He area of the region bounded by the parabola $y=x^2$ and $y=|x|$ is :
- A$3$
- B$\frac{1}{2}$
- ✓$\frac{1}{3}$
- D$2$
Answer: C.
View full solution →Question types
Application of Integrals question types
234 questions across 7 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.
234
Questions
7
Question groups
5
Question types
01
M.C.Q (1 Marks)
204 Q→02Assertion (A) & Reason (B) MCQ
6 Q→031 Marks Question
6 Q→042 Marks Questions
2 Q→053 Marks Question
4 Q→065 Marks Questions
2 Q→07Case study (4 Marks)
10 Q→Sample Questions
Application of Integrals questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
The area of the region bounded by the curve $x=y^2-2$ and $x=y$ is :
- A$\frac { 9 }{ 4 }$
- B$9$
- ✓$\frac { 9 }{ 2 }$
- D$\frac { 9 }{ 7 }$
Answer: C.
View full solution →The area of the ellipse $\frac{\text{x}2}{9}+\frac{\text{y}^2}{4}=1$ in first quadrant is $6\pi$ sq. units.
The ellipse is rotated about its centre in anti $-$ clockwise direction till its major axis coincides with $y-$ axis. Now the area of the ellipse in first Quadrant is $\pi$ sq. units.
The ellipse is rotated about its centre in anti $-$ clockwise direction till its major axis coincides with $y-$ axis. Now the area of the ellipse in first Quadrant is $\pi$ sq. units.
- A$2$
- ✓$4$
- C$6$
- D$8$
Answer: B.
View full solution →Area bounded by the lines $y = |x| - 2$ and $y = 1 - |x - 1|$ is equal to :
- ✓$4$ sq. units
- B$6$ sq. units
- C$2$ sq. units
- D$8$ sq. units
Answer: A.
View full solution →Area of the region bounded by y = |x – 1| and y = 1 is:
- A$2\text{ sq.}\text{ units}$
- ✓$1\text{ sq.}\text{ units}$
- C$\frac{1}{2}\text{ sq.}\text{ units}$
- D$\text{None of these}$
Answer: B.
View full solution →Directions: In these questions, a statement of Assertion is followed by a statement of Reason is given. Choose the correct answer out of the following choices:
Assertion: The area of the smaller region bounded by the ellipse $\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}=1$ the line $\frac{\text{x}}{3}+\frac{\text{y}}{2}=1$ is $\frac{3}{2}(\pi-2)\text{ sq.units}$
Reason: Formula to calculate the area of the smaller region bounded by the ellipse $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$ and the line $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$ is $\frac{\text{ab}}{4}(\pi-2) \text{ sq.units}$
Assertion: The area of the smaller region bounded by the ellipse $\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}=1$ the line $\frac{\text{x}}{3}+\frac{\text{y}}{2}=1$ is $\frac{3}{2}(\pi-2)\text{ sq.units}$
Reason: Formula to calculate the area of the smaller region bounded by the ellipse $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$ and the line $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$ is $\frac{\text{ab}}{4}(\pi-2) \text{ sq.units}$
- ✓Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
- BAssertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
- CAssertion is correct statement but Reason is wrong statement.
- DAssertion is wrong statement but Reason is correct statement.
Answer: A.
View full solution →Assertion $(A) :$ The area bounded by the curve $y=2 \cos x$ and the $x$-axis from $x=0$ to $x=2 \pi$ is $8$ sq. units.
Reason $(R) :$ Maximum value of the curve $y=2 \cos x$ is $2 .$
Reason $(R) :$ Maximum value of the curve $y=2 \cos x$ is $2 .$
- ABoth $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A).$
- ✓Both $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of $(A).$
- C$(A)$ is true but $(R)$ is false.
- D$(A)$ is false but $(R)$ is true.
Answer: B.
View full solution →Assertion $(A)$ : The area bounded by the curves $y^2=4 a^2(x-1)$ and lines $x=1$ and $y=4 a$ is $\frac{8 a}{3}$ sq. units.
Reason $(R)$ : The area enclosed between the parabola $y^2=49 x$ and its latus rectum $\frac{8 a^2}{3}$ sq. units.
Reason $(R)$ : The area enclosed between the parabola $y^2=49 x$ and its latus rectum $\frac{8 a^2}{3}$ sq. units.
- ABoth $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$.
- ✓Both $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of $(A).$
- C$(A)$ is true but $(R)$ is false.
- D$(A)$ is false but $(R)$ is true.
Answer: B.
View full solution →Assertion (A): The area bounded by the parabola $y^2=4 a x$ and the line $x=a$ and $x=4 a$ is $\frac{56 a^2}{3}$ sq. units.
Reason (R) : The area bounded by the parabola $y^2=49 x$ and $y=m x$ is $8 a^2 / 3 m^3$ sq. units.
Reason (R) : The area bounded by the parabola $y^2=49 x$ and $y=m x$ is $8 a^2 / 3 m^3$ sq. units.
- ABoth (A) and (R) are true and (R) is the correct explanation of (A).
- BBoth (A) and (R) are true but (R) is not the correct explanation of (A).
- ✓(A) is true but (R) is false.
- D(A) is false but (R) is true.
Answer: C.
View full solution →Assertion (A) : The area of the smaller region bounded by the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is $\frac{3}{2}(\pi-2)$ sq. units.
Reason (R) : Formula to calculate the area of the smaller region bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the line $\frac{x}{a}+\frac{y}{b}=1$ is $\frac{a b}{4}(\pi-2)$ sq. units.
Reason (R) : Formula to calculate the area of the smaller region bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the line $\frac{x}{a}+\frac{y}{b}=1$ is $\frac{a b}{4}(\pi-2)$ sq. units.
- ✓Both (A) and (R) are true and (R) is the correct explanation of (A).
- BBoth (A) and (R) are true but (R) is not the correct explanation of (A).
- C(A) is true but (R) is false.
- D(A) is false but (R) is true.
Answer: A.
View full solution →The area bounded by the curve y = x|x|, x-axis and the ordinates x = -1 and x = 1 is given by
Area bounded by the curve $y = x^3$, the x-axis and the ordinates $x = -2$ and $x = 1$ is
View full solution →Find the area under the given curves and given lines: $y = x^4, x = 1, x = 5$ and x-axis
View full solution →Find the area under the given curves and given lines: $y = x^2, x = 1, x = 2$ and x - axis.
View full solution →Area of the region bounded by the curve $y^2 = 4x$, y-axis and the line $y = 3$ is
View full solution →Find the area bounded by the curve y = cos x between x = 0 and x = 2$\pi$
View full solution →Find the area of the region bounded by the line y = 3x + 2, the x-axis and the ordinates x = - 1 and x = 1
Find the area bounded by the curve y = sin x between x = 0 and $x = 2\pi $
View full solution →Sketch the graph of y = |x + 3| and evaluate $\int_{ - 6}^0 {\left| {x + 3} \right|dx} $
View full solution →Find the area enclosed by the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
View full solution →Find the area enclosed by the circle $x^2 + y^2 = a^2$
View full solution →Find the area of the region bounded by the ellipse $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1$.
View full solution →Find the area of the region bounded by the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$
View full solution →A mirror in the shape of an ellipse represented by $\frac{\text{x}^2}{9}+-\frac{\text{y}^2}{4}=1$ was hanging on the wall. Arun and his sister were playing with ball inside the house, even their mother refused to do so. All of sudden, ball hit the mirror and got a scratch in the shape of line represented by $\frac{\text{x}}{3}+\frac{\text{y}}{2}=1$
Based on the above information, answer the following questions.
View full solution →Based on the above information, answer the following questions.
- Point(s) of intersection of ellipse and scratch (straight line) is (are).
- (0, 2), (3, 0)
- (2, 0), (3, 0)
- (2, 3), (0, 0)
- (0, 3), (3, 0)
- Area of smaller region bounded by the ellipse and line is represented by.
- The value of $\frac{2}{3}\int\limits_{0}^{3}\sqrt{9-\text{x}^2}\text{dx}$ is.
- $\frac{\pi}{2}$
- $\pi$
- $\frac{3\pi}{2}$
- $\frac{\pi}{4}$
- The value of $2\int\limits_{0}^{3}\bigg(1-\frac{\text{x}}{3}\bigg)\text{dx}$ is.
- 0
- 1
- 2
- 3
- Area of the smaller region bounded by the mirror and scratch is.
- $3\Big(\frac{\pi}{2}+1\Big)\text{ sq.units}$
- $\Big(\frac{\pi}{2}+1\Big)\text{ sq.units}$
- $\Big(\frac{\pi}{2}-1\Big)\text{ sq.units}$
- $3\Big(\frac{\pi}{2}-1\Big)\text{ sq.units}$
A child cut a pizza with a knife. Pizza is circular in shape which is represented by $x^2+y^2=4$ and sharp edge of knife represents a straight line given by $\text{x}=\sqrt{3\text{y}}$ Based on the above information, answer the following questions.
View full solution →
- The point(s) of intersection of the edge of knife (line) and pizza shown in the figure is (are).
- $(1, \sqrt{3}),(-1,-\sqrt{3})$
- $(\sqrt{3},1),(-\sqrt{3,}-1)$
- $(\sqrt{2,}0),(0,\sqrt{3})$
- $(-\sqrt{3,}),(1,-\sqrt{3})$
- Which of the following shaded portion represent the smaller area bounded by pizza and edge of knife in first quadrant?
- Value of area of the region bounded by circular pizza and edge of knife in first quadrant is.
- $\frac{\pi}{2}\text{ sq.units}$
- $\frac{\pi}{3}\text{ sq.units}$
- $\frac{\pi}{5}\text{ sq.units}$
- $\pi\text{ sq.units}$
- Area of each slice of pizza when child cut the pizza into 4 equal pieces is.
- $\pi\text{ sq.units}$
- $\frac{\pi}{2}\text{ sq.units}$
- $3\pi\text{ sq.units}$
- $2\pi\text{ sq.units}$
- Area of whole pizza is.
- $3\pi\text{ sq.units}$
- $2\pi\text{ sq.units}$
- $5\pi\text{ sq.units}$
- $4\pi\text{ sq.units}$
Graphs of two function $\text{f}(\text{x})=\text{sin}\text{ x}$ and $\text{(g)}\text{x}=\text{cos}\text{ x}$ is given below:
Based on the above information, answer the following questions.
View full solution →Based on the above information, answer the following questions.
- In $(0, \pi)$, the curves $\text{f}(\text{x})=\text{sin}\text{ x}$ and $\text{g}\text{ (x)}=\text{cos}\text{ x}$ at $\text{x}=$
- $\frac{\pi}{2}$
- $\frac{\pi}{3}$
- $\frac{\pi}{4}$
- ${\pi}$
- Value of $\int\limits_{0}^{\frac{\pi}{4}}\text{sin}\text{ x}\text{ dx}$ is.
- $1-\frac{1}{\sqrt{2}}$
- $1+\frac{1}{\sqrt{2}}$
- $2-\frac{1}{\sqrt{2}}$
- $2+\frac{1}{\sqrt{2}}$
- Value of $\int\limits_\frac{\pi}{4}^{\frac{\pi}{2}}\text{cos}\text{ x}\text{ dx}$ is.
- $1+\frac{1}{\sqrt{2}}$
- $1-\frac{1}{\sqrt{2}}$
- $2-\sqrt{2}$
- $2+\sqrt{2}$
- Value of $\int\limits_{0}^{\pi}\text{sin}\text{ x}\text{ dx}$ is.
- 0
- 1
- 2
- -2
- Value of $\int\limits_{0}^\frac{\pi}{2}\text{sin}\text{ x}\text{ dx}$ is.
- 0
- 1
- 3
- 4
Consider the following equations of curves $x^2 = y$ and $y = x.$
On the basis of above information, answer the following questions.
View full solution →On the basis of above information, answer the following questions.
- The point(s) of intersection of both the curves is (are).
- $(0, 0)(2, 2)$
- $(0, 0)(1, 1)$
- $(0, 0)(-1, -1)$
- $(0, 0)(-2, -2)$
- Area bounded by the curves is represented by which of the following graph?
- The value of the integral $\int\limits_{1}^{0}\text{x}\ \text{dx}$ is.
- $\frac{1}{4}$
- $\frac{1}{3}$
- $\frac{1}{2}$
- $1$
- The value of the integral $\int\limits_{0}^{1}\text{x}^2\ \text{dx}$ is.
- $\frac{1}{4}$
- $\frac{1}{3}$
- $\frac{1}{2}$
- $1$
- The value of area bounded by the curves $x^2 = y$ and $x = y$ is.
- $\frac{1}{6}\text{ sq}.\text{unit}$
- $\frac{1}{3}\text{ sq}.\text{unit}$
- $\frac{1}{2}\text{ sq}.\text{unit}$
- ${1}\text{ sq}.\text{unit}$
Location of three houses of a society is represented by the points A(-1, 0), B(1, 3) and C(3, 2) as shown in figure.
Based on the above information, answer the following questions
View full solution →Based on the above information, answer the following questions
- Equation of line AB is.
- $\text{y}=\frac{3}{2}(\text{x}+1)$
- $\text{y}=\frac{3}{2}(\text{x}-1)$
- $\text{y}=\frac{1}{2}(\text{x}+1)$
- $\text{y}=\frac{1}{2}(\text{x}-1)$
- Equation of line BC is.
- $\text{y}=\frac{1}{2}\text{x}-\frac{7}{2}$
- $\text{y}=\frac{3}{2}\text{x}-\frac{7}{2}$
- $\text{y}=\frac{-1}{2}\text{x}+\frac{7}{2}$
- $\text{y}=\frac{3}{2}\text{x}+\frac{7}{2}$
- Area of region ABCD is.
- 2 sq. units
- 4 sq. units
- 6 sq. units
- 8 sq. units
-
Area of $\triangle\text{ADC}$ is,
- 4 sq. units
- 8 sq. units
- 16 sq. units
- 32 sq. units
- Area of $\triangle\text{ABC}$ is.
- 3 sq. units
- 4 sq. units
- 5 sq. units
- 6 sq. units
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