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Class 12 Math Calculus Area

Find the area of the region enclosed by the curves y= x log x and

y = 2 x - 2 x^{2}

.

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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

Consider the function defined implicity by the equation

y^{2} - 2 y e^{s i n^{- 1} x} + x^{2} - 1 + [x] + e^{2 s i n^{- 1} x} = 0 (where [x] denotes the greatest integer function) .

The area of the region bounded by the curve and the line

x = - 1

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The area of the loop of the curve

a y^{2} = x^{2} (a - x)

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The area bounded by the two branches of curve

(y - x)^{2} = x^{3}

and the straight line x=1 is

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The area made by curve

f (x) = [x] + x - [x]

and x-axis when

0 \leq x \leq n (n \in N)

is equal to { where [x] is greatest integer function}

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Match the statements given in List I with the values given in List II.

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The area bounded by

y = 3 - ∣ 3 - x ∣ and y = \frac{6}{∣ x + 1 ∣}

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Find the area enclosed by

y = g (x),

x-axis, x=1 and x=37, where g(x) is inverse of

f (x) = x^{3} + 3 x + 1

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A_{i}

is the area bounded by

∣ x - a_{i} ∣ + ∣ y ∣ = b_{i}, I \in N, where a_{i + 1} = a_{i} + \frac{3}{2} b_{i} and b_{i + 1} = \frac{b _{i}}{2}, a_{1} = 0, = b_{1} = 32,

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Related Questions

Consider the function defined implicity by the equation

y^{2} - 2 y e^{s i n^{- 1} x} + x^{2} - 1 + [x] + e^{2 s i n^{- 1} x} = 0 (where [x] denotes the greatest integer function) .

The area of the region bounded by the curve and the line

x = - 1

play button image

The area of the loop of the curve

a y^{2} = x^{2} (a - x)

play button image

The area bounded by the two branches of curve

(y - x)^{2} = x^{3}

and the straight line x=1 is

play button image

The area made by curve

f (x) = [x] + x - [x]

and x-axis when

0 \leq x \leq n (n \in N)

is equal to { where [x] is greatest integer function}

play button image

Match the statements given in List I with the values given in List II.

play button image

The area bounded by

y = 3 - ∣ 3 - x ∣ and y = \frac{6}{∣ x + 1 ∣}

play button image

Find the area enclosed by

y = g (x),

x-axis, x=1 and x=37, where g(x) is inverse of

f (x) = x^{3} + 3 x + 1

play button image

A_{i}

is the area bounded by

∣ x - a_{i} ∣ + ∣ y ∣ = b_{i}, I \in N, where a_{i + 1} = a_{i} + \frac{3}{2} b_{i} and b_{i + 1} = \frac{b _{i}}{2}, a_{1} = 0, = b_{1} = 32,

play button image

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Copyright FILO EDTECH PVT. LTD. 2021