Find the area enclosed between the parabola y2=4ax and the line y=mx
The area enclosed between the parabola,y2=4ax,and the line,y=mx,is represented
by the shaded area OABO as
The points of intersection of both the curves are(0,0)and(4a/m2,4a/m).
We draw AC perpendicular to x-axis.
∴Area OABO=Area OCABO-Area(ΔOCA)
=
\[\int_{0}^{\frac{4a}{m^3}} 2\sqrt{ax} \,dx\]\[-\int_{0}^{\frac{4a}{m^2}} mx \,dx\]=2√a[\(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\)]4a/m20-m[\(\frac{x^2}{2}\)]4a/m2 0
=\(\frac 43\)√a\((\frac{4a}{m^2})^{3/2}\)-\(\frac m2\)[(\(\frac{4a}{m^2}\))2]
=\(\frac{32a^2}{3m^3}\)-\(\frac m2\)(\(\frac {16a^2}{m^4}\))
=\(\frac{32a^2}{3m^3}\)-\(\frac{8a^2}{3m^3}\)
=\(\frac{8a^2}{3m^3}\)units.
If 5f(x) + 4f (\(\frac{1}{x}\)) = \(\frac{1}{x}\)+ 3, then \(18\int_{1}^{2}\) f(x)dx is: