The correct options are
C b−a=2
D 1/2∫−1/2f(x) dx=π6−√34
Given : (1−x2)dydx=(√1−x2)(x4+2x)+xy
⇒dydx+xyx2−1=x4+2x√1−x2
For the square root to be defined,
1−x2>0⇒x∈(−1,1)
Comparing with dydx+P(x)y=Q(x)
I.F.=e∫P(x)dx
⇒∫P(x)dx=∫xx2−1dx
Assuming x2−1=t⇒2x dx=dt, then
∫P(x)dx=12∫dtt⇒∫P(x)dx=12ln|x2−1|
As x∈(−1,1), so
∫P(x)dx=12ln(1−x2)⇒I.F.=√1−x2
Solution is :
y√1−x2=∫(x4+2x) dx⇒y√1−x2=x55+x2+C∵f(0)=0, ∴C=0⇒y=f(x)=x55+x2√1−x2
Clearly, f(x) is neither even nor odd function.
f(x) is defined ∀ x∈(−1,1)
Therefore, b−a=2
I=1/2∫−1/2x55+x2√1−x2 dx⇒I=1/2∫−1/2x2√1−x2dx (∵x5 is odd function)⇒I=21/2∫0x2√1−x2dx
Taking x=sinθ⇒dx=cosθ dθ
I=2π/6∫0sin2θ dθ⇒I=π/6∫0(1−cos2θ) dθ⇒I=[θ−sin2θ2]π/60∴I=π6−√34