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Let $f (x) = m i n . (x + 1, \sqrt{1 - x})$ for all $x \leq 1$ . Then the area bounded by $y = f (x)$ and the $x$ -axis is

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Important Questions on Application of Integrals

HARD

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

Area of the region bounded by the curve

y^{2} = 4 x, y

-axis and the line

y = 3

HARD

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The parabola

y^{2} = 4 x + 1

divides the disc

x^{2} + y^{2} \leq 1

into regions with areas

A_{1}

and

A_{2} .

Then

|A_{1} - A_{2}|

equals

HARD

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

On the real line

R

, we define two functions

f

and

g

as follows:

f (x) = m i n {x - [x], 1 - x + [x]}

g (x) = m a x \{x - [x], 1 - x + [x]\}

Where

[x]

denotes the largest integer not exceeding

x

. The positive integer

n

for which

\int_{0}^{n} (g (x) - f (x)) d x = 100

HARD

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The area bounded by the curve

y = \cos x

, the line joining

(- \frac{π}{4}, \cos (- \frac{π}{4})) & (0, 2)

and the line joining

(\frac{π}{4}, \cos (\frac{π}{4})) & (0, 2)

HARD

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

Area of the region

{(x, y) \in R^{2} : y \geq \sqrt{|x + 3|},

5 y \leq x + 9 \leq 15}

is equal to

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The area of the region bounded by the lines

y = 2 x + 1, y = 3 x + 1

and

x = 4

EASY

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The area enclosed between the parabolas

y^{2} = 16 x

and

x^{2} = 16 y

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The area (in sq. units) of the region bounded by the curves

y = 2^{x}

and

y = |x + 1|

, in the first quadrant is

HARD

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

Smaller area enclosed by the circle

x^{2} + y^{2} = 4

and the line

x + y = 2

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

Let

I = \int_{a}^{b} (x^{4} - 2 x^{2}) d x .

I

is minimum then the ordered pair

(a, b)

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

Let

A = \{(x, y) : y^{2} \leq 4 x, y - 2 x \geq - 4\}

. The area of the region

A

in square units is

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The area of the region bounded by

x^{2} = 4 y, y = 1, y = 4

and the

y-

axis lying in the first quadrant is _______ square units.

HARD

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The area of the region

(

in square units

)

above the

x

-axis bounded by the curve

y = \tan x, 0 \leq x \leq \frac{π}{2}

and the tangent to the curve at

x = \frac{π}{4}

HARD

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The area (in sq. units) of the region

\{(x, y) : x \geq 0, x + y \leq 3, x^{2} \leq 4 y a n d y \leq 1 + \sqrt{x}\}

HARD

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The area (in

sq . units

) of the region

{(x, y) : y^{2} \geq 2 x

and

x^{2} + y^{2} \leq 4 x, x \geq 0, y \geq 0}

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The area (in sq. units) of the region

A = \{(x, y) : \frac{y^{2}}{2} \leq x \leq y + 4\}

is:

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The area (in sq. units) of the region

A = \{(x, y) : x^{2} \leq y \leq x + 2\}

HARD

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The area (in square units) bounded by the curves $y = \sqrt{x}, 2 y - x + 3 = 0$ , $X$ -axis and lying in the first quadrant is

MEDIUM

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The area (in sq. units) bounded by the parabola

y = x^{2} - 1,

the tangent at the point

(2, 3)

to it and the

y

-axis is

EASY

Mathematics>Integral Calculus>Application of Integrals>Area Under Simple Curves

The area (in sq. units) of the region

{x \in R : x \geq 0, y \geq 0, y \geq x - 2

and

y \leq \sqrt{x}}

Let fx=min.x+1,1-x for all x≤1. Then the area bounded by y=fx and the x-axis is

Important Questions on Application of Integrals

Learn and Prepare for any exam you want !

Let $f (x) = m i n . (x + 1, \sqrt{1 - x})$ for all $x \leq 1$ . Then the area bounded by $y = f (x)$ and the $x$ -axis is