If $\int _0^2x(2-x)^b\,dx = \frac{32}{7}$, where $b\in N$ then $b =$

Question

If the solution of \[ \left( 1 + 2e^\frac{x}{y} \right) dx + 2e^\frac{x}{y} \left( 1 - \frac{x}{y} \right) dy = 0 \] is \[ x + \lambda y e^\frac{x}{y} = c \quad \text{(where $c$ is an arbitrary constant), then $ \lambda $ is:} \]

Answer 1

Step 1: Use the Beta function:
$\int_0^2 x(2-x)^b dx$: put $x = 2t$. Then it equals $4\cdot2^b\int_0^1 t(1-t)^b dt = 2^{b+2}B(2, b+1)$.

Step 2: Evaluate B:
$B(2,b+1) = \dfrac{1!\,b!}{(b+2)!} = \dfrac{1}{(b+1)(b+2)}$.
So the integral is $\dfrac{2^{b+2}}{(b+1)(b+2)}$.

Step 3: Solve:
We need $\dfrac{2^{b+2}}{(b+1)(b+2)} = \dfrac{32}{7}$. Since 7 divides the denominator, $b+1 = 7$ or $b+2 = 7$. With $b = 6$: $\dfrac{256}{56} = \dfrac{32}{7}$. It works.

Final Answer:
$b = 6$, option (B). \[ \boxed{6 \text{ (B)}} \]

If \(\int _0^2x(2-x)^b\,dx = \frac{32}{7}\), where \(b\in N\) then \(b =\)

Show Hint

The Correct Option is B

Solution and Explanation

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