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:

${\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}={\left(\frac{f\left(x\right)}{x}\right)}^3\Rightarrow x^3{\displaystyle \underset{3}{\overset{x}{\int }}f\left(x\right)dx}=f^3\left(x\right),$ differentiating w.r.to x

$x^{3}f\left(x\right)+3x^2\frac{f^3\left(x\right)}{x^3}=3f^2\left(x\right)f\text{'}\left(x\right)\Rightarrow 3y^2\frac{dy}{dx}=x^{3}y=\frac{3y^3}{x}\Rightarrow 3xy\frac{dy}{dx}=x^4+3y^2$  

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

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

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

Given curve is $y=cosx$

$y=sinx$ for $0\le x\le \frac{\pi }{2}$

And $y-axis$

We know that $sinx=cosx$ at $x=\frac{\pi }{4}and<\frac{\pi }{4}<\frac{\pi }{2}$ i.e,  $cos\frac{\pi }{4}=sin\frac{\pi }{4}=1/\surd 2$

So the point of intersection is at $x=\frac{\pi }{4}$

The given equation of the lines are

$\begin{array}{l}2x+y=4------(1)\\ 3x-2y=6-----(2)\\ x-3y+5=0----(3)\end{array}$

$\therefore$ Area of $?ABC=area(PCBQ)-area(?APC)-area(?AQB)$

$\begin{array}{l}={\displaystyle \underset{1}{\overset{4}{\int }}{y}_{(3)}dx-}{\displaystyle \underset{1}{\overset{2}{\int }}{y}_{(1)}dx-{\displaystyle \underset{2}{\overset{4}{\int }}{y}_{(2)}dx-}}\\ ={\displaystyle \underset{1}{\overset{4}{\int }}\frac{\left(x+5\right)}{3}dx-{\displaystyle \underset{1}{\overset{2}{\int }}\left(4-2x\right)dx-{\displaystyle \underset{2}{\overset{4}{\int }}\left(\frac{3x-6}{2}dx\right)}}}\end{array}$

$\therefore$ The point of intersection of the circle and the parabola is . $A\left(\frac{1}{2},\right)\&C\left(\frac{1}{2},-\right)$

Taking in first quadrant

Area of $(OABO)=area(OADO)+area(BADB)$

The given vertices of the triangle are A(2,0),B(4,5)and C(6,3)

So, equation of line AB is $y-0=\frac{5-0}{4-2}\left(x-2\right)$

$=y=\frac{5}{2}\left(x-2\right)$

Similarly equation of BC is

$\begin{array}{l}y-5=\frac{3-5}{6-4}\left(x-4\right)\\ =y=5+\left(-x+4\right)\\ y=9-x\end{array}$

And equation of AC is $y-0=\frac{3-0}{6-2}\left(x-2\right)$

= $y=\frac{3}{4}\left(x-2\right)$

$\therefore$ Area of $?ABC$

= $area(?ABE)+area(BCDE)-area(?ACD)$

$\begin{array}{l}={\displaystyle \underset{2}{\overset{4}{\int }}{y}_{AB}dx+}{\displaystyle \underset{4}{\overset{6}{\int }}{y}_{BC}dx-{\displaystyle \underset{2}{\overset{6}{\int }}{y}_{BC}dx}}\\ ={\displaystyle \underset{2}{\overset{4}{\int }}\frac{5}{2}\left(x-2\right)dx+{\displaystyle \underset{4}{\overset{6}{\int }}\left(9-x\right)dx-{\displaystyle \underset{2}{\overset{6}{\int }}\frac{3}{4}}}}\left(x-2\right)dx\\ =\frac{5}{2}{\left[\frac{x^2}{2}-2x\right]}_2^4+{\left[9x-\frac{x^2}{2}\right]}_4^6-\frac{3}{4}{\left[\frac{x^2}{2}-2x\right]}_2^6\end{array}$

$\begin{array}{l}=\frac{5}{2}\left[\left(\frac{4^2}{2}-2.4\right)-\left(\frac{2^2}{2}-2.2\right)\right]+\left[\left(9.6-\frac{6^2}{2}\right)-\left(9.4-\frac{4^2}{2}\right)\right]-\frac{3}{4}\left[\left(\frac{6^2}{2}-2.6\right)-\left(\frac{2^2}{2}-2.2\right)\right]\\ =\frac{5}{2}\left[\left(8-8\right)-\left(2-4\right)\right]+\left[54-18-36+8\right]-\frac{3}{4}\left[18-12-2+4\right]\\ =5+8-6=7uni{t}^2\end{array}$