Question

find the area enclosed by y=x & curve y²=16x​

Answer 1

Answer:

y2=16x,y=2x

⇒4x=2x

x=2⇒x=4

Area=∫04(16x−2x)dx

=∫04(4x−2x)dx

=[234x23−x2]04

=38[8−0]−[16−0]

=364−16

=316sq.