Question

f(x)= kx^2, 0<x<1 find k​

Answer 1

Answer:

∴∫02f(x)dx=∫02kxdx=1

                   ⇒K[2x2]02=1

                   ⇒k[24−0]=1

                   ⇒k×2=1

                   ⇒k=21

ii) P(1<x<2)

                  ∫12f(x)dx=∫1221xdx Putting value of k=21

                 =21[2x2]12

                 =21[24−21]

                 =21×

Step-by-step explanation:

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Answer 2

Answer:

the value of a probability density function over the whole range is equal to

′

1

′

i) ∴∫

0

2

f(x)dx=∫

0

2

kxdx=1

⇒K[

2

x

2

]

0

2

=1

⇒k[

2

4

−0]=1

⇒k×2=1

⇒k=

2

1