Find the area bounded by the curves y=x^(3)-x and y=x^(2)+x. | Filo

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Class 12 Math Calculus Area

Find the area bounded by the curves

y = x^{3} - x and y = x^{2} + x .

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Similar topics

relations and functions

trigonometric functions

inverse trigonometric functions

application of derivatives

integrals

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relations and functions

trigonometric functions

inverse trigonometric functions

application of derivatives

integrals

Related Questions

The area in the first quadrant between

x^{2} + y^{2} = π^{2}

y = sin x

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The area of the region enclosed by the curves

y = x, x = e, y = \frac{1}{x}

and the positive x-axis is

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If y=f(x) is a monotonic function in (a,b), then the area bounded by the ordinates at

x = a, x = b, y = f (x) and y = f (c) (where c \in (a, b)) is minimum when c = \frac{a + b}{2}

Proof : A = a \int c (f (c) - f (x)) d x + c \int b (f (c)) d x

= f (c) (c - a) - a \int c (f (x)) d x + a \int b (f (x)) d x - f (c) (b - c)

\Rightarrow A = [2 c - (a + b)] f (c) + c \int b (f (x)) d x - a \int c (f (x)) d x

Differentiating w.r.t. c, we get

\frac{d A}{d c} = [2 c - (a + b)] f^{'} (c) + 2 f (c) + 0 - f (c) - (f (c) - 0)

For maxima and minima ,

\frac{d A}{d c} = 0

\Rightarrow f^{'} (c) [2 c - (a + b)] = 0 (a s f^{'} (c) \neq = 0)

c = \frac{a + b}{2}

Also for c < \frac{a + b}{2}, \frac{d A}{d c} < 0 and for c > \frac{a + b}{2}, \frac{d A}{d c} > 0

Hence, A is minimum when

c = \frac{a + b}{2}

.
The value of the parameter a for which the area of the figure bounded by the abscissa axis, the graph of the function

y = x^{3} + 3 x^{2} + x + a

, and the straight lines, which are parallel to the axis of ordinates and cut the abscissa axis at the point of extremum of the function, which is the least, is

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The area (in sq. units) of the region

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

play button image

Find the area bounded by the curve

x^{2} = y, x^{2} = - y, and y^{2} = 4 x - 3 .

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The area enclosed by the curve

C : y = x 9 - x^{2} (x \geq 0)

and the x-axis is___.

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Find the area of the region

{(x, y) : y^{2} \leq 4 x, 4 x^{2} + 4 y^{2} \leq 9} .

play button image

The area of the closed figure bounded by

x = - 1 x = 2,

y = {- x^{2} + 2, 2 x - 1, x \leq 1 x > 1

and the abscissa axis is

play button image

Related Questions

The area in the first quadrant between

x^{2} + y^{2} = π^{2}

y = sin x

play button image

The area of the region enclosed by the curves

y = x, x = e, y = \frac{1}{x}

and the positive x-axis is

play button image

If y=f(x) is a monotonic function in (a,b), then the area bounded by the ordinates at

x = a, x = b, y = f (x) and y = f (c) (where c \in (a, b)) is minimum when c = \frac{a + b}{2}

Proof : A = a \int c (f (c) - f (x)) d x + c \int b (f (c)) d x

= f (c) (c - a) - a \int c (f (x)) d x + a \int b (f (x)) d x - f (c) (b - c)

\Rightarrow A = [2 c - (a + b)] f (c) + c \int b (f (x)) d x - a \int c (f (x)) d x

Differentiating w.r.t. c, we get

\frac{d A}{d c} = [2 c - (a + b)] f^{'} (c) + 2 f (c) + 0 - f (c) - (f (c) - 0)

For maxima and minima ,

\frac{d A}{d c} = 0

\Rightarrow f^{'} (c) [2 c - (a + b)] = 0 (a s f^{'} (c) \neq = 0)

c = \frac{a + b}{2}

Also for c < \frac{a + b}{2}, \frac{d A}{d c} < 0 and for c > \frac{a + b}{2}, \frac{d A}{d c} > 0

Hence, A is minimum when

c = \frac{a + b}{2}

.
The value of the parameter a for which the area of the figure bounded by the abscissa axis, the graph of the function

y = x^{3} + 3 x^{2} + x + a

, and the straight lines, which are parallel to the axis of ordinates and cut the abscissa axis at the point of extremum of the function, which is the least, is

play button image

The area (in sq. units) of the region

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

play button image

Find the area bounded by the curve

x^{2} = y, x^{2} = - y, and y^{2} = 4 x - 3 .

play button image

The area enclosed by the curve

C : y = x 9 - x^{2} (x \geq 0)

and the x-axis is___.

play button image

Find the area of the region

{(x, y) : y^{2} \leq 4 x, 4 x^{2} + 4 y^{2} \leq 9} .

play button image

The area of the closed figure bounded by

x = - 1 x = 2,

y = {- x^{2} + 2, 2 x - 1, x \leq 1 x > 1

and the abscissa axis is

play button image

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Blog
Micro Class
PDFs
Become a Tutor
Download App

About Us
Contact Us
Privacy Policy and Terms & Conditions

Copyright FILO EDTECH PVT. LTD. 2021