Question

Using integration, find the area of the region enclosed by the curve y²= 4x and x² + y² = 16.​

Answer 1

Answer:

x2=4y                                y2=4x

x=2y                            =4×2y

                                            y2=8y

⇒y4−8y=0

then y=0,4

x2=4y

we get x=4,0

So, the points of intersection are (0,0) & (4,4)

Area =∫04∫2xx2/4dydx=∫04[y]2xx2/4dx

=∫04[4x2−2x]dx

=[12x3−34⋅x3/2]04

Step-by-step explanation:

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