ist quadrant
Divide the region in first quadrant into vertical strips each vertical strip has its lower end on x-axis and upper and on the parabola y = √8x. So the approximating rectangle has length |y| and width d x. This rectangle can move between x = 0 and x = 2.

(ii) Given eqn. of parabola be y² = 12 x represents a right handed parabola with vertex at (0,0). Here 4 a = 12
⇒ a = 3
∴ required area = 2 × area of region OABO

 

(iii) The given eqn. of parabola be
y² = 4 a(x + λ)
represents a right handed parabola with vertex (-λ, 0) it meets y-axis at A(0, 2√aλ)

Que-3: Draw the rough sketch of the curve y =√3x+4x and find the area under the curve, above the x-axis and between x = 0 and x = 4.

Sol: The eqn. of given curve y = √3x+4
The table of values is given as under :

x	0	4	2
y	2	4	√10

Plot the points (0, 2) ;(4, 4) and (2,√10) on graph paper and join then by free hand curve.
∴ Required area of region OABCO

= 2/9[(16)^3/2-(4)^3/2]
= 2/9[64 – 8]
= 112/9 sq. units

 

Que-4: Sketch the graph of y = |x + 1|. Evaluate  

4
∫
2

|x + 1| d x. What does the value of the integral represent on

Sol: For Graph ; For x < -1; y = f(x) = -(x + 1) For x = -1 ; y = 0 For x > -1 ; y = x + 1

= −1/2(0 – 9) + 1/2(9 – 0)
= 9/2 + 9/2 = 9

When x = -4 < -1 ∴ y = -(-4 + 1) = 3 i.e. point becomes (-4, 3). When x = 2 > -1
∴ y = 2 + 1 = 3 i.e. point becomes (2, 3).
Further, area of ∆ ABC = |1/2×(−3)×(3)| = 92 sq. units
and area of ∆CDE = |1/2×3×3| = 9/2 sq. units

= sum of areas of two ∆ from -4 to -1 and -1 to 2 .

Que-5: Find the area bounded by the curve y = x² and the line y = 16. [Fig. 25.41]

Sol: (i) The given curve y = x²
and the line y = 16 intersects when 16 = x²
⇒ x = ± 4
∴ Region R = {(x, y) ; x² ≤ y ≤ 10 ; 0 ≤ x ≤ 4}

= 64 – 64/3
= 128/3 sq. units

Que-6: Sketch the region lying in the first quadrant and bounded by y = 4 x², x = 0, y = 1 and y = 4. Find the area of the region, using integration.

Sol: The given curve is y = 4 x² be an upward parabola with vertex at (0, 0) and is symmetrical about y-axis.
∴ required area = area of region OABCO

= 1/3[8 – 1]
= 7/3 square units

Que-7: Find the gradients of the curve y = x²(2 – x) at the points (0, 0) and (2, 0) and sketch the part of the curve of which both x and y are positive. Find the area between this part of the curve and the axis.

Sol:

x	0	1	2	3/2
y	0	1	0	9/8

Que-8: Find the area included between the curve y = 9 – x² , the X-axis and the lines x = -2 and x = ± 2.

Sol:

Que-9: Find the area bounded by the curve y² = 4 x, the line y = 3, and the y-axis.

Sol:

Que-10: Find the area of the region bounded by y = -1, y = 2, x = y and x = 0.

Sol: Given lines are
y = -1 ……………….(1)
y = 2 ……………….(2)
x = y ……………….(3)
x = 0 …………………(4)
and
Since y = -1 and y = 2 are the lines are parallel to x-axis and x = 0 represents y-axis.
The line y = x passes through origin and making a slope of 45° with positive direction of x-axis.
Divide the region into horizontal strips as shown in shaded portion.


= −1/2(0 – 1) + 1/2 (4 – 0)
= 1/2 + 2 = 5/2 sq. units

Que-11: Find the area bounded by the parabola y² = 2 x and the ordinates x = 1 and x = 4.

Sol:

Que-12: Find the area bounded by the curve y² = 4 a² (x – 1) and the lines x = 1 and y = 4 a.

Sol:

Que-13: Evaluate the area of the region bounded by the curve y = 2√1−x², and the X-axis, after drawings rough sketch of the same.

Sol:

Que-14: Make a rough sketch of the curve 3 y = (2 x + 1)(2 – x), and calculate its gradient at the point where it meets the axis of y. Find the area bounded by the curve and the axis of x.

Sol:

Que-15: Given alongside figure shows a sketch of the curve y = (x – 1)(4 – x). If P is the point (2, 2), verify that the tangent at P passes through the origin O as shown. Calculate the area O A P enclosed between the tangent, the curve and the axis of x.

Sol:

Que-16: The curve y = a x² + b x + c passes through the points (1, 0),(2, 0) and its gradient at the point (2, 0) is 2 . Find the numerical value of the area included between the curve and the axis of x.

Sol: The eqn. of given curve y = a x² + b x + c eqn. (1) passes through the point (1, 0) and (2, 0).
∴ 0 = a + b + c
0 = 4 a + 2 b + c
Diff. eqn. (1) both sides w.r.t. x; ; we have
dy/dx = 2 a x + b
∴ gradient of curve at point (2, 0) = 2
⇒ (dy/dx) (2,0) = 2
⇒ 4 a + b = 2
eqn. (3) -2 x eqn. (4) ; we have
4 a + 2 b + c – 8 a – 2 b + 4 = 0
⇒ -4 a + c = -4
⇒ 4 a – c = 4
eqn. (4) – eqn. (2) ; we have
3a – c = 2
eqn. (5) – eqn. (6) gives; a = 2
∴ from (6) ; c = 6 – 2 = 4
∴ from (4); b = 2 – 4 a = 2 – 8 = -6
Thus from (1); we have
y = 2 x² – 6 x + 4
eqn. (7) meets x-axis when y = 0
∴ 2 x² – 6 x + 4 = 0
⇒ (x – 1)(2 x – 4) = 0
⇒ x = 1,2

= |2(8/3) – 6 + 4 – 1/3 + 3/2 – 2]|
= 2[7/3 + 3/2 – 4]
= |2(14+9−24)/6|
= 1/3 sq. units

Que-17: The line y = 2 x meets the curve y² = 4 x at the point O (the origin) and P, and P N is perpendicular to the y-axis. Prove that the area between the curve and O P is one-half the area enclosed by the lines O N, N P and the curve.

Sol: eqn. of given line be
y = 2 x

y² = 4 x
and eqn. of given curve be
eqn. (2) represents a right handed parabola with vertex at (0, 0) and line (1) meets curve (2) when 4 x² = 4 x
⇒ 4 x(x – 1) = 0
⇒ x = 0,1
∴ from (1); y = 0, 2
Thus the point of intersection are O(0, 0) and P(1, 2).
∴ area betwen the curve and OP

Que-18: Calculate the areas of the two parts into which the area enclosed by the x-axis and the curve y=18 x-3 x² is divided by the line x=4.

Sol: Given eqn. of curve be
y = 18 x – 3 x²
⇒ y = -3(x² – 6 x + 9 – 9)
⇒ y = -3(x – 3)² + 27
⇒ 3(x – 3)² = 27 – y
⇒ (x – 3)²
= -1/3(y – 27)
which represents a downward parabola with vertex (3, 27). Curve (1) meets x-axis at y = 0
∴18 x – 3 x² =0
⇒ 3 x(6 – x) = 0
⇒ x = 0, 6
i.e. at points (0, 0) and Q(6, 0)
The line x = 4 meets curve (1)
When y = 72 – 48 = 24 i.e. at P(4, 24)

Que-19: Draw the rough sketch of y² + 1 = x, x < 2 and find the area enclosed by the curve and the line x = 2.

Sol: Given equation of curve be y² + 1 = x
⇒ y² = x – 1 Clearly it represents a parabola (right handed) with vertex (1, 0) and does not meeting y-axis at any point.
Further the given curve meets the line x = 2 at y² + 1 = 2
⇒ y = ± 1 i.e. at points (2, + 1) and (2, – 1).

Que-20: Find the area bounded by the curve y = 4 – x² and the line y = 0 and y = 3.

Sol:

We want to find the area of region bounded by curves given below:
and
y = 4 – x²
y = 0, y = 3
eqn. (1) represents a parabola (downward) with vertex (0,4). y = 4 – x² meets y = 3 when 3 = 4 – x²
⇒ x² = 1
⇒ x = ± 1
∴ points of intersection are ( ± 1,3).
∴ required area = 2 × area of region OABCO
Divide the region into small horizontal strips. Each strip has left end on y-axis and right end on curve y = 4 – x².

Que-21: Make a rough sketch of the following curves. Also, find the area enclosed between the curves and the axes.
(i) y = cos x, 0 ≤ x ≤ π/2
(ii) y = cos 2 x, 0 ≤ x ≤ π/4
(iii) y = sin x, 0 ≤ x ≤ π/2
(iv) y = cos ^2 x, 0 ≤ x ≤ π/2

Sol:  (i) Given eqn. of curve be y = cos x ; 0 ≤ x ≤ π/2
Table of values is given as under:

x	0	π/2	π/3	π/2
y	1	12	0	0

(ii) Given eqn. of curve be
y = cos 2 x ; 0 ≤ x ≤ π/4

x	0	π/6	π/4
y	1	1/2	0

(iii) Equation of given curve be
y = sin x ; 0 ≤ x ≤ π/2

x	0	π/6	π/4	π/3	π/2
y	0	1/2	1/√2	√3/2	1

(iv) Given curve is y = cos² x
For rough sketch we construct a table of values as under :

x	0	π/6	π/4	π/3	π/2
y	0	3/4	1/2	1/4	1

Que-22: Draw a rough sketch of the curve y = cos² x in [0, π] and find the area enclosed by the the lines x = 0, x = π and the x-axis.

Sol: Given eqn. of curve be y = cos² x in [0, π]

x	0	π/6	π/4	π/3	π/2	2π/3	π
y	1	3/4	1/2	1/4	0	1/4	1

Que-23: Draw a rough sketch of the curve y = 1/2 π + 2 sin² x and find the area between the x-axis, the curve and the ordinates x = 0 and x = π

Sol: Given eqn. of curve be y = π/2 + 2 sin² x

x	0	π/6	π/4	π/3	π/2	2π/3	3π/4	5π/6	π
y	π/2	π/2+1/2	π/2+1	π/2+3/2	3.57	3.07	2.57	2.07	1.57

The rough sketch for given curve is given as under.
Divide the region into vertical strips. Each vertical strip has lower end on x-axis and upper end on given curve.
Thus the corresponding rectangle has length |y| and width d x and this rectangle move from x = 0 and x = π.

= π²/2 + [π – 0 – 0 – 0]
= π²/2  + π
=π/2  (π + 2) sq. units

 

Que-24: Show that the area included between the x-axis and the curve a² y = x² (x + a) is a²/12

Sol: Clearly the curve a² y = x² (x + a)
meets x-axis i.e. y = 0, where x² (x + a) = 0
⇒ x = 0,-a
∴ curve meets x-axis at (0,0) and (-a, 0).
∴ required area

Que-25: (i) Calculate the area bounded by the curve y = x² – 1, the x-axis and the line y = 8.
(ii). Calculate the approximate increase in this area, if the line y = 8 is changed to y = 8.01.

Sol:

Que-26: Find the area of the region bounded by y = -1, y = 2, x = y³ and x = 0.

Sol:

Que-27: Find the area bounded by x = a t² , y = 2 a t between the ordinates corresponding to t = 1 and t = 2.

Sol:

–: End of Application of Integrals Class 12 OP Malhotra Exe-25A ISC Maths Solutions Ch-25 :–

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