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Application of Integrals Ncert Solutions Maths class 12th Maths Class 12th Application of Integrals

16. Find the area of the region bounded by the curves y = x²+2, y = x,

x = 0 and x = 3

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Vishal Baghel | Contributor-Level 10

4 months ago

The equation of the curve is $y = x^{2} + 2 \Rightarrow x^{2} = y - 2$ - (1) and

lines are

$y = x$ - (2)

$x = 0$ - (3)

$x = 3$ - (4)

Equation (1)is a parabola with vertex (0,2)

Equation (2)is a straight line passing origin with shape = $t a n θ = 1 = θ = 4 5^{?}$

$∴$ The required area enclosed OBCDO = area (ODCAO)-area (OBAO)

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Let a cure y = y(x) pass through the point (3, 3) and the area of the origin under this curve, above the x-axis and between the abscissae 3 and x (>3) be ${(\frac{y}{x})}^{3}$ . If the curve also passes through the point $(α, 6 \sqrt[]{10})$ in the first quadrant, then a is equal to………….

Raj Pandey

$\int_{3}^{x} f (x) d x = {(\frac{f (x)}{x})}^{3} \Rightarrow x^{3} \int_{3}^{x} f (x) d x = f^{3} (x),$ differentiating w.r.to x

$x^{3} f (x) + 3 x^{2} \frac{f^{3} (x)}{x^{3}} = 3 f^{2} (x) f' (x) \Rightarrow 3 y^{2} \frac{d y}{d x} = x^{3} y = \frac{3 y^{3}}{x} \Rightarrow 3 x y \frac{d y}{d x} = x^{4} + 3 y^{2}$

After solving we get $y^{2} = \frac{x^{4}}{3} + c x^{2}$ also curve passes through (3, 3) Þ c = -2

$∴ y^{2} = \frac{x^{4}}{3} - 2 x^{2}$ which passes through $(α, 6 \sqrt[]{10})$ $∴ \frac{α^{4} - 6 α^{2}}{3} = 360 \Rightarrow α = 6$

The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = y^a is $\frac{364}{3},$ is equal to:

Raj Pandey

Since a is a odd natural number then $| \int_{1}^{3} y^{a} d y | = \frac{364}{3} \Rightarrow | {(\frac{y^{a + 1}}{a + 1})}_{1}^{3} | = \frac{364}{3} \Rightarrow \frac{3^{a + 1}}{a + 1} = \frac{364}{3}$

Þ a = 5

The area bounded by the curves y² = x and x² = -y is equal to

Vishal Baghel

$y^{2} = x, a = \frac{1}{4}$

$x^{2} = - y, b = \frac{1}{4}$

$A = \frac{16 | a b |}{3} = \frac{1}{3}$

limₓ→∞ [∫₀ˣ tan⁻¹t dt] / √x²+1 is equal to:

Vishal Baghel

lim (x→∞) (∫? ^ (√x²+1) tan? ¹t dt) / x = lim (x→∞) (tan? ¹ (√x²+1) * (x/√ (x²+1) = lim (x→∞) (tan? ¹ x) * (x/√ (x²+1) = π/2

Area of the region bounded by x = 0, y = 0, x = 2, y = 2, y ≤ e? and y ≥ ln x, is:

Vishal Baghel

A = ∫? ² lnx dx = 2ln2 – 1
A' = 4 - 2 (2ln2 – 1) = 6 – 4ln2

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16. Find the area of the region bounded by the curves y = x²+2, y = x,

x = 0 and x = 3

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16. Find the area of the region bounded by the curves y = x2 +2, y = x, x = 0 and x = 3

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16. Find the area of the region bounded by the curves y = x²+2, y = x,

x = 0 and x = 3