Please refer to Application of Integrals Class 12 Mathematics Exam Questions provided below. These questions and answers for Class 12 Mathematics have been designed based on the past trend of questions and important topics in your Class 12 Mathematics books. You should go through all Class 12 Mathematics Important Questions provided by our teachers which will help you to get more marks in upcoming exams.
Class 12 Mathematics Exam Questions Application of Integrals
Class 12 Mathematics students should read and understand the important questions and answers provided below for Application of Integrals which will help them to understand all important and difficult topics.
Very Short Answer Type Questions
Question. Find the area of the ellipse
Answer. Since, area of the ellipse
∴ Required area = p × 4 × 9 = 36p sq. units.
Question. Find the area enclosed by the curve 2 y = x and the line y = 16
Answer.
Question. Find the area bounded by y = x2 ,ℎ𝑒𝑥 − axis and the lines 𝑥 = −1 and 𝑥 = 1.
Answer.
Question. Find the area of the region bounded by y = |x|, x ≤ 5 in the first quadrant.
Answer. 3. We have, y = –x, if x < 0 … (i)
y = x, if x ≥ 0 …(ii)
Question. Find the area of the region bounded by the parabola y2 = x and the straight line 2y = x
Answer. (4/3) sq. units
Question. Find the area of the region bounded by the curve x2 + y2 = 1.
Answer. π sq. units
Question. Using integration find the area enclosed by the curve
Answer.
Question. Using integration, find the area of the region enclosed by the curves y = log x, x-axis and ordinates x = 1, x = 2.
Answer. 12π
Question. Find the area of the region bounded by the parabola y2 = 8x and the line 𝑥 = 2
Answer. (32/3) sq. units
= 2log 2 – 1 = log 4 – log e
= (4/e ) log sq. units
Question. Find the area bounded by the curve y = 3x , x-axis and the ordinates x=1 and x=3.
Answer.
Question. Find the area of the region bounded by the curve y2 = 4x and the line y=2.
Answer. 2/3
Question. Using integration find the area enclosed by the curve
Answer. 20π sq. units
Question. Find the area of the smaller region bounded by x2 + y2 = 9 and the line x = 1.
Answer. We have, x2 + y2 = 9
and x = 1.
Question. Find the area of the region bounded by the curve y = x + 1 and the lines x = 2 and x = 3.
Answer. We have, y = x + 1, which is a straight line
Question. Find the area between the curve y = 4 + 3x – x2 and x-axis
Answer. We have, y = 4 + 3x – x2, a parabola with vertex at
Putting y = 0, we get x2 – 3x – 4 = 0
⇒ (x – 4)(x + 1) = 0 ⇒ x = –1 or x = 4
Question. Find the area of the region bounded by the curve x = 2y + 3 and the lines y = 1 and y = –1.
Answer. We have x = 2y + 3, a straight line
Question. Find the area bounded by the curve y2 = 9x and the lines x = 1, x = 4 and y = 0 in the first quadrant
Answer. We have, y2 = 9x and lines x = 1, x = 4
Question. Find the area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2.
Answer.
Question. Find the area bounded by the curves y = sin x, the line x = 0 and the line x = 2π.
Answer.
= −cosπ + cos0+|−cos2π + cosπ|
= 1 + 1 + |–1 –1| = 2 + |– 2| = 2 + 2 = 4 sq. units
Short Answer Type Questions -I
Question. Find the area of the region bounded by the curve y2 = x and the lines x = 1, x = 4 and the x-axis.
Answer. Since the given parabola y2 = x is symmetrical about positive x-axis
Question. If y = 2 sin x + sin 2x for 0 ≤ x ≤ 2 π, then find the area enclosed by the curve and x-axis.
Answer. We have, y = 2sin x + sin 2x, 0 ≤ x ≤ 2π
Putting y = 0, we get 2sin x + sin 2x = 0
⇒ 2sin x + 2sin x cos x = 0
⇒ 2sin x (1 + cos x) = 0
⇒ sin x = 0 or cos x = –1
⇒ x = 0, p, 2π
∴ Required area
Question. Find the area bounded by the lines y = ||x| – 1| and the x-axis.
Answer. We have, y = |x – 1|, if x ≥ 0
Question. Find the area of triangle whose two vertices formed from the x-axis and line y = 3 – |x|.
Answer. We have, y = 3 – |x|
⇒ y = 3 + x, ” x < 0 …(i)
and y = 3 – x, ” x ≥ 0 …(ii)
∴ Required area = area of shaded region
Question. Find the area of region bounded by the curve y2 = 4x and the lines x = 2, x = 4 and the x-axis.
Answer. Since the given curve represented by the equation y2 = 4x is a parabola
∴ Required area
Question. Draw the region lying in first quadrant and bounded by y = 9x2, x = 0, y = 1 and y = 4. Also, find the area of region using integration.
Answer. The rough sketch of the curve y = 9x2, x = 0, y = 1 and y = 4 is as shown in the figure.
Question. Find the area of the region bounded by the curve
Answer.
Question. Find the area of the region bounded by y2 = 9x, x = 2, x = 4 and the x-axis in the first quadrant.
Answer. The given parabola is y2 = 9x. It is symmetrical about positive x-axis.
Question. Using integration, find the area of region bounded between the line x = 2 and the parabola y2 = 8x.
Answer. The rough sketch of the parabola y2 = 8x and line x = 2 is as shown in the figure.
Question. Find the area bounded by the curve y = x |x| , x-axis and the lines x = –3 and x = 3.
Answer. The equation of the curve is
∴ Required area = 2(Area of region shaded in first quadrant)
Short Answer Type Questions -II
Question. Draw the graph of curve y = |x + 1|. Hence
Answer. We have, y = |x + 1|
Question. Find the area of the region bounded by the curve
line y = x and the positive x-axis.
Answer. We have, curve
and line y = x
The rough sketch of the curve and line y = x is shown in the figure.
Question. Find the area of region bounded by y = √x and y = x.
Answer. We have, curves y = √x and y = x.
The points of intersection of y = √x and y = x are O(0, 0) and A(1, 1).
Question. Find the area bounded by the ellipse
and the ordinates x = ae and x = 0, where b2 = a2(1 – e2) and e < 1.
Answer. The rough sketch of ellipse
and the line x = 0 and x = ae is shown in the figure.
Question. Find the area of the region bounded by the parabola y2 = 2x + 1 and the line x – y – 1 = 0.
Answer. The rough sketch of the parabola y2 = 2x + 1 and line x – y – 1 = 0 is as shown in the figure.
The intersection points of y2 = 2x + 1 and x – y = 1 are (0, –1) and (4, 3).
The required area of shaded region
Question. Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32.
Answer. We have, x2 + y2 = 32 …(i)
and y = x …(ii)
Solving (i) and (ii), the intersection points are O(0, 0) and A(4, 4) in first quadrant.
The rough sketch of the circle x2 + y2 = 32 and line y = x is shown in the figure.
Question. Find the area of smaller region bounded by the ellipse
Answer.
Question. Find the area of the triangle formed by the tangent and normal at the point (1,√3) on the circle x2 + y2 = 4 and the x-axis.
Answer. The tangent on x 2 + y 2 = 4 at (1, √3) is x + √3y = 4
and equation of normal at (1,√3) is y = x 3.
∴ Required area
Question. Draw the region bounded by y = 2x – x2 and x-axis and find its area using integration.
Answer. We have, y = 2x – x2 ⇒ – y = x2 – 2x
⇒ –y + 1 = x2 – 2x + 1 ⇒ –(y – 1) = (x – 1)2
Clearly it represents a parabola opening downwards whose vertex is (1, 1) and cuts x-axis at (0, 0) and (2, 0).
The rough sketch of the curve is given below :
Question. Find the area bounded by the curve y = 2x – x2 and the straight line y = –x.
Answer. The curve y = 2x – x2 represents a parabola opening downwards and cutting x-axis at (0, 0) and (2, 0). Clearly, y = –x represents a line passing through the origin and making 135° with x-axis. A rough sketch of the two curves is shown in the figure. The region whose area is to be found is shaded in figure. The two curves intersect each other at (0, 0) and (3, –3).
Question. Determine the area under the curve
included between the lines x = 0 and x = a.
Answer. We have given the equation of curve
⇒ y2 + x2 = a2, a circle with centre (0, 0) and radius a
Thus the required area = area of the shaded region
Question. AOB is a positive quadrant of the ellipse
where OA = a, OB = b. Find the area between the arc AB and chord AB of the ellipse.
Answer. Required area
Question. Find the area of the region bounded by y = |x – 1| and y = 1.
Answer. We have, y = x – 1, if x – 1 ≥ 0
y = – x + 1, if x – 1 < 0
Question. If the area bounded the curve y2 = 16x and line y = mx is 2/3, then find the value of m.
Answer. We have, y2 = 16x, a parabola with vertex (0, 0) and line y = mx.
∴ Required area
Question. Find the area enclosed between the parabola 4y = 3x2 and the straight line 3x – 2y + 12 = 0.
Answer.
Long Answer Type Questions
Question. Using integration, find the area of the region bounded by the curves :
y = |x + 1| + 1, x = – 3, x = 3 and y = 0.
Answer. Given, y = |x + 1| +1
We now draw the lines : y = 0, x = 3, x = –3 and
y = x + 2 if x ≥ – 1 …(i)
y = – x if x < – 1 …(ii)
Lines (i) and (ii) intersect each other at (– 1, 1)
Question. Find the area bounded by the circle x2 + y2 = 16 and the line √3y = x in the first quadrant, using integration.
Answer.
Curves (i) and (ii) intersect each other at (2 √3, 2) and (−2 √3, − 2).
∴ Required area = Area of region OBAO
= area DOBC + area of region BCAB
Question. Using integration, find the area bounded by the curve x2 = 4y and the line x = 4y – 2.
Answer. The given curve is x2 = 4y …(i)
and line is x = 4y – 2 …(ii)
Solving (i) and (ii), we get (x + 2) = x2
⇒ x2 – x – 2 = 0 ⇒ (x – 2) (x + 1) = 0 ⇒ x = 2, –1
Thus the points of intersection of the given curve and
Question. Find the area bounded by lines y = 4x + 5, y = 5 – x and 4y = x + 5.
Answer. We have, y = 4x + 5, y = 5 – x and 4y = x + 5
The rough sketch of the lines is shown in the figure.
Here, equation of line AB is y = 4x + 5,equation of line AC is y = 5 – x and equation of line BC is y = x+5/4
The intersection point of line AB and AC is at A(0, 5).
Similarly, the intersection point of line AB and line BC is at B(–1, 1) and the intersection point of line AC and BC is at C(3, 2).
Question. Find the area of the region bounded by the parabola y = x2 and y = |x|.
Answer. The given curves are y = x2 …(i)
Their points of intersection are A(1, 1), O(0, 0) and B(–1, 1).
∴ Required area
