The area bounded by the curve $y=\sqrt{x}, Y$-axis and between the lines $y=0$ and $y=3$ is
View full solution →Question types
Application of Integrals question types
59 questions across 7 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.
59
Questions
7
Question groups
5
Question types
01
M.C.Q (1 Marks)
30 Q→02Assertion (A) & Reason (B) MCQ
5 Q→031 Marks Question
7 Q→042 Marks Questions
3 Q→053 Marks Question
4 Q→065 Marks Questions
2 Q→07Case study (4 Marks)
8 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.
Shown below is the curve defined by the equation $y=\log (x+1)$ for $x \geq 0$.
Which of these is the area of the shaded region?
View full solution →Which of these is the area of the shaded region?
If the area bounded by the curve $2 x^2+y^2=2$ is $A$. Then which of the following is the value of $A$ ?
View full solution →Which of these is equal to the area enclosed between the curve $x^2+y^2=16$ and the coordinate axes in the first quadrant?
View full solution →Area bounded by the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ is
View full solution →Assertion $(A) :$ The area of the region bounded by the curve $y^2=4 x$ and the line $x=3$ is $8 \sqrt{3}$ sq. units.
Reason $(R):$ The area is symmetric about $x$ and $y$ axes.
View full solution →Reason $(R):$ The area is symmetric about $x$ and $y$ axes.
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 $2 .$
View full solution →Reason $(R) :$ Maximum value of the curve $y=2 \cos x$ is $2 .$
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.
View full solution →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.
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.
View full solution →Reason $(R) :$ The area enclosed between the parabola $y^2=49 x$ and its latus rectum $\frac{8 a^2}{3}$ sq. units.
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.
View full solution →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.
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 of the region bounded by two parabolas $y=x^2$ and $y^2=x$.
View full solution →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.
- $3\Big(\frac{\pi}{2}+1\Big)\text{ sq.units}$
- $\Big(\frac{\pi}{2}+1\Big)\text{ sq.units}$
- $\Big(\frac{\pi}{2}-1\Big)\text{ sq.units}$
- $3\Big(\frac{\pi}{2}-1\Big)\text{ sq.units}$
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 →- 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
In a classroom, teacher explains the properties of a particular curve by saying that this particular curve has beautiful up and downs. It starts at 1 and heads down until $\pi$ radian, and then heads up again and closely related to sine function and both follow each other, exactly $\frac{\pi}{2}$ radians apart 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.
- Name the curve, about which teacher explained in the classroom.
- Cosine
- Sine
- Tangent
- Cotangent
- Area of curve explained in the passage from 0 to $\frac{\pi}{2}$ is.
- $\frac{1}{3}\text{ sq.unit}$
- $\frac{1}{2}\text{ sq.unit}$
- ${1}\text{ sq.unit}$
- ${2}\text{ sq.units}$
- Area of curve discussed in classroom from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$ is.
- -2 sq. units
- 2 sq. units
- 3 sq. units
- -3 sq. units
- Area of curve discussed in classroom from $\frac{3\pi}{2}$ to $2\pi$ is.
- 1 sq. unit
- 2 sq. units
- 3 sq. units
- 4 sq. units
- Area of explained curve from 0 to $2\pi$ is.
- 1 sq. unit
- 2 sq. units
- 3 sq. units
- 4 sq. units
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