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 <math> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mi>y</mi> </mrow> <mrow> <mi>x</mi> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </math> . If the curve also passes through the point <math> <mrow> <mrow> <mo>(</mo> <mrow> <mi>α</mi> <mo>,</mo> <mtext> </mtext> <mtext> </mtext> <mn>6</mn> <mroot> <mrow> <mn>1</mn> <mn>0</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mo>)</mo> </mrow> </mrow> </math> in the first quadrant, then a is equal to………….

∫ 3 x f ( x ) d x = ( f ( x ) x ) 3 ⇒ x 3 ∫ 3 x f ( x ) d x = f 3 ( x ) , differentiating w.r.to x x 3 f ( x ) + 3 x 2 f 3 ( x ) x 3 = 3 f 2 ( x ) f ' ( x ) ⇒ 3 y 2 d y d x = x 3 y = 3 y 3 x ⇒ 3 x y d y d x = x 4 + 3 y 2 After solving we get y 2 = x 4 3 + c x 2 also curve passes through (3, 3) Þ c = -2 ∴ y 2 = x 4 3 − 2 x 2 which passes through ( α , 6 1 0 ) ∴ α 4 − 6 α 2 3 = 3 6 0 ⇒ α = 6

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Maths NCERT Exemplar Solutions Class 11th Chapter Twelve Application of Integrals Maths Integrals Maths Class 12th

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………….

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Raj Pandey | Contributor-Level 9

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$\int_{3}^{x} f (x) d x = {(\frac{f (x)}{x})}^{3} \Rightarrow x^{3} \int_{3}^{x} f (x) d x = f^{3} (x),$ 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 $(α, 6 \sqrt[]{10})$ $∴ \frac{α^{4} - 6 α^{2}}{3} = 360 \Rightarrow α = 6$

<math> <mrow> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mn>3</mn> </mrow> <mrow> <mi>x</mi> </mrow> </munderover> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mstyle> <mo>=</mo> <msup> <mrow> <mrow> <mo>(</mo> <mrow> <mfrac> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mi>x</mi> </mrow> </mfrac> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mo>⇒</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mstyle displaystyle="true"> <mrow> <munderover> <mo>∫</mo> <mrow> <mn>3</mn> </mrow> <mrow> <mi>x</mi> </mrow> </munderover> <mrow> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>d</mi> <mi>x</mi> </mrow> </mrow> </mstyle> <mo>=</mo> <msup> <mrow> <mi>f</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>,</mo> </mrow> </math> differentiating w.r.to x <math> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mi>f</mi> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>+</mo> <mn>3</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mfrac> <mrow> <msup> <mrow> <mi>f</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mn>3</mn> <msup> <mrow> <mi>f</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mi>f</mi> <mo>'</mo> <mrow> <mo>(</mo> <mrow> <mi>x</mi> </mrow> <mo>)</mo> </mrow> <mo>⇒</mo> <mn>3</mn> <msup> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mfrac> <mrow> <mi>d</mi> <mi>y</mi> </mrow> <mrow> <mi>d</mi> <mi>x</mi> </mrow> </mfrac> <mo>=</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> <mi>y</mi> <mo>=</mo> <mfrac> <mrow> <mn>3</mn> <msup> <mrow> <mi>y</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <mi>x</mi> </mrow> </mfrac> <mo>⇒</mo> <mn>3</mn> <mi>x</mi> <mi>y</mi> <mfrac> <mrow> <mi>d</mi> <mi>y</mi> </mrow> <mrow> <mi>d</mi> <mi>x</mi> </mrow> </mfrac> <mo>=</mo> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msup> <mo>+</mo> <mn>3</mn> <msup> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>  After solving we get  <math> <mrow> <msup> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>+</mo> <mi>c</mi> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math>  also curve passes through (3, 3) Þ c = -2 <math> <mrow> <mo>∴</mo> <msup> <mrow> <mi>y</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>=</mo> <mfrac> <mrow> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>−</mo> <mn>2</mn> <msup> <mrow> <mi>x</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </math> which passes through <math> <mrow> <mrow> <mo>(</mo> <mrow> <mi>α</mi> <mo>,</mo> <mn>6</mn> <mroot> <mrow> <mn>1</mn> <mn>0</mn> </mrow> <mrow></mrow> </mroot> </mrow> <mo>)</mo> </mrow> </mrow> </math> <math> <mrow> <mo>∴</mo> <mfrac> <mrow> <msup> <mrow> <mi>α</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msup> <mo>−</mo> <mn>6</mn> <msup> <mrow> <mi>α</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> <mo>=</mo> <mn>3</mn> <mn>6</mn> <mn>0</mn> <mo>⇒</mo> <mi>α</mi> <mo>=</mo> <mn>6</mn> </mrow> </math>

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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:

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

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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

39. Choose the correct answer:

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Vishal Baghel

Given curve is $y = c o s x$

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34. Using the method of integration, find the area of the region bounded by the lines: 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0

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33. Using the method of integration, find the area of the triangle whose vertices are A (2, 0), B (4, 5) and C (6, 3).

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$\begin{array}{l} = \int_{2}^{4} y_{A B} d x + \int_{4}^{6} y_{B C} d x - \int_{2}^{6} y_{B C} d x \\ = \int_{2}^{4} \frac{5}{2} (x - 2) d x + \int_{4}^{6} (9 - x) d x - \int_{2}^{6} \frac{3}{4} (x - 2) d x \\ = \frac{5}{2} {[\frac{x^{2}}{2} - 2 x]}_{2}^{4} + {[9 x - \frac{x^{2}}{2}]}_{4}^{6} - \frac{3}{4} {[\frac{x^{2}}{2} - 2 x]}_{2}^{6} \end{array}$

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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………….

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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 ( y x ) 3 . If the curve also passes through the point ( α , 6 1 0 ) in the first quadrant, then a is equal to………….

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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………….