Question
Find the area between the parabolas $y^2 = 5x$ and $x^2 = 5y$
✓
Answer
Given equations of the parabolas are $y^2 = 5x ......(i)$
and $x^2 = 5y.$
$\therefore y =\frac{x^2}{5}$
From (i), we get
$y =\sqrt{5 x}$ (iii) $[\because \ln$ first quadrant, $y>0]$
Find the points of intersection of $y^2=5 x$ and $x^2=5 y$.
Substituting (ii) in (i), we get
$ \left(\frac{x^2}{5}\right)^2=5 x$
$\therefore x^4=125 x$
$\therefore x^4-125 x=0$
$\therefore x\left(x^3-125\right)=0$
$\therefore x=0 \text { or } x^3=125=5^3$
$\therefore x=0 \text { or } x=5 $
When $x=0, y=0$ and when $x=5, y=5$
$\therefore$ The points of intersection of $y^2=5 x$ and $x^2=5 y$ are $O(0,0)$ and $B(5,5)$.
Draw $BD \perp OX$
Required area $=$ area of the region $OABCO$
$=$ area of the region $O D B C O-$ area of the region $ODBAO$
$=$ area under the parabola $y^2=5 x-$ area under the parabola $x^2=5 y$
$=\int_0^5 \sqrt{5 x} d x-\int_0^5 \frac{x^2}{5} d x \ldots . .[$ From (iii) and (ii)]
$=\sqrt{5} \int_0^5 x^{\frac{1}{2}} d x-\frac{1}{5} \int_0^5 x^2 d x$
$=\sqrt{5}\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^5-\frac{1}{5}\left[\frac{x^3}{3}\right]_0^5$
$=\frac{2 \sqrt{5}}{3}\left[(5)^{\frac{3}{2}}-0\right]-\frac{1}{15}\left[(5)^3-0\right]$
$=\frac{2 \sqrt{5}}{3}(5 \sqrt{5})-\frac{1}{15}(125)$
$=\frac{50}{3}-\frac{25}{3}$
$=\frac{25}{3} \text { sq.units }$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Calculate Value Index Number for the following using Simple Aggregate Method
| Commodity | Base Year | Current Year | ||
| Price | Quantity | Price | Quantity | |
| A | 30 | 13 | 40 | 15 |
| B | 40 | 15 | 70 | 20 |
| C | 10 | 12 | 60 | 22 |
| D | 50 | 10 | 90 | 18 |
| E | 20 | 14 | 100 | 16 |
If $A=\left[\begin{array}{ccc}-1 & 2 & 1 \\ -3 & 2 & -3\end{array}\right]$ and $B=\left[\begin{array}{cc}2 & 1 \\ -3 & 2 \\ -1 & 3\end{array}\right]$, prove that $\left(A+B^{\top}\right)^{\top}=$
$
A^{\top}+B
$
→$
A^{\top}+B
$
Two samples from bivariate populations have $15$ observations each. The sample means of $X$ and $Y$ are $25$ and $18$ respectively. The corresponding sum of squares of deviations from means are $136$ and $148$ respectively. The sum of product of deviations from respective means is $122.$ Obtain the regression equation of $x$ on $y$
→| x | y | $x-\bar{x}$ | $y-\bar{y}$ | $(x-\bar{x})(y-\bar{y})$ | $(x-\bar{x})^2$ | $(y-\bar{y})^2$ |
| 1 | 5 | -2 | -4 | 8 | 4 | 16 |
| 2 | 7 | -1 | -2 | $\square$ | 1 | 4 |
| 3 | 9 | 0 | 0 | 0 | 0 | 0 |
| 4 | 11 | 1 | 2 | 2 | 4 | 4 |
| 5 | 13 | 2 | 4 | 8 | 1 | 16 |
| Total = 15 | Total = 45 | Total = 0 | Total = 0 | Total =$\square$ | Total = 10 | Total = 40 |
Mean of $y =\bar{y}=$$\square$
$ b_{x y}=\frac{\square}{\square}$
$b_{y x}=\frac{\square}{\square} $
Regression equation of $x$ on $y$ is $(x-\bar{x})= b _{x y}(y-\bar{y})$
$\therefore$ Regression equation $x$ on $y$ is$\square$
Regression equation of $y$ on $x$ is $(y-\bar{y})= b _{y x}(x-\bar{x})$
$\therefore$ Regression equation of $y$ on $x$ is$\square$
For manufacturing $x$ units, labour cost is $150-54 x$ and processing cost is $x^2$. Price of each unit is $p=10800-4 x^2$. Find the values of $x$ for which
(i) total cost is decreasing
(ii) revenue is increasing.
→(i) total cost is decreasing
(ii) revenue is increasing.
Following table shows the all India infant mortality rates (per ‘000) for years 1980 to 2010
Fit a trend line by the method of least squares
Solution: Let us fit equation of trend line for above data.
Let the equation of trend line be $y=a+b x$$\ldots(i)$
Here $n =7$ (odd), middle year is $\square$ and $h =5$
The normal equations are
$\Sigma y=n a+b \Sigma x$
As, $\Sigma x=0, a=\square$
Also, $\Sigma x y=a \Sigma x+b \Sigma x^2$
As, $\Sigma x=0, b=\square$
$\therefore$ The equation of trend line is $y=\square$
→| Year | 1980 | 1985 | 1990 | 1995 |
| IMR | 10 | 7 | 5 | 4 |
| Year | 2000 | 2005 | 2010 | |
| IMR | 3 | 1 | 0 |
Solution: Let us fit equation of trend line for above data.
Let the equation of trend line be $y=a+b x$$\ldots(i)$
Here $n =7$ (odd), middle year is $\square$ and $h =5$
| Year | IMR (y) | x | $x^2$ | x.y |
| 1980 | 10 | – 3 | 9 | – 30 |
| 1985 | 7 | – 2 | 4 | – 14 |
| 1990 | 5 | – 1 | 1 | – 5 |
| 1995 | 4 | 0 | 0 | 0 |
| 2000 | 3 | 1 | 1 | 3 |
| 2005 | 1 | 2 | 4 | 2 |
| 2010 | 0 | 3 | 9 | 0 |
| Total | 30 | 0 | 28 | – 44 |
$\Sigma y=n a+b \Sigma x$
As, $\Sigma x=0, a=\square$
Also, $\Sigma x y=a \Sigma x+b \Sigma x^2$
As, $\Sigma x=0, b=\square$
$\therefore$ The equation of trend line is $y=\square$
Solve the following differential equation
$\sec ^2 x \tan y d x+\sec ^2 y \tan x d y=0$
Solution: $\sec ^2 x \tan y d x+\sec ^2 y \tan x d y=0$
$\therefore \frac{\sec ^2 x}{\tan x} d x+\square=0$
Integrating, we get
$\square+\int \frac{\sec ^2 y}{\tan y} d y=\log c$
Each of these integral is of the type
$\int \frac{ f ^{\prime}(x)}{ f (x)} d x=\log | f ( x )|+\log C$
$\therefore$ the general solution is
$\square+\log |\tan y|=\log c$
$\therefore \log |\tan x \cdot \tan y|=\log c$
$\square$
This is the general solution.
→$\sec ^2 x \tan y d x+\sec ^2 y \tan x d y=0$
Solution: $\sec ^2 x \tan y d x+\sec ^2 y \tan x d y=0$
$\therefore \frac{\sec ^2 x}{\tan x} d x+\square=0$
Integrating, we get
$\square+\int \frac{\sec ^2 y}{\tan y} d y=\log c$
Each of these integral is of the type
$\int \frac{ f ^{\prime}(x)}{ f (x)} d x=\log | f ( x )|+\log C$
$\therefore$ the general solution is
$\square+\log |\tan y|=\log c$
$\therefore \log |\tan x \cdot \tan y|=\log c$
$\square$
This is the general solution.
A dealer deals in two products X and Y. He has ₹ 1,00,000/- to invest and space to store 80 pieces. Product X costs ₹ 2500/- and product Y costs ₹ 1000/- per unit. He can sell the items X and Y at respective profits of ₹ 300 and ₹ 90. Construct the LPP and find the number of units of each product to be purchased to maximize its profit
Solve the differential equation $\frac{ d y}{ d x}+\frac{x-2 y}{2 x-y}=0$
→Given the following table, find Walsh’s Price Index Number by completing the activity.
Walsh's price Index Number is
$ P _{01}( W )=\frac{\square}{\sum p _0 \sqrt{ q _0 q _1}} \times 100$
$=\frac{510}{\square} \times 100$
$=\square $
→| Commodity | $p_0$ | $q_0$ | $p_1$ | $q_1$ | $q_0q_1$ | $\sqrt{ q _0 q _1}$ | $p _0 \sqrt{q_0 q_1}$ | $p _1 \sqrt{ q _0 q _1}$ |
| I | 20 | 9 | 30 | 4 | 36 | $\square$ | $\square$ | 180 |
| II | 10 | 5 | 50 | 5 | $\square$ | 5 | 50 | $\square$ |
| III | 40 | 8 | 10 | 2 | 16 | $\square$ | 160 | $\square$ |
| IV | 30 | 4 | 20 | 1 | $\square$ | 2 | $\square$ | 40 |
| Total | – | – | – | – | 390 | $\square$ |
Walsh's price Index Number is
$ P _{01}( W )=\frac{\square}{\sum p _0 \sqrt{ q _0 q _1}} \times 100$
$=\frac{510}{\square} \times 100$
$=\square $