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
386 questions across 8 question groups — pick any mix to generate a Maths paper with step-by-step answer keys.
386
Questions
8
Question groups
5
Question types
01
M.C.Q (1 Marks)
174 Q→02Assertion (A) & Reason (B) MCQ
3 Q→031 Marks Question
3 Q→042 Marks Questions
1 Q→053 Marks Question
6 Q→065 Marks Questions
179 Q→07Case study (4 Marks)
10 Q→08Fill In The Blanks[1 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 $6\pi\ \text{sq. units}.$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 $\pi\ \text{sq. units}.$
- 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$\frac{1}{2}\text{ sq.}\text{ units}$
- DNone 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 →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 region bounded by the curve $y^2 = 4x$ and the line $x = 3$ is $8\sqrt{3} \text{sq. units.}$
Reason : The area of the region bounded by the curve $x^2 = 4y$ and the line $x = 4y - 2$ is $\frac{9}{8} \text{sq. units}.$
Assertion : The area of the region bounded by the curve $y^2 = 4x$ and the line $x = 3$ is $8\sqrt{3} \text{sq. units.}$
Reason : The area of the region bounded by the curve $x^2 = 4y$ and the line $x = 4y - 2$ is $\frac{9}{8} \text{sq. units}.$
- AAssertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
- ✓Assertion 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: 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 bounded by the curves $\text{y}^2 = 4\text{a}^2(\text{x} - 1) $ and lines $\text{x}=1$ and $\text{y}=4 a$ is $\frac{8\text{a}}{3}\text{sq.units}$
Reason : The area enclosed between the parabola $\text{y}=\text{x}^2-\text{x}+2$ and the line $\text{y}=\text{x+2}$ is $\frac{4}{3}\text{ sq.units}$
Assertion : The area bounded by the curves $\text{y}^2 = 4\text{a}^2(\text{x} - 1) $ and lines $\text{x}=1$ and $\text{y}=4 a$ is $\frac{8\text{a}}{3}\text{sq.units}$
Reason : The area enclosed between the parabola $\text{y}=\text{x}^2-\text{x}+2$ and the line $\text{y}=\text{x+2}$ is $\frac{4}{3}\text{ sq.units}$
- AAssertion 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.
- ✓Assertion is wrong statement but Reason is correct statement.
Answer: D.
View full solution →The area of the region bounded by the circle $x^2 + y^2 = 1$ is :
- A$2\pi\text{ sq.}\text{ units}$
- ✓$\pi\text{ sq.}\text{ units}$
- C$3\pi\text{ sq.}\text{ units}$
- D$4\pi\text{ sq.}\text{ units}$
Answer: B.
View full solution →Sketch the graph of $\text{y}=\sqrt{\text{x}+1}$ in [0, 4] and determine the area of the region enclosed by the curve, the x-axis and the lines x = 0, x = 4.
View full solution →Using integration, find area of the bounded between the line $x = 2$ and the parabola $y^2 = 8x$.
View full solution →If $\text{A}=\begin{bmatrix}2&3\\5&7\end{bmatrix},\text{ B}=\begin{bmatrix}-1&0&2\\3&4&1\end{bmatrix},\text{C}=\begin{bmatrix}-1&2&3\\2&1&0\end{bmatrix},$ find
A + B and B + C
View full solution →A + B and B + C
Solve the following differential equation:$x \cos \text{y dy} = ( xe^{x} \log x + e^{x}) dx$
View full solution →Find the area bounded by the curve $\text{y}=\sin\text{x}$ between x = 0 and $\text{x}=2\pi.$
View full solution →Calcualte the area under the curve $\text{y}=2\sqrt{\text{x}}$ included between the lines x = 0 and x = 1.
View full solution →Find the intervals in which the function f given by $\text{f}\text{(x)}=\text{x}^3+\frac{1}{\text{x}^3},\text{x}\neq0\text{ is}$ (i) increasing (ii) decreasing.
View full solution →Draw a rough sketch of the curve $\text{y}=\sqrt{\text{x}-1}$ in the interval [1, 5]. Find the area under the curve and between the lines x = 1 and x = 5.
View full solution →Prove that the curves $\text{y}^{2} = 4x \text{ and } x^{2} = 4y $ divided the area of square bounded by $x = 0, x = 4,y=4 \text{ and } y = 0$ into three equal parts.
View full solution →Tangent to the circle $\text{x}^{2} + \text{y}^{2} = 4$ at any point on it in the first quadrant makes intercepts OA and OB on x and y axes respectively, O being the centre of the circle. Find the minimum value of (OA + OB).
View full solution →If the area bounded by the parabola $\text{y}^{2} = 16\text{ax}$ and the line $\text{y = 4 mx}$ is $\frac{\text{a}^{2}}{12}$ sq. units, then using integration, find the value of m.
View full solution →Using integration, find the area of the region bounded by the triangle whose vertices are (–1, 2), (1, 5) and (3, 4).
Prove that the curves $\text{y}^{2} = 4x \text{ and } x^{2} = 4y $ divided the area of square bounded by $x = 0, x = 4,y=4 \text{ and } y = 0$ into three equal parts.
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
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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