The region represented by x - y ≤ 2 and x + y ≤ 2 is bounded by a

square of side length 2 2 units.

rhombus of area 8 2 sq . units .

rhombus of side length 2 units.

EASY

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The area bounded by the lines $y = 2, x = 1$ , $x = a$ and the curve $y = f (x),$ which cuts the lines $x = 1, x = a$ and above the line $y = 2$ for all $a \geq 1,$ is equal to $\frac{2}{3} [{(2 a)}^{3 / 2} - 3 a + 3 - 2 \sqrt{2}] .$ Then, $f (x) =$

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Important Questions on Area under Curves

HARD

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

Let

g (x) = \cos x^{2}, f (x) = \sqrt{x},

and

α, β (α < β)

be the roots of the quadratic equation

18 x^{2} - 9 π x + π^{2} = 0

. Then the area (in sq. units) bounded by the curve

y = (g o f) (x)

and the lines

x = α, x = β

and

y = 0,

HARD

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

The figure shows a portion of the graph $y = 2 x - 4 x^{3} .$ The line $y = c$ is such that the areas of the regions marked $I$ and $II$ are equal. If $a, b$ are the $x$ -coordinates of $A, B$ respectively, then $a + b$ equals-

Question Image

HARD

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

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}

MEDIUM

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

The area (in sq. units) of the region

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

is:

HARD

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

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>Area under Curves>Area under the Curve

The area (in sq. units) of the region

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

MEDIUM

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

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>Area under Curves>Area under the Curve

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

y^{2} = 4 λ x

and the line

y = λ x, λ > 0,

\frac{1}{9}

, then

λ

is equal to

EASY

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

The area of region bounded by the lines

y = m x, x = 1

and

x = 2

and the

x

-axis is

7.5

sq. units, then

m

HARD

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

The area of the region bounded by the curve

y = |x^{3} - 4 x^{2} + 3 x|

and the

x

-axis,

0 \leq x \leq 3,

MEDIUM

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

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

y = 2^{x}

and

y = |x + 1|

, in the first quadrant is

MEDIUM

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

Let

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

I

is minimum then the ordered pair

(a, b)

HARD

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

If the area (in sq. units) of the region

\{(x, y) : y^{2} \leq 4 x, x + y \leq 1, x \geq 0, y \geq 0\}

a \sqrt{2} + b,

then

a - b

is equal to

HARD

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

Let

S (α) = \{(x, y) : y^{2} \leq x, 0 \leq x \leq α\}

and

A (α)

is area of the region

S (α) .

If for a

λ, 0 < λ < 4, A (λ) : A (4) = 2 : 5,

then

λ

equals:

MEDIUM

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

The region represented by

|x - y| \leq 2

and

|x + y| \leq 2

is bounded by a

EASY

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

The area enclosed by the curve

x = 3 cosθ, y = 2 sinθ, 0 \leq θ \leq π

, is (in square units)

HARD

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

The area (in sq units) of the quadrilateral formed by the tangents at the endpoints of the latus rectum to the ellipse $\frac{x^{2}}{9} + \frac{y^{2}}{5} = 1$ is?

MEDIUM

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

The area( in sq. units) bounded by

y = \sin^{2} x, x = \frac{π}{2}

and

x = π

MEDIUM

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

The area of the region

A = {(x, y) : 0 \leq y \leq x |x| + 1 and - 1 \leq x \leq 1}

in sq. units, is

HARD

Mathematics>Integral Calculus>Area under Curves>Area under the Curve

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

[(x, y) : y^{2} \leq 2 x and y \geq 4 x - 1]

The area bounded by the lines y=2, x=1, x=a and the curve y=fx, which cuts the lines x=1, x=a and above the line y=2 for all a≥1, is equal to 232a3/2-3a+3-22.Then, fx=

Important Questions on Area under Curves

Learn and Prepare for any exam you want !

The area bounded by the lines $y = 2, x = 1$ , $x = a$ and the curve $y = f (x),$ which cuts the lines $x = 1, x = a$ and above the line $y = 2$ for all $a \geq 1,$ is equal to $\frac{2}{3} [{(2 a)}^{3 / 2} - 3 a + 3 - 2 \sqrt{2}] .$ Then, $f (x) =$