1. Find the equation of tangent line at point (2,20).
[tex]y'=(5x^2)'=5\cdot 2x=10x,\\ \\y'(2)=10\cdot 2=20.[/tex]
The equation of the tangent line is
[tex]y=20(x-2)+20,\\ \\y=20x-20.[/tex]
2. Express x:
[tex]y=20x-20\Rightarrow x=\dfrac{y}{20}+1;\\ \\y=5x^2\Rightarrow x=\sqrt{\dfrac{y}{5}}.[/tex]
3. Find the area of bounded region:
[tex]A=\int\limits^{20}_0 {\left(\dfrac{y}{20}+1-\sqrt{\dfrac{y}{5}} } \right)\, dy=\left(\dfrac{y^2}{40}+y-\dfrac{2\sqrt{y^3} }{3\sqrt{5} } \right)\big|^{20}_0=[/tex]
[tex]=\dfrac{400}{40}+20-\dfrac{2\sqrt{20^3} }{3\sqrt{5} }=30-\dfrac{80}{3}=\dfrac{10}{3}\ sq. un.[/tex]
Answer: [tex]\dfrac{10}{3}\ sq. un.[/tex]
