Question

Area under the curve v =32" + 4x between the limits - to - is 1) 8 2) 16 4 2​

Answer 1

Answer:

Area =∫04∫2x x2/4dydx=∫04[y]2x x2/4dx. =∫04[4x2−2x ]dx. =[12x3−34⋅x3/2]04. =[1243−34(4)3/2]. =316 sq.

Answer 2

Answer:

Correct option is b

3

16

x

2

=4y y

2

=4x

x=2

y

=4×2

y

y

2

=8

y

⇒y

4

−8y=0

then y=0,4

x

2

=4y

we get x=4,0

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

Area =∫

0

4

∫

2

x

x

2

/4

dydx=∫

0

4

[y]

2

x

x

2

/4

dx

=∫

0

4

[

4

x

2

−2

x

]dx

=[

12

x

3

−

3

4

⋅x

3/2

]

0

4

=[

12

4

3

−

3

4

(4)

3/2

]

=

3

16

sq. units.