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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Draw a rough sketch of the curve  $y = \sqrt[]{x - 1}$ in the interval $[1, 5]$ . Find the area under the curve and between the lines  $x = 1$ and $x = 5$ .

8 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} H e r e, w e h a v e y = \sqrt[]{x - 1} \\ \Rightarrow y^{2} = x - 1 (P a r a b o l a) \\ A r e a o f t h e required r e g i o n = \int_{1}^{5} \sqrt[]{x - 1} d x \\ = \frac{2}{3} {[{(x - 1)}^{3 / 2}]}_{1}^{5} \\ = \frac{2}{3} [{(5 - 1)}^{3 / 2} - 0] = \frac{2}{3} \times {(4)}^{3 / 2} \\ = \frac{2}{3} \times 8 = \frac{16}{3} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{16}{3} s q . u n i t s . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Using integration, find the area of the region bounded by the line $2 y = 5 x + 7$ , the x-axis, and the lines $x = 2$ and $x = 8$ .

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t 2 y = 5 x + 7, x - a x i s, x = 2 a n d x = 8 \\ L e t u s d r a w t h e g r a p h o f 2 y = 5 x + 7 \Rightarrow y = \frac{5 x + 7}{2} \\ \begin{matrix} x & 1 \\ - 1 & y \\ 6 & 1 \end{matrix} \\ A r e a o f t h e required s h a d e d r e g i o n = \int_{2}^{8} (\frac{5 x + 7}{2}) d x \\ = \frac{1}{2} {[\frac{5 x^{2}}{2} + 7 x]}_{2}^{8} \\ = \frac{1}{2} [\frac{5}{2} (64 - 4) + 7 (8 - 2)] \\ = \frac{1}{2} [\frac{5}{2} \times 60 + 7 \times 6] = \frac{1}{2} [150 + 42] = \frac{1}{2} \times 192 = 96 s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = 96 s q . u n i t s . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Calculate the area under the curve $y = 2 \sqrt[]{x}$ included between the lines  $x = 0$ and $x = 1$ .

8 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h e c u r v e s y = 2 \sqrt[]{x}, x = 0 a n d x = 1 . \\ y = 2 \sqrt[]{x} \Rightarrow y^{2} = 4 x (P a r a b o l a) \\ Required a r e a = \int_{0}^{1} (2 \sqrt[]{x}) d x \\ = 2 \times \frac{2}{3} {[x^{3 / 2}]}_{0}^{1} \\ = \frac{4}{3} [{(1)}^{3 / 2} - 0] = \frac{4}{3} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{4}{3} s q . u n i t s . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Sketch the region ${(x, 0) : y = \sqrt[]{4 - x^{2}}}$ and the x-axis. Find the area of the region using integration.

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e a r e g i v e n t h a t {(x, 0) : y = \sqrt[]{4 - x^{2}}} \\ \Rightarrow y^{2} = 4 - x^{2} \\ \Rightarrow x^{2} + y^{2} = 4 w h i c h i s a c i r c l e . \\ Required a r e a = 2 . \int_{0}^{2} \sqrt[]{4 - x^{2}} d x \\ [Since c i r c l e i s s y m m e t r i c a l a b o u t y - a x i s .] \\ = 2 . \int_{0}^{2} \sqrt[]{{(2)}^{2} - x^{2}} d x \\ = 2 . {[\frac{x}{2} \sqrt[]{4 - x^{2}} + \frac{4}{2} s i n^{- 1} \frac{x}{2}]}_{0}^{2} \\ = 2 [(\frac{2}{2} - \sqrt[]{4 - 4} + 2 s i n^{- 1} (1)) - (0 + 0)] = 2 [2 . \frac{π}{2}] = 2 π s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = 2 π s q . u n i t s . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area of the region bounded by the line $x = 2$ and the parabola $y^{2} = 8 x$ .

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e a r e g i v e n t h a t y^{2} = 8 x a n d x = 2 \\ y^{2} = 8 (2) = 16 \\ ∴ y = \pm 4 \\ Required a r e a = \int_{0}^{2} \sqrt[]{8 x} d x = 2 \times 2 \sqrt[]{2} \int_{0}^{2} \sqrt[]{x} d x \\ = 4 \sqrt[]{2} \times \frac{2}{3} {[x^{3 / 2}]}_{0}^{2} \\ = \frac{8 \sqrt[]{2}}{3} [{(2)}^{3 / 2} - 0] = \frac{8 \sqrt[]{2}}{3} \times 2 \sqrt[]{2} = \frac{32}{3} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{32}{3} s q . u n i t s . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area of the region enclosed by the parabola $x^{2} = y$ and the line $y = x + 2$ .

7 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} H e r e, x^{2} = y a n d y = x + 2 \\ ∴ x^{2} = x + 2 \\ \Rightarrow x^{2} - x - 2 = 0 \\ \Rightarrow x^{2} - 2 x + x - 2 = 0 \\ \Rightarrow x (x - 2) + 1 (x - 2) = 0 \\ \Rightarrow (x - 2) (x + 1) = 0 ∴ x = - 1, x = 2 \\ G r a p h o f y = x + 2 \\ \begin{matrix} x & 0 \\ - 2 & y \\ 2 & 0 \end{matrix} \\ A r e a o f t h e required r e g i o n = \int_{- 1}^{2} (x + 2) d x - \int_{- 1}^{2} x^{2} d x \\ = {[\frac{x^{2}}{2} + 2 x]}_{- 1}^{2} - \frac{1}{3} {[x^{3}]}_{- 1}^{2} \\ = [(\frac{4}{2} + 4) - (\frac{1}{2} - 2)] - \frac{1}{3} [8 - (- 1)] \\ = (6 + \frac{3}{2}) - \frac{1}{3} (9) = \frac{15}{2} - 3 = \frac{9}{2} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{9}{2} s q . u n i t s . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area of the region included between the curves  $y^{2} = 9 x$ and $y = x$ .

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e a r e g i v e n t h a t y^{2} = 9 x \dots (i) \\ a n d y = x \dots (i i) \\ S o l v i n g e q n . (i) a n d (i i) w e h a v e \\ \Rightarrow x^{2} = 9 x \Rightarrow x^{2} - 9 x = 0 \\ x (x - 9) = 0 ∴ x = 0, x = 9 \\ Required a r e a = \int_{0}^{9} \sqrt[]{9 x} d x - \int_{0}^{9} x d x = 3 \int_{0}^{9} \sqrt[]{x} d x - \int_{0}^{4} x d x \\ = 3 . \frac{2}{3} {[x^{3 / 2}]}_{0}^{9} - \frac{1}{2} {[x^{2}]}_{0}^{9} \\ = 2 [{(9)}^{3 / 2} - 0] - \frac{1}{2} [{(9)}^{2} - 0] \\ = 2 (27) - \frac{1}{2} (81) = 54 - \frac{81}{2} = \frac{108 - 81}{2} \\ = \frac{27}{2} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{27}{2} s q . u n i t s . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area of the region bounded by the curves  $y^{2} = 4 x$ and $x^{2} = 4 y$ .

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e a r e g i v e n t h a t y^{2} = 4 x a n d x^{2} = 4 y \\ y = \frac{x^{2}}{4} \\ \Rightarrow {(\frac{x^{2}}{4})}^{2} = 4 x \\ \Rightarrow \frac{x^{4}}{16} = 4 x \\ \Rightarrow x^{4} = 64 x \Rightarrow x^{4} - 64 x = 0 \\ \Rightarrow x (x^{3} - 64) = 0 \\ ∴ x = 0, x = 4 \\ Required a r e a = \int_{0}^{4} \sqrt[]{4 x} d x - \int_{0}^{4} \frac{x^{2}}{4} d x = 2 \int_{0}^{4} \sqrt[]{x} d x - \frac{1}{4} \int_{0}^{4} x^{2} d x \\ = 2 . \frac{2}{3} {[x^{3 / 2}]}_{0}^{4} - \frac{1}{4} . \frac{1}{3} {[x^{3}]}_{0}^{4} \\ = \frac{4}{3} [{(4)}^{3 / 2} - 0] - \frac{1}{12} [{(4)}^{3} - 0] \\ = \frac{4}{3} [8] - \frac{1}{12} [64] = \frac{1}{12} [64] = \frac{32}{3} - \frac{16}{3} = \frac{16}{3} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{16}{3} s q . u n i t s . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area of the region bounded by the curves $y = x^{3}$ , $y = x + 6$ , and $x = 0$ .

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e a r e g i v e n t h a t y = x^{3}, y = x + 6 a n d x = 0 \\ S o l v i n g y = x^{3} a n d y = x + 6, w e g e t \\ x + 6 = x^{3} \\ \Rightarrow x^{3} - x - 6 = 0 \\ \Rightarrow x^{2} (x - 2) + 2 x (x - 2) + 3 (x - 2) = 0 \\ \Rightarrow (x - 2) (x^{2} + 2 x + 3) = 0 \\ x^{2} + 2 x + 3 h a s n o r e a l r o o t s . ∴ x = 2 \\ Required a r e a o f t h e s h a d e d r e g i o n = \int_{0}^{2} (x + 6) d x - \int_{0}^{2} x^{3} d x \\ = {[\frac{x^{2}}{2} + 6 x]}_{0}^{2} - \frac{1}{4} {[x^{4}]}_{0}^{2} \\ = (\frac{4}{2} + 12) - (0 + 0) - \frac{1}{4} [{(2)}^{4} - 0] \\ = 14 - \frac{1}{4} \times 16 = 14 - 4 = 10 s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = 10 s q . u n i t s . \end{array}$

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New answer posted•a year ago

Exemplar Solution Maths NCERT Exemplar Solutions Class 12th Chapter Eight

Find the area of the region bounded by the parabolas $y^{2} = 2 p x$ and $x^{2} = 2 p y$ .

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

This is a short answer type question as classified in NCERT Exemplar

$\begin{array}{l} W e a r e g i v e n t h a t x^{2} = 2 p y \dots (i) \\ a n d y^{2} = 2 p x \dots (i i) \\ F r o m e q n . (i) w e g e t y = \frac{x^{2}}{2 p} \\ P u t t i n g t h e t h e o f y i n e q n . (i i), w e h a v e \\ {(\frac{x^{2}}{2 p})}^{2} = 2 p x \Rightarrow \frac{x^{4}}{4 p^{2}} = 2 p x \\ \Rightarrow x^{4} = 8 p^{3} x \Rightarrow x^{4} - 8 p^{3} x = 0 \\ \Rightarrow x (x^{3} - 8 p^{3}) = 0 \\ ∴ x = 0, 2 p \\ Required a r e a = A r e a o f t h e r e g i o n (O C B A - O D B A) \\ = \int_{0}^{2 p} \sqrt[]{2 p x} d x - \int_{0}^{2 p} \frac{x^{2}}{2 p} d x \\ = \sqrt[]{2 p} \int_{0}^{2 p} \sqrt[]{x} d x - \frac{1}{2 p} \int_{0}^{2 p} x^{2} d x = \sqrt[]{2 p} . \frac{2}{3} {[x^{3 / 2}]}_{0}^{2 p} - \frac{1}{2 p} . \frac{1}{3} {[x^{3}]}_{0}^{2 p} \\ = \frac{2 \sqrt[]{2}}{3} [{(2 p)}^{3 / 2} - 0] - \frac{1}{6 p} [{(2 p)}^{3} - 0] \\ = \frac{2 \sqrt[]{2}}{3} \sqrt[]{p} . 2 \sqrt[]{2} p^{\frac{3}{2}} - \frac{1}{6 p} . 8 p^{3} \\ = \frac{8}{3} . p^{2} - \frac{8}{6} p^{2} = \frac{8}{6} p^{2} = \frac{4}{3} p^{2} s q . u n i t s \\ H e n c e, t h e r e q u i r e d a r e a = \frac{4}{3} p^{2} s q . u n i t s . \end{array}$

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