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

$\begin{array}{l}Giventhecurvesy=2\sqrt[]{x},x=0andx=1.\\ y=2\sqrt[]{x}\Rightarrow y^2=4x\left(Parabola\right)\\ Requiredarea={\displaystyle \underset{0}{\overset{1}{\int }}\left(2\sqrt[]{x}\right)dx}\\ =2\times \frac{2}{3}{\left[x^{3/2}\right]}_0^1\\ =\frac{4}{3}\left[{\left(1\right)}^{3/2}-0\right]=\frac{4}{3}sq.units\\ Hence,therequiredarea=\frac{4}{3}sq.units.\end{array}$

International students must submit LIU application form before the application deadlines. They must make sure to complete teh application by verifying to meet the admission requirements and submitting mandatory admission documents before paying the LIU application fee and submitting it. The LIU application fee for undergraduate programs is USD 50 (INR 4.3 K),  and for postgraduate programs the application fee varies with the program. Hence, applicants must check the application fee for PG programs from the official website according to their programme of choice.

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

$\begin{array}{l}Wearegiventhat\left\{\left(x,0\right):y=\sqrt[]{4-x^2}\right\}\\ \Rightarrow y^2=4-x^2\\ \Rightarrow x^2+y^2=4whichisacircle.\\ Requiredarea=2.{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{4-x^2}dx}\\ \left[Sincecircleissymmetricalabouty-axis.\right]\\ =2.{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{{\left(2\right)}^2-x^2}dx}\\ =2.{\left[\frac{x}{2}\sqrt[]{4-x^2}+\frac{4}{2}si{n}^{-1}\frac{x}{2}\right]}_0^2\\ =2\left[\left(\frac{2}{2}-\sqrt[]{4-4}+2si{n}^{-1}\left(1\right)\right)-\left(0+0\right)\right]=2\left[2.\frac{\pi }{2}\right]=2\pi sq.units\\ Hence,therequiredarea=2\pi sq.units.\end{array}$

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

$\begin{array}{l}Wearegiventhat{y}^2=8xandx=2\\ y^2=8\left(2\right)=16\\ \therefore y=\pm 4\\ Requiredarea={\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{8x}dx=2\times 2\sqrt[]{2}{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{x}dx}}\\ =4\sqrt[]{2}\times \frac{2}{3}{\left[x^{3/2}\right]}_0^2\\ =\frac{8\sqrt[]{2}}{3}\left[{\left(2\right)}^{3/2}-0\right]=\frac{8\sqrt[]{2}}{3}\times 2\sqrt[]{2}=\frac{32}{3}sq.units\\ Hence,therequiredarea=\frac{32}{3}sq.units.\end{array}$

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

$\begin{array}{l}Here,x^2=yandy=x+2\\ \therefore x^2=x+2\\ \Rightarrow x^2-x-2=0\\ \Rightarrow x^2-2x+x-2=0\\ \Rightarrow x\left(x-2\right)+1\left(x-2\right)=0\\ \Rightarrow \left(x-2\right)\left(x+1\right)=0\therefore x=-1,x=2\\ Graphofy=x+2\\ \begin{array}{cc}\hfill x\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill y\hfill \\ \hfill 2\hfill & \hfill 0\hfill \end{array}\\ Areaoftherequiredregion={\displaystyle \underset{-1}{\overset{2}{\int }}\left(x+2\right)dx-}{\displaystyle \underset{-1}{\overset{2}{\int }}{x}^{2}dx}\\ ={\left[\frac{x^2}{2}+2x\right]}_{-1}^2-\frac{1}{3}{\left[x^3\right]}_{-1}^2\\ =\left[\left(\frac{4}{2}+4\right)-\left(\frac{1}{2}-2\right)\right]-\frac{1}{3}\left[8-\left(-1\right)\right]\\ =\left(6+\frac{3}{2}\right)-\frac{1}{3}\left(9\right)=\frac{15}{2}-3=\frac{9}{2}sq.units\\ Hence,therequiredarea=\frac{9}{2}sq.units.\end{array}$

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

$\begin{array}{l}Wearegiventhat{y}^2=9x\dots \left(i\right)\\ andy=x\dots \left(ii\right)\\ Solvingeqn.\left(i\right)and\left(ii\right)wehave\\ \Rightarrow x^2=9x\Rightarrow x^2-9x=0\\ x\left(x-9\right)=0\therefore x=0,x=9\\ Requiredarea={\displaystyle \underset{0}{\overset{9}{\int }}\sqrt[]{9x}dx-}{\displaystyle \underset{0}{\overset{9}{\int }}xdx}=3{\displaystyle \underset{0}{\overset{9}{\int }}\sqrt[]{x}dx-}{\displaystyle \underset{0}{\overset{4}{\int }}xdx}\\ =3.\frac{2}{3}{\left[x^{3/2}\right]}_0^9-\frac{1}{2}{\left[x^2\right]}_0^9\\ =2\left[{\left(9\right)}^{3/2}-0\right]-\frac{1}{2}\left[{\left(9\right)}^2-0\right]\\ =2\left(27\right)-\frac{1}{2}\left(81\right)=54-\frac{81}{2}=\frac{108-81}{2}\\ =\frac{27}{2}sq.units\\ Hence,therequiredarea=\frac{27}{2}sq.units.\end{array}$

Yes, submitting GMAT score can be considered extremely beneficial for admission in the MBA programs at the Long Island University. LIU generally requires GMAT or GRE scores for admission in MBA programs. A minimum score of 400 in GMAT is considered for admission. Moreover, if a candidate has a very string profile and work experience, there is a chance that the GMAT score could be waived. But, it is always beneficial to provide the score, as it enhances a candidates profile and increase chances of acceptance.

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

$\begin{array}{l}Wearegiventhat{y}^2=4xand{x}^2=4y\\ y=\frac{x^2}{4}\\ \Rightarrow {\left(\frac{x^2}{4}\right)}^2=4x\\ \Rightarrow \frac{x^4}{16}=4x\\ \Rightarrow x^4=64x\Rightarrow x^4-64x=0\\ \Rightarrow x\left(x^3-64\right)=0\\ \therefore x=0,x=4\\ Requiredarea={\displaystyle \underset{0}{\overset{4}{\int }}\sqrt[]{4x}dx-}{\displaystyle \underset{0}{\overset{4}{\int }}\frac{x^2}{4}dx}=2{\displaystyle \underset{0}{\overset{4}{\int }}\sqrt[]{x}dx-\frac{1}{4}}{\displaystyle \underset{0}{\overset{4}{\int }}{x}^{2}dx}\\ =2.\frac{2}{3}{\left[x^{3/2}\right]}_0^4-\frac{1}{4}.\frac{1}{3}{\left[x^3\right]}_0^4\\ =\frac{4}{3}\left[{\left(4\right)}^{3/2}-0\right]-\frac{1}{12}\left[{\left(4\right)}^3-0\right]\\ =\frac{4}{3}\left[8\right]-\frac{1}{12}\left[64\right]=\frac{1}{12}\left[64\right]=\frac{32}{3}-\frac{16}{3}=\frac{16}{3}sq.units\\ Hence,therequiredarea=\frac{16}{3}sq.units.\end{array}$

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

$\begin{array}{l}Wearegiventhaty=x^3,y=x+6andx=0\\ Solvingy=x^{3}andy=x+6,weget\\ x+6=x^3\\ \Rightarrow x^3-x-6=0\\ \Rightarrow x^2\left(x-2\right)+2x\left(x-2\right)+3\left(x-2\right)=0\\ \Rightarrow \left(x-2\right)\left(x^2+2x+3\right)=0\\ x^2+2x+3hasnorealroots.\therefore x=2\\ Requiredareaoftheshadedregion={\displaystyle \underset{0}{\overset{2}{\int }}\left(x+6\right)dx-}{\displaystyle \underset{0}{\overset{2}{\int }}{x}^{3}dx}\\ ={\left[\frac{x^2}{2}+6x\right]}_0^2-\frac{1}{4}{\left[x^4\right]}_0^2\\ =\left(\frac{4}{2}+12\right)-\left(0+0\right)-\frac{1}{4}\left[{\left(2\right)}^4-0\right]\\ =14-\frac{1}{4}\times 16=14-4=10sq.units\\ Hence,therequiredarea=10sq.units.\end{array}$